sup{x + y: x in S1, y in S2} = sup{S1} + sup{S2}
ame...@my-deja.com
Please read through my proof and see if it is makes sense. Thanks.
If S1 and S2 are nonempty subsets of R that are bounded above, prove
that sup{x + y: x in S1, y in S2} = sup{S1} + sup{S2}
Proof.
Since both S1 and S2 are nonempty subsets of R, both are bounded above
and have supremums. So by definition,
sup{S1} >= x for each x in S1 and sup{S1} =< any upperbound of S1 and
sup{S2} >= y for each y in S2 and sup{S2} =< any upperbound of S2.
Which implies 1.) sup{S1} + sup{S2} >= x + y for each x in S1 and y in
S2 or {x + y: x in S1, y in S2}. Hence the set
{x + y: x in S1, y in S2} is bounded above by sup{S1} + sup{S2} and
consequently it has a supremum. So
2.) sup{S1} + sup{S2} >= sup{x + y: x in S1, y in S2} and
3.) sup{x + y: x in S1, y in S2} >= {x + y: x in S1, y in S2}. By
definition,
4.) sup{S1} + sup{S2} =< any upperbound of {S1} + any upperbound of
{S1}. Since sup{x + y: x in S1, y in S2} qualifies as any upperbound
of {S1} + {S2} then sup{S1} + sup{S2} =< sup{x + y: x in S1, y in S2}.
After combining this with 2, we have
sup{S1} + sup{S2} = sup{x + y: x in S1, y in S2}. QED
Sent via Deja.com http://www.deja.com/
Before you buy.
Robin Chapman
ame...@my-deja.com wrote:
>
>
> Please read through my proof and see if it is makes sense. Thanks.
>
> If S1 and S2 are nonempty subsets of R that are bounded above, prove
> that sup{x + y: x in S1, y in S2} = sup{S1} + sup{S2}
>
> Proof.
>
> Since both S1 and S2 are nonempty subsets of R, both are bounded above
> and have supremums. So by definition,
> sup{S1} >= x for each x in S1 and sup{S1} =< any upperbound of S1 and
> sup{S2} >= y for each y in S2 and sup{S2} =< any upperbound of S2.
> Which implies 1.) sup{S1} + sup{S2} >= x + y for each x in S1 and y in
> S2 or {x + y: x in S1, y in S2}. Hence the set
> {x + y: x in S1, y in S2} is bounded above by sup{S1} + sup{S2} and
> consequently it has a supremum. So
OK
> 2.) sup{S1} + sup{S2} >= sup{x + y: x in S1, y in S2} and
OK
> 3.) sup{x + y: x in S1, y in S2} >= {x + y: x in S1, y in S2}. By
> definition,
Well I would say this is a category error. The left side is a number
and the right side is a set. I'd write sup{x + y: x in S1, y in S_2}
>= x + y for each x in S1 and y in S2.
> 4.) sup{S1} + sup{S2} =< any upperbound of {S1} + any upperbound of
> {S1}.
The last S1 should be S2. Also I would write upper bound of S1
not upperbound of {S1} etc.
> Since sup{x + y: x in S1, y in S2} qualifies as any upperbound
> of {S1} + {S2} then sup{S1} + sup{S2} =< sup{x + y: x in S1, y in S2}.
I presume {S1} + {S2} denotes the set {x + y: x in S1, y in S2} used
above. I would make this explicit and write S1 + S2 instead.
Alas, though, this statement is a nonsequitur. You have not
shown that sup{x + y: x in S1, y in S2} has the form
(upper bound of S1) + (upper bound of S2).
> After combining this with 2, we have
> sup{S1} + sup{S2} = sup{x + y: x in S1, y in S2}. QED
You need to show that no number less than sup(S1) + sup(S2)
is an upper bound for S1 + S2. Assume such exists and aim for
a contradiction.
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
vaillant
> prove that sup{x + y: x in S1, y in S2} = sup{S1} + sup{S2}
> Proof.
> Since both S1 and S2 are nonempty subsets of R, both are bounded
> above
My advice is to work in [-oo,+oo] where any subset *has* a supremum,
(it may be unbounded above, then sup is +oo, it may be empty, then sup
is -oo.
However, a sum x+y in [-oo,+oo] may not be meaningful (one needs to
avoid (-oo) + (+oo)), Similarly, you want sup(S1) + sup(S2) to be
meaningful. (Otherwise you ll never prove the above)
Assume S1 and S2 are non-empty subsets of ]-oo,+oo] (then x+y and
supS1+supS2 will always be meaningful and we have a good chance of the
above being true)
Let A=sup{x+y: x in S1, y in S2}
For all x in S1 and y in S2, we have x+y<=supS1 + supS2
Hence A<=supS1+supS2
For all x in S1 and y in S2, we have x+y<=A. Let y being fixed. If y is
real, then x<=A-y for all x in S1, and therefore supS1<=A-y, i.e.
y+supS1<=A. If y=+oo, then A=+oo and y+supS1<=A is still true. If supS1
is real then y<=A-supS1 for all y in S2, and therefore supS2<=A-supS1,
i.e. supS1+supS2<=A. If supS1=+oo, then A=+oo and supS1+supS2<=A is
still true.
We have proved that A=supS1+supS2.
I hope this is correct.
Regards. Noel.
* Sent from RemarQ http://www.remarq.com The Internet's Discussion Network *
The fastest and easiest way to search and participate in Usenet - Free!
Simon
and sets, don't be surprised if your instructors give you low marks!
You have a large logical gap in the late part of the proof too, but I
guess you knew that was the hard bit.
OK, the idea is there are things x in S1 arbitrarily near sup{S1} and
things y in S2
arbitrarily near sup{S2} so you can find x + y arbitrarily near sup{S1}
+ sup{S2}. Now can you translate that into your version of maths?
ame...@my-deja.com wrote:
>
> Please read through my proof and see if it is makes sense. Thanks.
>
> If S1 and S2 are nonempty subsets of R that are bounded above, prove
> that sup{x + y: x in S1, y in S2} = sup{S1} + sup{S2}
>
> Proof.
>
> Since both S1 and S2 are nonempty subsets of R, both are bounded above
> and have supremums. So by definition,
> sup{S1} >= x for each x in S1 and sup{S1} =< any upperbound of S1 and
> sup{S2} >= y for each y in S2 and sup{S2} =< any upperbound of S2.
> Which implies 1.) sup{S1} + sup{S2} >= x + y for each x in S1 and y in
> S2 or {x + y: x in S1, y in S2}. Hence the set
> {x + y: x in S1, y in S2} is bounded above by sup{S1} + sup{S2} and
> consequently it has a supremum. So
> 2.) sup{S1} + sup{S2} >= sup{x + y: x in S1, y in S2} and
> 3.) sup{x + y: x in S1, y in S2} >= {x + y: x in S1, y in S2}. By
> definition,
> 4.) sup{S1} + sup{S2} =< any upperbound of {S1} + any upperbound of
> {S1}. Since sup{x + y: x in S1, y in S2} qualifies as any upperbound
> of {S1} + {S2} then sup{S1} + sup{S2} =< sup{x + y: x in S1, y in S2}.
> After combining this with 2, we have
> sup{S1} + sup{S2} = sup{x + y: x in S1, y in S2}. QED
>
> Sent via Deja.com http://www.deja.com/
> Before you buy.
--
Dr Simon Fitzpatrick, Shenton Park, Western Australia
Mathematician and International Correspondence Chess Master
http://www.q-net.net.au/~dsf/Simon.html
Fred Galvin
> In article <86j6f4$j11$1...@nnrp1.deja.com>,
> ame...@my-deja.com wrote:
> >
> > 3.) sup{x + y: x in S1, y in S2} >= {x + y: x in S1, y in S2}. By
> > definition,
>
> Well I would say this is a category error. The left side is a
> number and the right side is a set.
This seems a bit harsh: the use of "a < B" to mean "a < b for each
element b of the set B" is common enough in the literature. Maybe this
notation should be avoided in elementary courses, so as not to confuse
the young. Still, it seems no more objectionable than the standard
notation for cosets, e.g., t+X where t is a vector and X is a set of
vectors.
David C. Ullrich
Fred Galvin wrote:
> On Tue, 25 Jan 2000, Robin Chapman wrote:
>
> > In article <86j6f4$j11$1...@nnrp1.deja.com>,
> > ame...@my-deja.com wrote:
> > >
> > > 3.) sup{x + y: x in S1, y in S2} >= {x + y: x in S1, y in S2}. By
> > > definition,
> >
> > Well I would say this is a category error. The left side is a
> > number and the right side is a set.
>
> This seems a bit harsh: the use of "a < B" to mean "a < b for each
> element b of the set B" is common enough in the literature.
That's in a context where the writer has reason to think that the
reader will know exactly what's meant, and the writer is trying
to explain something to the reader that the reader presumably
doesn't already know. A context like the present is very different:
The reader presumably already understands what's what, and the
writer is trying to show that _he_ understands what he's writing,
that he understands various aspects of proof, that he understands
exactly what the notation means, etc. So it's appropriate to be
much more careful - a person shouldn't use "a < B" to mean
"a < b for each element b of the set B" unless this usage has been
explained explicitly.
Really. This was a homework problem. When I see "a < B"
used this way in a paper or a talk there's no problem (there's
also almost always at least one mumble explaining the
meaning, btw), but if I saw this on a homework problem I'd
complain about it. Seems likely or at least possible that the
guy grading the poster's homework is going to feel the same
way.
Fred Galvin
> Fred Galvin wrote:
>
> > On Tue, 25 Jan 2000, Robin Chapman wrote:
> >
> > > In article <86j6f4$j11$1...@nnrp1.deja.com>,
> > > ame...@my-deja.com wrote:
> > > >
> > > > 3.) sup{x + y: x in S1, y in S2} >= {x + y: x in S1, y in S2}. By
> > > > definition,
> > >
> > > Well I would say this is a category error. The left side is a
> > > number and the right side is a set.
> >
> > This seems a bit harsh: the use of "a < B" to mean "a < b for each
> > element b of the set B" is common enough in the literature.
>
> That's in a context where the writer has reason to think that the
> reader will know exactly what's meant, and the writer is trying
> to explain something to the reader that the reader presumably
> doesn't already know. A context like the present is very different:
> The reader presumably already understands what's what, and the
> writer is trying to show that _he_ understands what he's writing,
> that he understands various aspects of proof, that he understands
> exactly what the notation means, etc. So it's appropriate to be
> much more careful - a person shouldn't use "a < B" to mean
> "a < b for each element b of the set B" unless this usage has been
> explained explicitly.
Yeah, sure; like I said, it might be better to avoid this kind of
notation in elementary courses. But I still think "error" is too
strong a word for that more-or-less-standard notation, and I have no
way of knowing whether or not it has been explained explicitly by that
guy's instructor.
Karl Smith
over email. You might want to try working with Sup{S1} + y > x + y. If
you can see that route easily go with it. Otherwise, go with what you've
got
ame...@my-deja.com wrote:
> Please read through my proof and see if it is makes sense. Thanks.
>
> If S1 and S2 are nonempty subsets of R that are bounded above, prove
> that sup{x + y: x in S1, y in S2} = sup{S1} + sup{S2}
>
> Proof.
>
> Since both S1 and S2 are nonempty subsets of R, both are bounded above
> and have supremums. So by definition,
> sup{S1} >= x for each x in S1 and sup{S1} =< any upperbound of S1 and
> sup{S2} >= y for each y in S2 and sup{S2} =< any upperbound of S2.
> Which implies 1.) sup{S1} + sup{S2} >= x + y for each x in S1 and y in
> S2 or {x + y: x in S1, y in S2}. Hence the set
> {x + y: x in S1, y in S2} is bounded above by sup{S1} + sup{S2} and
> consequently it has a supremum. So
> 2.) sup{S1} + sup{S2} >= sup{x + y: x in S1, y in S2} and
> 3.) sup{x + y: x in S1, y in S2} >= {x + y: x in S1, y in S2}. By
> definition,
David C. Ullrich
Fred Galvin wrote:
> [...]
>
> Yeah, sure; like I said, it might be better to avoid this kind of
> notation in elementary courses. But I still think "error" is too
> strong a word for that more-or-less-standard notation, and I have no
> way of knowing whether or not it has been explained explicitly by that
> guy's instructor.
If the guy was getting the easy bits right, or even avoiding
other notational errors I might agree. But since this is actually
not the _only_ fishy bit about the notation it seems more likely
that we haven't quite got the _absolutely_ standard parts straight
yet, and we should do that first.
(What's the sup of {S1} supposed to be anyway? In the
present context it's only sets of reals that have sups, and
{S1} is not a set of reals. This doesn't seem important
enough to bother pointing out for its own sake, but it
does make me suspicious about the other issue...)
ame...@my-deja.com
> If the guy was getting the easy bits right, or even avoiding
> other notational errors I might agree. But since this is actually
> not the _only_ fishy bit about the notation it seems more likely
> that we haven't quite got the _absolutely_ standard parts straight
> yet, and we should do that first.
>
> (What's the sup of {S1} supposed to be anyway?
It's supposed the Least Upper Bound of the set S1 or the Supremum of
S1. The reason I wrote it as sup{S1} is to avoid confusion, but I
guess I made more confusion.
> In the
> present context it's only sets of reals that have sups, and
> {S1} is not a set of reals.
I stated very clearly that S1 and S2 are nonempty subsets of real
numbers. I didn't know that some might think that {S1} is a set with
only one element in it. How could it possibly be anything else in
light of what I said about S1 and S2 in the beginning?
ame...@my-deja.com
Karl Smith <kwsm...@unity.ncsu.edu> wrote:
> It looks good. I could really follow the notation at the end. It's
> hard
> over email.
Thank you.
>You might want to try working with Sup{S1} + y > x + y.
Since sup S1 >= y, then sup S2 + sup S2 >= sup S1 + y > x + y
=> sup S1 + sup S2 >= x + y. Then you need to show that
x + y >= sup S1 + sup S2. How do you that?
ame...@my-deja.com
Robin Chapman <r...@maths.ex.ac.uk> wrote:
> > Since sup{x + y: x in S1, y in S2} qualifies as any upperbound
> > of {S1} + {S2} then sup{S1} + sup{S2} =< sup{x + y: x in S1, y in
> > S2}.
>
> I presume {S1} + {S2} denotes the set {x + y: x in S1, y in S2} used
> above. I would make this explicit and write S1 + S2 instead.
You mean since S1 = {x: x in S1} and S2 = {y: y in S2}, then
S1 + S2 = {x: x in S1} + {y: y in S2} or {x + y: x in S1, y in S2}?
Since these sets are the same, their supremums or least upper bounds
are the same. Therefore sup (S1 + S2) = sup {x + y: x in S1, y in S2}
and sup S1 = sup {x: x in S1} and sup S2 = sup {y: y in S2} =>
sup S1 + sup S2 = sup {x: x in S1} + sup {y: y in S2}. I don't know if
this would complicate or or simplify matters.
>
> Alas, though, this statement is a nonsequitur. You have not
> shown that sup{x + y: x in S1, y in S2} has the form
> (upper bound of S1) + (upper bound of S2).
sup S1 =< a where a is any upper bound of S1 and sup S2 =< b where b is
any upperbound of S2 => sup S1 + sup S2 =< a + b =>
sup {x: x in S1} + sup {y: y in S2} =< a + b.
>
> > After combining this with 2, we have
> > sup{S1} + sup{S2} = sup{x + y: x in S1, y in S2}. QED
>
> You need to show that no number less than sup(S1) + sup(S2)
> is an upper bound for S1 + S2. Assume such exists and aim for
> a contradiction.
Suppose there is an upper bound c of S1 + S2 such that
c < sup S1 + sup S2. Since c is an upper bound for S1 + S2,
c >= x + y for each x in S1 and y in S2.
I'm stuck at this point, I don't see where it's going. I understand my
approach better and contrary to what other posters have said, I don't
see any flawed logic in my proof. The only problem with my original
proof was when I wrote the sets as if they were one number. Thanks for
pointing this out.
Robin Chapman
ame...@my-deja.com wrote:
> In article <86jpvh$11o$1...@nnrp1.deja.com>,
> Robin Chapman <r...@maths.ex.ac.uk> wrote:
>
> > > Since sup{x + y: x in S1, y in S2} qualifies as any upperbound
> > > of {S1} + {S2} then sup{S1} + sup{S2} =< sup{x + y: x in S1, y in
> > > S2}.
> >
> > I presume {S1} + {S2} denotes the set {x + y: x in S1, y in S2} used
> > above. I would make this explicit and write S1 + S2 instead.
>
> You mean since S1 = {x: x in S1} and S2 = {y: y in S2}, then
> S1 + S2 = {x: x in S1} + {y: y in S2} or {x + y: x in S1, y in S2}?
No, I mean that the string "S1 + S2" (or "{S1} + {S2}") is meaningless
per se. I reconstructed what I thought might be your meaning.
If you use such notation you might explicitly indicate what you
mean by it.
> Since these sets are the same, their supremums or least upper bounds
> are the same. Therefore sup (S1 + S2) = sup {x + y: x in S1, y in S2}
> and sup S1 = sup {x: x in S1} and sup S2 = sup {y: y in S2} =>
> sup S1 + sup S2 = sup {x: x in S1} + sup {y: y in S2}. I don't know
if
> this would complicate or or simplify matters.
>
> >
> > Alas, though, this statement is a nonsequitur. You have not
> > shown that sup{x + y: x in S1, y in S2} has the form
> > (upper bound of S1) + (upper bound of S2).
>
> sup S1 =< a where a is any upper bound of S1 and sup S2 =< b where b
is
> any upperbound of S2 => sup S1 + sup S2 =< a + b =>
> sup {x: x in S1} + sup {y: y in S2} =< a + b.
>
> >
> > > After combining this with 2, we have
> > > sup{S1} + sup{S2} = sup{x + y: x in S1, y in S2}. QED
This is a non-sequitur. To infer this you would need to show that
sup{x + y: x in S1, y in S2} had the form a + b where a
is an upper bound of S1 and b is an upper bound of S2. Nowhere
in your original posting nor in this latest followup have you
demonstrated this.
> I'm stuck at this point, I don't see where it's going. I understand
my
> approach better and contrary to what other posters have said, I don't
> see any flawed logic in my proof. The only problem with my original
> proof was when I wrote the sets as if they were one number. Thanks
for
> pointing this out.
There is a non-sequitur in your argument, which I have now pointed out
twice. Thus, your argument is not (as you describe it) a proof.
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
Dave L. Renfro
[sci.math Sat, 29 Jan 2000 00:13:02 GMT]
<http://forum.swarthmore.edu/epigone/sci.math/twermalwoi>
wrote (in part)
> I stated very clearly that S1 and S2 are nonempty subsets
> of real numbers. I didn't know that some might think that
> {S1} is a set with only one element in it. How could it
> possibly be anything else in light of what I said about S1
> and S2 in the beginning?
I think everyone knew what you meant by {S1}. Their point
was that *details do matter*, and that {S1} is--by any
definition you'll find in a math book--the set whose only
element is S1. Part of the task of learning how to formally
prove things is to learn the proper use of standard notation,
and it is this that likely caused the comments you got.
In case you're thinking "what else could {S1} mean", then
I would ask you how you would denote the set whose only element
is S1? One definition of compactness is in terms of sets of sets
of real numbers, and it is quite possible that in this context
one would want to consider {S1}.
Dave L. Renfro
ame...@my-deja.com
Robin Chapman <r...@maths.ex.ac.uk> wrote:
>
> This is a non-sequitur. To infer this you would need to show that
> sup{x + y: x in S1, y in S2} had the form a + b where a
> is an upper bound of S1 and b is an upper bound of S2.
OK. Let's start at sup S1 + sup S2 =< a + b; I know I'm right up to
this point. Since {x: x in S1} = S1 and {y: y in S2} = S2 then
{x + y: x in S1, y in S2} = S1 + S2. So
{x + y: x in S1, y in S2} is bounded above by a + b and since S1 + S2
is a nonempty set of real numbers, {x + y: x in S1, y in S2} is as
well. Therefore, by the least upper bound theorem,
{x + y: x in S1, y in S2} has a supremum. From the statement
sup S1 + sup S2 >= x + y for each x in S1 and y in S2, we know that
{x + y: x in S1, y in S2} is bounded above by sup S1 + sup S2. So
sup S1 + sup S2 >= sup {x + y: x in S1, y in S2}.
{x + y: x in S1, y in S2} = S1 + S2 => sub {x + y: x in S1, y in S2} is
an upper bound of S1 + S1 => sub {x + y: x in S1, y in S2} is an upper
bound of S1 + an upper bound of S2. Thus,
sup S1 + sup S2 =< a + b =>
sup S1 + sup S2 =< sup S1 + sup S2 =< a + b. This completes the
proof.
If I left anything out again or overlooked something please let me know.
Thanks
ame...@my-deja.com
ame...@my-deja.com wrote:
> In article <86vj04$gf2$1...@nnrp1.deja.com>,
> Robin Chapman <r...@maths.ex.ac.uk> wrote:
>
> >
> > This is a non-sequitur. To infer this you would need to show that
> > sup{x + y: x in S1, y in S2} had the form a + b where a
> > is an upper bound of S1 and b is an upper bound of S2.
>
> OK. Let's start at sup S1 + sup S2 =< a + b; I know I'm right up to
> this point. Since {x: x in S1} = S1 and {y: y in S2} = S2 then
> {x + y: x in S1, y in S2} = S1 + S2. So
> {x + y: x in S1, y in S2} is bounded above by a + b and since S1 + S2
> is a nonempty set of real numbers, {x + y: x in S1, y in S2} is as
> well. Therefore, by the least upper bound theorem,
> {x + y: x in S1, y in S2} has a supremum. From the statement
> sup S1 + sup S2 >= x + y for each x in S1 and y in S2, we know that
> {x + y: x in S1, y in S2} is bounded above by sup S1 + sup S2. So
> sup S1 + sup S2 >= sup {x + y: x in S1, y in S2}.
> {x + y: x in S1, y in S2} = S1 + S2 => sub {x + y: x in S1, y in S2}
is
> an upper bound of S1 + S1 => sub {x + y: x in S1, y in S2} is an upper
> bound of S1 + an upper bound of S2. Thus,
> sup S1 + sup S2 =< a + b =>
sup S1 + sup S2 =< sup {x + y: x in S1, y in S2}.
Robin Chapman
> {x + y: x in S1, y in S2} = S1 + S2 => sub {x + y: x in S1, y in S2}
is
> an upper bound of S1 + S1 => sub {x + y: x in S1, y in S2} is an upper
> bound of S1 + an upper bound of S2.
Presumably "sub" in the first line should read "sup". Also "S1 + S1"
should presumably read "S1 + S2". The final assertion is that
sup{x + y: x in S1, y in S2} has the form (upper bound of S1)
+ (upper bound of S2). No evidence is given for this assertion.
It's a non-sequitur.
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
ame...@my-deja.com
Robin Chapman <r...@maths.ex.ac.uk> wrote:
> The final assertion is that
> sup{x + y: x in S1, y in S2} has the form (upper bound of S1)
> + (upper bound of S2). No evidence is given for this assertion.
Assume S1 and S2 are the same as defined in the original posting, then
{x:x in S1} = S1 and {y: y in S2} = S2 =>
{x + y: x in S1, y in S2} = S1 + S2 right? Since both S1 and S2 are
bounded above, S1 + S2 =< (upper bound of S1) + (upper bound of S2) =>
{x + y: x in S1, y in S2} =< (upper bound of S1) + (upper bound of
S2). The set on the left is bounded above and nonempty so it has a
supremum. Since
{x:x in S1} + {y: y in S2} = {x + y: x in S1, y in S2} then
1.)sup{x:x in S1} + sup{y: y in S2} = sup{x + y: x in S1, y in S2}
hence sup{x + y: x in S1, y in S2} has the form (upper bound of S1)
+ (upper bound of S2). Is this what you're talking about? What's
wrong with concluding the proof at #1? That's what I set out to do.
Robin Chapman
ame...@my-deja.com wrote:
> In article <872799$7pu$1...@nnrp1.deja.com>,
> Robin Chapman <r...@maths.ex.ac.uk> wrote:
> > The final assertion is that
> > sup{x + y: x in S1, y in S2} has the form (upper bound of S1)
> > + (upper bound of S2). No evidence is given for this assertion.
>
> Assume S1 and S2 are the same as defined in the original posting, then
> {x:x in S1} = S1 and {y: y in S2} = S2 =>
> {x + y: x in S1, y in S2} = S1 + S2 right? Since both S1 and S2 are
Are you advancing this as a *definition* of "S1 + S2"?
> bounded above, S1 + S2 =< (upper bound of S1) + (upper bound of S2) =>
Again, an inequality between a set and a number. I presume this
means that each element of S1 + S2 is <= each number of the form
(upper bound of S1) + (upper bound of S2).
> {x + y: x in S1, y in S2} =< (upper bound of S1) + (upper bound of
> S2). (*)
So this is the same as the previous statement.
> The set on the left is bounded above and nonempty so it has a
> supremum. Since
> {x:x in S1} + {y: y in S2} = {x + y: x in S1, y in S2} then
> 1.)sup{x:x in S1} + sup{y: y in S2} = sup{x + y: x in S1, y in S2}
This is a non-sequitur, a statement given without justification.
A special case of (*) is that x + y <= sup(S1) + sup(S2)
for all x in S1, y in S2. Thus sup(S1) + sup(S2) is an upper bound
of S1 + S2 and so, by definition of supremum,
sup(S1) + sup(S2) >= sup(S1 + S2). But at this point you assert
equality, but give no reason for doing so.
ame...@my-deja.com
Robin Chapman <r...@maths.ex.ac.uk> wrote:
> > bounded above, S1 + S2 =< (upper bound of S1) + (upper bound of S2)
>
> Again, an inequality between a set and a number.
I don't know why I wrote that, but it's not what I meant. One of the
best ways to learn is by making mistakes.
> I presume this
> means that each element of S1 + S2 is <= each number of the form
> (upper bound of S1) + (upper bound of S2).
x + y <= a + b where x is in S1, y is in S2 and a is any upper bound of
S1 and b is any upper bound of S2.(**)
> > {x + y: x in S1, y in S2} =< (upper bound of S1) + (upper bound of
> > S2). (*)
>
> So this is the same as the previous statement.
>
> > The set on the left is bounded above and nonempty so it has a
> > supremum. Since
> > {x:x in S1} + {y: y in S2} = {x + y: x in S1, y in S2} then
> > 1.)sup{x:x in S1} + sup{y: y in S2} = sup{x + y: x in S1, y in S2}
>
> This is a non-sequitur, a statement given without justification.
> A special case of (*) is that x + y <= sup(S1) + sup(S2)
> for all x in S1, y in S2.
(**) implies x + y <= sup(S1) + sup(S2) for all x in S1, y in S2.
> Thus sup(S1) + sup(S2) is an upper bound
> of S1 + S2 and so, by definition of supremum,
> sup(S1) + sup(S2) >= sup(S1 + S2).
Which is equivalent to
sup(S1) + sup(S2) >= sup{x + y: x in S1, y in S2} or do I need to
justify that as well?
Now all there is left to do is to prove that
sup(S1) + sup(S2) <= sup{x + y: x in S1, y in S2} which I already did
before.
Robin Chapman
ame...@my-deja.com wrote:
> Which is equivalent to
> sup(S1) + sup(S2) >= sup{x + y: x in S1, y in S2} or do I need to
> justify that as well?
Yes.
> Now all there is left to do is to prove that
> sup(S1) + sup(S2) <= sup{x + y: x in S1, y in S2} which I already did
> before.
I don't recall ever seeing any proof of this in your previous posts.
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
ame...@my-deja.com
Robin Chapman <r...@maths.ex.ac.uk> wrote:
> In article <873vop$e9q$1...@nnrp1.deja.com>,
> ame...@my-deja.com wrote:
>
> > Which is equivalent to
> > sup(S1) + sup(S2) >= sup{x + y: x in S1, y in S2} or do I need to
> > justify that as well?
>
> Yes.
This is justified because x + y <= a + b where x, y, a, and b are the
same as defined before implies x + y <= sup(S1) + sup(S2) implies that
{x + y: x in S1, y in S2} is bounded above by sup(S1) + sup(S2)
therefore it has a supremum. So definition of a least upper bound
sup(S1) + sup(S2) >= sup{x + y: x in S1, y in S2}
> > Now all there is left to do is to prove that
> > sup(S1) + sup(S2) <= sup{x + y: x in S1, y in S2} which I already
did
> > before.
>
> I don't recall ever seeing any proof of this in your previous posts.
Starting from sup(S1) + sup(S2) <= a + b as defined before, I said that
sup{x + y: x in S1, y in S2} => sup(S1) + sup(S2) because
sup{x + y: x in S1, y in S2} qualifies as any upper bound of S1 + S2.
I suppose I didn't give enough justification for that. Since this
statement is true, there must be some way of proving that
a + b <= sup{x + y: x in S1, y in S2} or better yet
a + b = sup{x + y: x in S1, y in S2}. I'll try to prove the latter.
Proof.
a + b have been defined before as
(any upper bound of S1) + (any upper bound of S1). This can be
rewritten as (any upper bound of S1) + (any upper bound of S2) which is
equivalent to any upper bound of (S1 + S2) which is equivalent to
any upper bound of ({x: x in S1} + {y: y in S2}) which is equivalent to
any upper bound of ({x + y: x in S1, y in S2}). This upper bound is
arbitrary, so pick sup{x + y: x in S1, y in S2}.
This I think should conclude to prove. As always, tell me if I
overlooked something.
Robin Chapman
ame...@my-deja.com wrote:
> This can be
> rewritten as (any upper bound of S1) + (any upper bound of S2) which
is
> equivalent to any upper bound of (S1 + S2)
Again, the same nonsequitur. The assertion, without any evidence,
that each upper bound of S1 + S2 has the form a + b where a and
b are upper bounds of S1 and S2 respectively.
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
ame...@my-deja.com
Robin Chapman <r...@maths.ex.ac.uk> wrote:
> In article <8759dm$ebh$1...@nnrp1.deja.com>,
> ame...@my-deja.com wrote:
> > This can be
> > rewritten as (any upper bound of S1) + (any upper bound of S2) which
> is
> > equivalent to any upper bound of (S1 + S2)
>
> Again, the same nonsequitur. The assertion, without any evidence,
> that each upper bound of S1 + S2 has the form a + b where a and
> b are upper bounds of S1 and S2 respectively.
Naturally a => x, b => y implies a + b => x + y where x is in S1 and y
is in S2 and a and b are upper bounds of S1 and S2 respectively.
That's as close as you can come to showing that S1 + S2 has the form
a + b. If that's not it, I don't see what you're driving at
Suppose z is a supremum of the S where S is a subset of real numbers.
By definition of a least upper bound two conditions must hold to be a
supremum:
1.) z is an upper bound for the set S.
2.) If a is any upper bound for S, z <= a.
#1 tells us that z => a which implies that both a and z are supremums
of the same set, hence equal.
This is the same line of reasoning which I used in proving what I set
out to do in the original problem. If I didn't do this in my proof,
please show me exactly where I went wrong.
Robin Chapman
ame...@my-deja.com wrote:
> In article <8764ql$q6$1...@nnrp1.deja.com>,
> Robin Chapman <r...@maths.ex.ac.uk> wrote:
> > In article <8759dm$ebh$1...@nnrp1.deja.com>,
> > ame...@my-deja.com wrote:
> > > This can be
> > > rewritten as (any upper bound of S1) + (any upper bound of S2)
which
> > is
> > > equivalent to any upper bound of (S1 + S2)
> >
> > Again, the same nonsequitur. The assertion, without any evidence,
> > that each upper bound of S1 + S2 has the form a + b where a and
> > b are upper bounds of S1 and S2 respectively.
>
> Naturally a => x, b => y implies a + b => x + y where x is in S1 and y
> is in S2 and a and b are upper bounds of S1 and S2 respectively.
> That's as close as you can come to showing that S1 + S2 has the form
> a + b. If that's not it, I don't see what you're driving at
What does "S1 + S2 has the form a + b" mean. I understood that
S1 + S2 was a short notation for the *set* {x + y: x in S1, y in S2},
so that you seem to be asserting that a set (S1 + S2) is a number
(a + b).
> Suppose z is a supremum of the S where S is a subset of real numbers.
>
> By definition of a least upper bound two conditions must hold to be a
> supremum:
>
> 1.) z is an upper bound for the set S.
>
> 2.) If a is any upper bound for S, z <= a.
>
> #1 tells us that z => a which implies that both a and z are supremums
> of the same set, hence equal.
>
> This is the same line of reasoning which I used in proving what I set
> out to do in the original problem. If I didn't do this in my proof,
> please show me exactly where I went wrong.
I have done so on numerous occasions. You show that sup(S1) + sup(S2)
is an upper bound of the set S1 + S2, yet never show that it is
the *least* upper bound of that set.
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
ame...@my-deja.com
Robin Chapman <r...@maths.ex.ac.uk> wrote:
> What does "S1 + S2 has the form a + b" mean.
I don't know. You tell me, you brought it up in the first place.
> I understood that
> S1 + S2 was a short notation for the *set* {x + y: x in S1, y in S2},
> so that you seem to be asserting that a set (S1 + S2) is a number
> (a + b).
No. Other posters pointed this out before. I know the difference
between a set and a point, I just wrote it wrong at first that's all.
>
> > Suppose z is a supremum of the S where S is a subset of real
> >numbers.
> >
> > By definition of a least upper bound two conditions must hold to be
> > a
> > supremum:
> >
> > 1.) z is an upper bound for the set S.
> >
> > 2.) If a is any upper bound for S, z <= a.
> >
> > #1 tells us that z => a which implies that both a and z are
> > supremums
> > of the same set, hence equal.
> >
> > This is the same line of reasoning which I used in proving what I
> > set
> > out to do in the original problem. If I didn't do this in my proof,
> > please show me exactly where I went wrong.
>
> I have done so on numerous occasions. You show that sup(S1) + sup(S2)
> is an upper bound of the set S1 + S2, yet never show that it is
> the *least* upper bound of that set.
Oh, but I think I did. You keep talking about how I need to show that
S1 + S2 is in the form a + b as a critical step of the proof, I have
yet to understand what you are driving at. Either you are trying to
confuse me or you think I'm trying to do something that I'm not. The
statement S1 + S2 is in the form a + b is vacuous.
Robin Chapman
ame...@my-deja.com wrote:
> In article <876o3m$efa$1...@nnrp1.deja.com>,
> Robin Chapman <r...@maths.ex.ac.uk> wrote:
>
> > What does "S1 + S2 has the form a + b" mean.
>
> I don't know. You tell me, you brought it up in the first place.
I have no idea. You said it, so you should be able to explain it.
> > I understood that
> > S1 + S2 was a short notation for the *set* {x + y: x in S1, y in
S2},
> > so that you seem to be asserting that a set (S1 + S2) is a number
> > (a + b).
>
> No. Other posters pointed this out before. I know the difference
> between a set and a point, I just wrote it wrong at first that's all.
So, what did you mean to write?
> >
> > > Suppose z is a supremum of the S where S is a subset of real
> > >numbers.
> > >
> > > By definition of a least upper bound two conditions must hold to
be
> > > a
> > > supremum:
> > >
> > > 1.) z is an upper bound for the set S.
> > >
> > > 2.) If a is any upper bound for S, z <= a.
> > >
> > > #1 tells us that z => a which implies that both a and z are
> > > supremums
> > > of the same set, hence equal.
> > >
> > > This is the same line of reasoning which I used in proving what I
> > > set
> > > out to do in the original problem. If I didn't do this in my
proof,
> > > please show me exactly where I went wrong.
> >
> > I have done so on numerous occasions. You show that sup(S1) +
sup(S2)
> > is an upper bound of the set S1 + S2, yet never show that it is
> > the *least* upper bound of that set.
>
> Oh, but I think I did.
You are at liberty to think whatever you like. However it's still the
case that, as far as I can see, you have not shown this.
> You keep talking about how I need to show that
> S1 + S2 is in the form a + b as a critical step of the proof,
I have *never* talked about this.
> I have
> yet to understand what you are driving at. Either you are trying to
> confuse me or you think I'm trying to do something that I'm not. The
> statement S1 + S2 is in the form a + b is vacuous.
Not "vacuous" but meaningless.
You have successfully shown that sup(S1 + S2) <= sup(S1) + sup(S2).
You have frequently asserted the reverse inequality, yet never proved
it.
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
ame...@my-deja.com
Robin Chapman <r...@maths.ex.ac.uk> wrote:
> You have successfully shown that sup(S1 + S2) <= sup(S1) + sup(S2).
> You have frequently asserted the reverse inequality, yet never proved
> it.
In attempting to prove the reverse inequality I choose the direct
method because I thought it would be easier than proof by
contradiction, apparently it's not easier. So I'll see if I can prove
it by contradiction as you suggested before.
By definition of a least upper bound there is no upper bound of S1 + S2
less than it's supremum.
Assume there is an upper bound a + b < sup(S1 + S2) where a and b are
any upper bounds of S1 and S2 respectivly. This argument implies a + b
< x + y where x and y are any elements of S1 and S2 respectivly. Since
a and b are ANY upper bounds of S1 + S2, assume sup S1 = a and sup S2 =
b, so
sup S1 + sup S2 < x + y where x and y are the same as defined before.
So sup S1 + sup S2 < sup(S1 + S2) which is a contradiction, so
sup S1 + sup S2 =< sup(S1 + S2)?
Robin Chapman
ame...@my-deja.com wrote:
> In article <878olq$u7s$1...@nnrp1.deja.com>,
> Robin Chapman <r...@maths.ex.ac.uk> wrote:
> > You have successfully shown that sup(S1 + S2) <= sup(S1) + sup(S2).
> > You have frequently asserted the reverse inequality, yet never
proved
> > it.
>
> In attempting to prove the reverse inequality I choose the direct
> method because I thought it would be easier than proof by
> contradiction, apparently it's not easier. So I'll see if I can prove
> it by contradiction as you suggested before.
>
> By definition of a least upper bound there is no upper bound of S1 +
S2
> less than it's supremum.
its
The following argument is incoherent.
> Assume there is an upper bound a + b < sup(S1 + S2) where a and b are
> any upper bounds of S1 and S2 respectivly.
An upper bound of what?
To show that sup(S1) + sup(S2) is the supremum (the least upper bound)
of S1 + S2 you need to show that (i) it is an upper bound of S1 + S2
(which you have done) and (ii) that it is the least upper bound,
in other words there is no number c such that c < sup(S1) + sup(S2)
and c is an upper bound for S1 + S2. In aiming for a proof by
contradiction one would assume that there is a c with these properties
and then derive an absurdity. Presumably my c is your a + b, so that
you are only assuming that such a c is the sum of an upper bound
of S1 (viz. a) and an upper bound of S2 (viz. b). You never give any
reason why this is so. It's the same non-sequitur as always :-(
> This argument implies a + b
> < x + y where x and y are any elements of S1 and S2 respectivly.
No it doesn't.
> Since
> a and b are ANY upper bounds of S1 + S2, assume sup S1 = a and sup S2
=
> b, so
> sup S1 + sup S2 < x + y where x and y are the same as defined before.
> So sup S1 + sup S2 < sup(S1 + S2) which is a contradiction, so
> sup S1 + sup S2 =< sup(S1 + S2)?
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
ame...@my-deja.com
Robin Chapman <r...@maths.ex.ac.uk> wrote:
> To show that sup(S1) + sup(S2) is the supremum (the least upper bound)
> of S1 + S2 you need to show that (i) it is an upper bound of S1 + S2
> (which you have done) and (ii) that it is the least upper bound,
> in other words there is no number c such that c < sup(S1) + sup(S2)
> and c is an upper bound for S1 + S2.
Or the least upper bound is less than or equal to any upper bound of
the set in which it's the least upper bound. What is wrong with me
using the argument sup(S1) + sup(S2) =< a + b to justify (ii)? As
before, a and b are any upper bounds of the sets S1 and S2
respectively, therefore (a + b) is an upper bound of S1 + S2. Hence by
definition of a least upper bound, sup(S1) + sup(S2) is the supremum of
S1 + S2 i.e.
sup(S1) + sup(S2) = sup (S1 + S2) = sup {x + y: x in S1, y in S2}.
> In aiming for a proof by
> contradiction one would assume that there is a c with these properties
> and then derive an absurdity.
Is there anything wrong with assuming sup(S1) + sup(S2) > sub (S1 + S2)
and then reaching an absurdity that way?
Assume sup(S1) + sup(S2) > sup (S1 + S2),
then sup(S1) + sup(S2) > x + y where x and y are any elements of S1 and
S2 respectively. By definition, sup(S1) >= x and sup(S2) >= y, but
this is not possible when sup(S1) + sup(S2) > x + y so
sup(S1) + sup(S2) =< sup (S1 + S2).
Robin Chapman
ame...@my-deja.com wrote:
> In article <879c8n$cc5$1...@nnrp1.deja.com>,
> Robin Chapman <r...@maths.ex.ac.uk> wrote:
>
> > To show that sup(S1) + sup(S2) is the supremum (the least upper
bound)
> > of S1 + S2 you need to show that (i) it is an upper bound of S1 + S2
> > (which you have done) and (ii) that it is the least upper bound,
> > in other words there is no number c such that c < sup(S1) + sup(S2)
> > and c is an upper bound for S1 + S2.
>
> Or the least upper bound is less than or equal to any upper bound of
> the set in which it's the least upper bound.
OK.
> What is wrong with me
> using the argument sup(S1) + sup(S2) =< a + b to justify (ii)? As
> before, a and b are any upper bounds of the sets S1 and S2
> respectively, therefore (a + b) is an upper bound of S1 + S2.
You would have to show that sup(S1) + sup(S2) <= d where d is
*any* upper bound of S1 + S2. You are only doing this when d
has the form a + b where a, b are upper bounds of S1, S2 respectively.
It's the same non-sequitur as I have repeatedly pointed out:
how do you know that there cannot be an upper bound for S1 + S2
which is not of this form?
> Is there anything wrong with assuming sup(S1) + sup(S2) > sub (S1 +
S2)
> and then reaching an absurdity that way?
No. (But I presume you mean sup(S1 + S2)).
> Assume sup(S1) + sup(S2) > sup (S1 + S2),
> then sup(S1) + sup(S2) > x + y where x and y are any elements of S1
and
> S2 respectively.
OK
> By definition, sup(S1) >= x and sup(S2) >= y, but
OK
> this is not possible when sup(S1) + sup(S2) > x + y so
It certainly is possible.
> sup(S1) + sup(S2) =< sup (S1 + S2).
A new non-sequitur!
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
ame...@my-deja.com
Robin Chapman <r...@maths.ex.ac.uk> wrote:
> You would have to show that sup(S1) + sup(S2) <= d where d is
> *any* upper bound of S1 + S2.
From (i) in a previous post, the set (S1 + S2) is bounded above and
nonempty, hence it has a supremum. Let d be any upper bound of this
set such that (a + b) =< d (where a and b are the same as defined
previously). Hence sup(S1) + sup(S2) =< (a + b) =< d implies sup(S1) +
sup(S2) =< d. Therefore by definition of a least upper bound, sup(S1)
+ sup(S2) is the supremum of the set (S1 + S2) or sup(S1) + sup(S2) =
sup {x + y: x in S1, y in S2}.
Sent via Deja.com http://www.deja.com/
Before you buy.
Robin Chapman
ame...@my-deja.com wrote:
> In article <87begt$ub3$1...@nnrp1.deja.com>,
> Robin Chapman <r...@maths.ex.ac.uk> wrote:
>
> > You would have to show that sup(S1) + sup(S2) <= d where d is
> > *any* upper bound of S1 + S2.
>
> From (i) in a previous post, the set (S1 + S2) is bounded above and
> nonempty, hence it has a supremum. Let d be any upper bound of this
> set such that (a + b) =< d (where a and b are the same as defined
> previously).
So what are a and b. Let me guess.... upper bounds of S1 and S2
respectively? So your subsequent argument is only valid for d
which satisfy d >= a + b for some such a and b. Now what happens
if d is not of this form? Can such a d exist? If not, why not?
You don't address these questions, so your argument is a non-sequitur.
> Hence sup(S1) + sup(S2) =< (a + b) =< d implies sup(S1) +
> sup(S2) =< d. Therefore by definition of a least upper bound, sup(S1)
> + sup(S2) is the supremum of the set (S1 + S2) or sup(S1) + sup(S2) =
> sup {x + y: x in S1, y in S2}.
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
ame...@my-deja.com
Robin Chapman <r...@maths.ex.ac.uk> wrote:
> In article <87dis5$e4m$1...@nnrp1.deja.com>,
> ame...@my-deja.com wrote:
> > In article <87begt$ub3$1...@nnrp1.deja.com>,
> > Robin Chapman <r...@maths.ex.ac.uk> wrote:
> >
> > > You would have to show that sup(S1) + sup(S2) <= d where d is
> > > *any* upper bound of S1 + S2.
> >
> > From (i) in a previous post, the set (S1 + S2) is bounded above and
> > nonempty, hence it has a supremum. Let d be any upper bound of this
> > set such that (a + b) =< d (where a and b are the same as defined
> > previously).
>
> So what are a and b. Let me guess.... upper bounds of S1 and S2
> respectively? So your subsequent argument is only valid for d
> which satisfy d >= a + b for some such a and b.
Scratch that and just let d be any upper bound of the set
(S1 + S2).
> Now what happens
> if d is not of this form? Can such a d exist? If not, why not?
Now d could be less than, equal to or greater than (a + b). (a + b)
may or may not be in (S1 + S2), but we do know that it's an upper bound
of the set. Hence any upper bound of (S1 + S2) will be greater than or
equal to sup(S1) + sup(S2).
Robin Chapman
ame...@my-deja.com wrote:
> In article <87e1vk$nn5$1...@nnrp1.deja.com>,
> Robin Chapman <r...@maths.ex.ac.uk> wrote:
> > In article <87dis5$e4m$1...@nnrp1.deja.com>,
> > ame...@my-deja.com wrote:
> > > In article <87begt$ub3$1...@nnrp1.deja.com>,
> > > Robin Chapman <r...@maths.ex.ac.uk> wrote:
> > >
> > > > You would have to show that sup(S1) + sup(S2) <= d where d is
> > > > *any* upper bound of S1 + S2.
> > >
> > > From (i) in a previous post, the set (S1 + S2) is bounded above
and
> > > nonempty, hence it has a supremum. Let d be any upper bound of
this
> > > set such that (a + b) =< d (where a and b are the same as defined
> > > previously).
> >
> > So what are a and b. Let me guess.... upper bounds of S1 and S2
> > respectively? So your subsequent argument is only valid for d
> > which satisfy d >= a + b for some such a and b.
>
> Scratch that and just let d be any upper bound of the set
> (S1 + S2).
Good.
> > Now what happens
> > if d is not of this form? Can such a d exist? If not, why not?
>
> Now d could be less than, equal to or greater than (a + b).
Again, what are a and b?
> (a + b)
> may or may not be in (S1 + S2), but we do know that it's an upper
bound
> of the set. Hence any upper bound of (S1 + S2) will be greater than
or
> equal to sup(S1) + sup(S2).
This is a non-sequitur. Why must a typical upper bound d of S1 + S2
be >= sup(S1) + sup (S2)? You don't tell us :-(
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
ame...@my-deja.com
Robin Chapman <r...@maths.ex.ac.uk> wrote:
> > Scratch that and just let d be any upper bound of the set
> > (S1 + S2).
Since d is any upper bound of (S1 + S2), d >= x + y where x and y are
any elements of S1 and S2 respectively, d >= sup(S1) + sup(S2) and the
second condition for a supremum of the set (S1 + S2) has been
satified. Therefore sup(S1) + sup(S2) = sup(S1 + S2).
Robin Chapman
ame...@my-deja.com wrote:
>
> > > Scratch that and just let d be any upper bound of the set
> > > (S1 + S2).
>
> Since d is any upper bound of (S1 + S2), d >= x + y where x and y are
> any elements of S1 and S2 respectively,
OK
>d >= sup(S1) + sup(S2)
Now why is this? Note that sup(S1) need not be an element of S1,
so this statement is not necessarily an instance of the
previous (justified) statement that d >= x + y for any x in S1
and y in S2.
> and the
> second condition for a supremum of the set (S1 + S2) has been
> satified. Therefore sup(S1) + sup(S2) = sup(S1 + S2).
There's still a gap in the argument :-(
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
ame...@my-deja.com
Robin Chapman <r...@maths.ex.ac.uk> wrote:
> > Since d is any upper bound of (S1 + S2), d >= x + y where x and y
are
> > any elements of S1 and S2 respectively,
>
> OK
>
> >d >= sup(S1) + sup(S2)
>
From (i) in a previous post, sup(S1) + sup(S2) >= x + y for any x in S1
and y in S2. Therefore the set (S1 + S2) is bounded above by sup(S1) +
sup(S2). By definition of a least upper bound,
1.) sup(S1) + sup(S2) >= sup(S1 + S2)
2.) sup(S1 + S2) >= z where z is any element of the set (S1 + S2).
#2 implies that the set (S1 + S2) is bounded above by sup(S1 + S2) so
we have sup(S1 + S2) >= x + y for any x in S1 and y in S2. From (i),
we have sup(S1) + sup(S2) >= x + y.
Hence sup(S1 + S2) - sup(S1) - sup(S2)>= 0 implies
sup(S1 + S2) >= sup(S1) + sup(S2) which concludes the proof.
William Hale
> In article <87eje4$3oe$1...@nnrp1.deja.com>,
> Robin Chapman <r...@maths.ex.ac.uk> wrote:
>
> > > Scratch that and just let d be any upper bound of the set
> > > (S1 + S2).
>
> Since d is any upper bound of (S1 + S2), d >= x + y where x and y are
> any elements of S1 and S2 respectively, d >= sup(S1) + sup(S2) and the
> second condition for a supremum of the set (S1 + S2) has been
> satified. Therefore sup(S1) + sup(S2) = sup(S1 + S2).
Here you are trying to prove that any upper bound d of the set S1+S2
is greater than or equal to sup(S1) + sup(S2). That is, you are
trying to prove:
Claim1: Let d be a real number.
Then, d is an upper bound of S1+S2 implies d >= sup(S1) + sup(S2).
Rather then proving the implication, you can try to prove the contrapositive
of it:
Claim2: Let d be a real number.
Then,d not >= sup(S1) + sup(S2) implies
d is not an upper bound of S1+S2.
Claim1 and claim2 are equivalent to one another, both being the
contrapositive of the other. However, claim2 is easier to prove.
Try to prove claim2.
Note 1: Proving claim2 is a direct prove. If you try to prove claim1
by contradiction (which was recommended by Mr. Chapman), you would
essentially be proving claim2. I could go into more detail about
this comment if you wish.
Note 2: I beleive that the problem that you are having with the proof
of claim 1 is that you are trying to prove it in the same manner that
you proved that sup(S1) + sup(S2) is an upper bound of S1+S2.
However, the proof of claim1 (or of claim2) is different from that
proof in several ways, which you have avoided by trying to make
the two proofs similar. To recall, here is your proof that
sup(S1) + sup(S2) is an upper bound of S1+S2:
Claim: sup(S1) + sup(S2) is an upper bound of S1+S2
Proof: For any x in S1, x <= sup(S1).
For any y in S2, y <= sup(S2).
Hence, x+y <= sup(S1) + sup(S2).
Let z be any element of S1+S2.
Then, z = x + y for some x in S1 and for some y in S2.
Hence, z = x + y <= sup(S1) + sup(S2).
That is, sup(S1)+sup(S2) is an upper bound of S1+S2.
The proof of claim1 (or of claim2) differs in at least two ways
from the above proof. I believe that your hangup in proving
claim1 (or claim2) is that you are trying to make it the same
type of proof as above.
Note 3: You need to use the fact that sup(S1) is the *least*
upper bound of S1 somewhere in the proof. Note that you have
only used the fact that sup(S1) is an *upper* bound of S1.
--
Bill Hale
ame...@my-deja.com
ha...@mailhost.tcs.tulane.edu (William Hale) wrote:
> Here you are trying to prove that any upper bound d of the set S1+S2
> is greater than or equal to sup(S1) + sup(S2). That is, you are
> trying to prove:
>
> Claim1: Let d be a real number.
> Then, d is an upper bound of S1+S2 implies d >= sup(S1) +
sup(S2).
>
> Rather then proving the implication, you can try to prove the
contrapositive
> of it:
>
> Claim2: Let d be a real number.
> Then,d not >= sup(S1) + sup(S2) implies
> d is not an upper bound of S1+S2.
Suppose d is a real number such that d < sup(S1) + sup(S2), then
d - sup(S1) < sup(S2). As a consequence of the definition of least
upper bounds, there exists a y in S2 such that d - sup(S1) < y implies
d < y + sup(S1). Therefore, d cannot be an upper bound of the set
(S1 + S2). This concludes the proof of claim2.
By the contrapositive law, d is an upper bound for (S1 + S2)
and d >= sup(S1) + sup(S2). Also, is has been shown earlier that
sup(S1) + sup(S2) is an upper bound for the set (S1 + S2) thus
sup(S1) + sup(S2) = sup(S1 + S2). QED
Robin Chapman
ame...@my-deja.com wrote:
> In article <87i0vr$gjh$1...@nnrp1.deja.com>,
> Robin Chapman <r...@maths.ex.ac.uk> wrote:
> > > Since d is any upper bound of (S1 + S2), d >= x + y where x and y
> are
> > > any elements of S1 and S2 respectively,
> >
> >
> > >d >= sup(S1) + sup(S2)
> >
>
S1
> and y in S2. Therefore the set (S1 + S2) is bounded above by sup(S1)
+
> sup(S2). By definition of a least upper bound,
>
> 1.) sup(S1) + sup(S2) >= sup(S1 + S2)
> 2.) sup(S1 + S2) >= z where z is any element of the set (S1 + S2).
>
> #2 implies that the set (S1 + S2) is bounded above by sup(S1 + S2) so
> we have sup(S1 + S2) >= x + y for any x in S1 and y in S2. From (i),
> we have sup(S1) + sup(S2) >= x + y.
> Hence sup(S1 + S2) - sup(S1) - sup(S2)>= 0
Non-sequitur time again! Where on earth does this inequality come from?
> implies
> sup(S1 + S2) >= sup(S1) + sup(S2) which concludes the proof.
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
Robin Chapman
ame...@my-deja.com wrote:
> In article <hale-05020...@dialup04.tcs.tulane.edu>,
> ha...@mailhost.tcs.tulane.edu (William Hale) wrote:
>
> > Here you are trying to prove that any upper bound d of the set S1+S2
> > is greater than or equal to sup(S1) + sup(S2). That is, you are
> > trying to prove:
> >
> > Claim1: Let d be a real number.
> > Then, d is an upper bound of S1+S2 implies d >= sup(S1) +
> sup(S2).
> >
> > Rather then proving the implication, you can try to prove the
> contrapositive
> > of it:
> >
> > Claim2: Let d be a real number.
> > Then,d not >= sup(S1) + sup(S2) implies
> > d is not an upper bound of S1+S2.
>
> Suppose d is a real number such that d < sup(S1) + sup(S2), then
> d - sup(S1) < sup(S2). As a consequence of the definition of least
> upper bounds, there exists a y in S2 such that d - sup(S1) < y implies
> d < y + sup(S1).
OK An advance at last. But
> Therefore, d cannot be an upper bound of the set
> (S1 + S2).
Now why cannot d be an upper bound of S1 + S2? Were sup(S1) to be
an element of S1, then y + sup(S1) would be an element of S1 + S2
and we would get a contradction between d being an upper bound
of S1 + S2 and d < y + sup(S1). But sup(S1) need not be an element
of S1 ....
William Hale
<r...@maths.ex.ac.uk> wrote:
> In article <87j911$av4$1...@nnrp1.deja.com>,
> ame...@my-deja.com wrote:
> > In article <hale-05020...@dialup04.tcs.tulane.edu>,
> > ha...@mailhost.tcs.tulane.edu (William Hale) wrote:
> >
> > > Here you are trying to prove that any upper bound d of the set S1+S2
> > > is greater than or equal to sup(S1) + sup(S2). That is, you are
> > > trying to prove:
> > >
> > > Claim1: Let d be a real number.
> > > Then, d is an upper bound of S1+S2 implies d >= sup(S1) +
> > sup(S2).
> > >
> > > Rather then proving the implication, you can try to prove the
> > contrapositive
> > > of it:
> > >
> > > Claim2: Let d be a real number.
> > > Then,d not >= sup(S1) + sup(S2) implies
> > > d is not an upper bound of S1+S2.
> >
> > Suppose d is a real number such that d < sup(S1) + sup(S2), then
> > d - sup(S1) < sup(S2). As a consequence of the definition of least
> > upper bounds, there exists a y in S2 such that d - sup(S1) < y implies
> > d < y + sup(S1).
>
> OK An advance at last.
Yes.
In fact, this approach is much better than the approach I had in mind.
I said previously that the method of proof of claim2 differs in two ways
from the method of proof that sup(S1) + sup(S2) is an upper bound of
the set S1+S2. One way I thought that it differed was that the proof
of claim2 involved calculation rather than being just "verbal". My
proof of claim2 uses calculation, but ameen2 has nicely avoided
those calculations in the above approach of his. Nice going, ameen2!!
Thus, a proof of claim2 differs in one way from the proof of the
upper bound property of sup(S1) + sup(S2). What I had in mind was
that the proof of the upper bound property uses "for any" a lot,
while the proof of the least property uses "for some" or "there
exists" a lot. You have got off the track of trying to use the
universal quantifier "for any" or "for all", and you are now using
the existential quantifier "for some" or "there exists" or "find a y
such that ...".
--
Bill Hale
ame...@my-deja.com
> > > Claim2: Let d be a real number.
> > > Then,d not >= sup(S1) + sup(S2) implies
> > > d is not an upper bound of S1+S2.
> >
> > Suppose d is a real number such that d < sup(S1) + sup(S2), then
> > d - sup(S1) < sup(S2). As a consequence of the definition of least
> > upper bounds, there exists a y in S2 such that d - sup(S1) < y
implies
> > d < y + sup(S1).
>
> OK An advance at last. But
>
> > Therefore, d cannot be an upper bound of the set
> > (S1 + S2).
>
> Now why cannot d be an upper bound of S1 + S2?
> Were sup(S1) to be
> an element of S1, then y + sup(S1) would be an element of S1 + S2
Then one would be able to find an element of the set (S1 + S2) greater
than d.
"cut"
> But sup(S1) need not be an element
> of S1 ....
Anything less than sup(S1) will have an element in S1 greater than that
element.
Robin Chapman
Are you saying that since d - y < sup(S1) then there is x in S1 with
d - y < x .... ?
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
ame...@my-deja.com
Robin Chapman <r...@maths.ex.ac.uk> wrote:
> > Anything less than sup(S1) will have an element in S1 greater than
> that
> > element.
>
> Are you saying that since d - y < sup(S1) then there is x in S1 with
> d - y < x .... ?
Yes, this follows from the definition of a least upper bound. This
looks more definitive than what I did. Although my last statement
looks a bit uglier than this, it's still sufficient. I simply choose
to stop at an earlier point without that detail.