Why do irrationals have measure 1 on [0, 1] and rationals have measure 0?
Archie
An interval [0, 1] has measure 1. The rationals on this interval have
measure 0 (and the set of the rationals in general). A given interval
(a, b) has outer measure b - a. So it seems to me that any interval of
all real numbers has measure > 0. I get that the rationals have
measure 0 because each point can be thought of as a disjoint set, and
each point can be contained in an interval < epsilon for any
arbitrarily small epsilon. Between any two rationals there is an
irrational number.
But between any two irrationals there is a rational number. What makes
the irrationals different? There are "more" of them? How does this
make sense?
Thanks.
Dirk Van de moortel
9f6a60cc-228c-4395...@r3g2000vbp.googlegroups.com
The why-question in mathematics is usually answered by looking at
and understanding proofs.
There are "more" irrationals than rationals in the sense that we can
make a one-to-one relation between the naturals ( {0,1,2,3,4,...} )
and the rationals, but we cannot make such a relation between the
naturals and the irrationals.
Google for:
infinity cantor "set theory" diagonal
Dirk Vdm
Archie
<dirkvandemoor...@nospAm.hotmail.com> wrote:
> Archie <kohmover...@yahoo.com> wrote in message
>
> 9f6a60cc-228c-4395-9b55-49899a833...@r3g2000vbp.googlegroups.com
Thanks, but isn't it true that the Cantor set is uncountable but still
has measure 0? So the elements of the Cantor set cannot be put into
one-to-one correspondence with the naturals, right?
Mariano Suárez-Alvarez
Indeed.
It is true that a countable set always have measure zero,
but it is not true that all uncountable setshave positive
measure. You need to know how the points are actually
laid out in the line to know the measure.
With the Cantor set, what happens is that the set has so
many holes and theses holes are so big when compared with
the set itself that the measure is zero. You can construct
variations of the Cantor set (by taking away different
intervals of varying width) to get sets which are of the same
cardinal and of the same rough shape as the original Cantor
set but which have positive measure.
-- m
fishfry
<129a41c3-cc4a-4a7b...@r36g2000vbr.googlegroups.com>,
Archie <kohmo...@yahoo.com> wrote:
Yes, true. Any countable set has measure 0. The Cantor set is a
counterexample to the converse. That is, the proposition "Any set of
measure 0 must be countable" is shown to be false by the existence of
the Cantor set.
Virgil
<9f6a60cc-228c-4395...@r3g2000vbp.googlegroups.com>,
Archie <kohmo...@yahoo.com> wrote:
For one thing, the rationals can be ennumerated (put into a list with a
first, second, etc., so that all of them are accounted for) but the
irrationals cannot (a secondary result of the Cantor "diagonal" theorem
showing that the set of all binary infinite seqeunces cannot be
ennumerated).
Thus even though each is dense in the other, there are "more"
irrationals than rationals.
It can also be shown that the (countable) set of ratinoals in any
interval has an open cover of arbitrarily small total measure.
Virgil
Right!
rabbits77
singleton of measure 0 and add up countably many of them and they
all have measure 0!
Peter Webb
Some things are just denser than others. It is a paradoxical in terms of our
finite intuition, but so are a lot of things in this area.
Bill
"Peter Webb" <webbf...@DIESPAMDIEoptusnet.com.au> wrote in message
news:4a07a7d3$0$7377$afc3...@news.optusnet.com.au...
Note too that there "A LOT MORE" irrational numbers than rational numbers.
David C. Ullrich
Yes. Go back to Dirk's first sentence: Sometimes the reason
something's true is really given by the proof and that's just
the way it is.
Forget the irrationals. The fact that an open interval (0,1)
has outer measure 1 really isn't obvious, at least not to me.
It may be obvious that that's what we _want_ the outer
measure to be, but given that we can cover the rationals
by that union of intervals of small total length, which
sounds implausible at first, how do we know for sure
that we can't cover an open interval by a union of open
intervals of small total length?
I know how to prove that, it's not too hard. But if you
have a good explanation for why it's obviously true
let me know.
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Dave L. Renfro
> Forget the irrationals. The fact that an open interval (0,1)
> has outer measure 1 really isn't obvious, at least not to me.
> It may be obvious that that's what we _want_ the outer
> measure to be, but given that we can cover the rationals
> by that union of intervals of small total length, which
> sounds implausible at first, how do we know for sure
> that we can't cover an open interval by a union of open
> intervals of small total length?
>
> I know how to prove that, it's not too hard. But if you
> have a good explanation for why it's obviously true
> let me know.
Indeed ... If Archie is interested, a few minutes ago I
posted some notes of mine from a course I taught back in
Fall 2001 where this result about intervals (first prove
it for a non-degenerate closed and bounded interval ...)
is proved (pp. 13-15 in the notes). I found this result
possibly the most difficult to write out all the details
for in comparison to all the other results we covered during
at least the first couple of months of the course.
For the notes I'm talking about, see the .pdf file that is
attached to the following Math Forum archived sci.math post.
(Some of the natural page breaks in the original LaTeX notes
didn't "break" correctly during the .pdf conversion.)
http://mathforum.org/kb/message.jspa?messageID=6704953
Dave L. Renfro
David C. Ullrich
<renf...@cmich.edu> wrote:
>David C. Ullrich wrote (in part):
>
>> Forget the irrationals. The fact that an open interval (0,1)
>> has outer measure 1 really isn't obvious, at least not to me.
>> It may be obvious that that's what we _want_ the outer
>> measure to be, but given that we can cover the rationals
>> by that union of intervals of small total length, which
>> sounds implausible at first, how do we know for sure
>> that we can't cover an open interval by a union of open
>> intervals of small total length?
>>
>> I know how to prove that, it's not too hard. But if you
>> have a good explanation for why it's obviously true
>> let me know.
>
>Indeed ... If Archie is interested, a few minutes ago I
>posted some notes of mine from a course I taught back in
>Fall 2001 where this result about intervals (first prove
>it for a non-degenerate closed and bounded interval ...)
>is proved (pp. 13-15 in the notes). I found this result
>possibly the most difficult to write out all the details
>for in comparison to all the other results we covered during
>at least the first couple of months of the course.
It's not _that_ hard, especially if you cheat:
The letter I will denote an interval, with length |I|.
One inequality is clear; for the other direction
we need to show that if
(a,b) subset union_j I_j
then
sum |I_j| >= b - a.
So it's enough to show that
sum |I_j| >= b - a - 2 epsilon
for any epsilon > 0. Now K = [a+epsilon, b-epsilon]
is contained in the union of finitely many I_j:
K subset union_j=1^N I_j
and we only need to show that
sum_j=1^N |I_j| >= |K|.
Since we're only talking about finitely many I_j now
that last inequality is amenable to a little bit of
fussing around. Or you can cheat: The last inequality
is immediate if you know the basic properties of the
Riemann integral! (In particular the needed fussing
around is contained in the proof that the Riemann
integral is additive.)
>For the notes I'm talking about, see the .pdf file that is
>attached to the following Math Forum archived sci.math post.
>(Some of the natural page breaks in the original LaTeX notes
>didn't "break" correctly during the .pdf conversion.)
>
>http://mathforum.org/kb/message.jspa?messageID=6704953
>
>Dave L. Renfro
David C. Ullrich