Prove Sup a_n=Inf b_n for (a_n,b_n) sequence of nested intervals

[rabbits77]

Let (a_n,b_n) be a sequence of nested open intervals. Assume
that intersection of all (a_n, b_n) is empty.
Show that Sup a_n=Inf b_n

Proof:
{a_n} is an increasing sequence bounded above
and {b_n} is a decreasing sequence bounded below.
We know from previous theorems that since {a_n} and {b_n}
are monotonically increasing or decreasing
and lim b_n=inf b_n
and lim a_n=sup a_n
Need to show lim(a_n - b_n)=0
To show this...
b - a = (b - b_n) + (b_n - a_n) + (a_n - a)
|b - a| <= |b - b_n| + |b_n - a_n| + |a_n - a|
for any e > 0 we can find N such that for all n > N
|b - b_e| < e/3
|b_n - a_n| < e/3
|a_n - a| < e/3
and so |b - a| < e. Thus b - a=0 and a=b.

Is my proof correct?

[William Elliot]

On Mon, 6 Oct 2008, rabbits77 wrote:

> Let (a_n,b_n) be a sequence of nested open intervals. Assume

Of course, they are all not empty, ie no j with bj <= aj.

> that intersection of all (a_n, b_n) is empty.
> Show that Sup a_n=Inf b_n
>

a = sup_j aj <= b = inf b_j.
If a < b, then
nonnul (a,b) subset /\_j (aj, bj).

> Proof:
> {a_n} is an increasing sequence bounded above
> and {b_n} is a decreasing sequence bounded below.

Generalize to an uncountable nest of open intervals.

[TheGist]

rabbits77 wrote:
> Let (a_n,b_n) be a sequence of nested open intervals. Assume
> that intersection of all (a_n, b_n) is empty.
> Show that Sup a_n=Inf b_n
>

I think I previously made things harder and more awkward than necessary.
I think we can say that by the definition of nested intervals
lim(b_n - a_n)=0 since the distance between a_n,b_n approaches zero as
n-> oo.

Proof:
{a_n} is an increasing sequence bounded above
and {b_n} is a decreasing sequence bounded below.
We know from previous theorems that since {a_n} and {b_n}
are monotonically increasing or decreasing

then lim b_n=inf b_n
and lim a_n=sup a_n
Since lim(b_n - a_n)=0 we have b - a = lim b_n - lim a _n = lim(b_n -
a_n)=0 . Thus inf b_n = sup a_n .

I think this is a pretty straightforward and simple proof now.
Unless I made nay errors. Have I?

[rabbits77]

Here is another attempt...

Let (a_n,b_n) be a sequence of nested open intervals. Assume
that intersection of all (a_n, b_n) is empty.
Show that Sup a_n=Inf b_n

Proof:

{a_n} is an increasing sequence bounded above
and {b_n} is a decreasing sequence bounded below.
We know from previous theorems that since {a_n} and {b_n}
are monotonically increasing or decreasing
then lim b_n=inf b_n
and lim a_n=sup a_n

lim b_n - lim a_n=lim(b_n - a_n)= lim L/2^(n-1)=0
where L is the distance of I_0 and the length of I_n is L/2^(n-1).
Since, for example, we know(from a previous theorem) that for 0 < a < 1
we have a^n -> 0
Also, we know that this distance
approaches zero as (a_n,b_n) becomes a smaller and smaller interval as
a_n and b_n get closer and closer for each successive nested interval.

Is that correct?

If so then it follows that lim b_n - lim a _n = lim(b_n - a_n)=0 . Thus

[rabbits77]

rabbits77 wrote:
> Here is another attempt...
> Let (a_n,b_n) be a sequence of nested open intervals. Assume
> that intersection of all (a_n, b_n) is empty.
> Show that Sup a_n=Inf b_n
>
> Proof:
> {a_n} is an increasing sequence bounded above
> and {b_n} is a decreasing sequence bounded below.
> We know from previous theorems that since {a_n} and {b_n}
> are monotonically increasing or decreasing
> then lim b_n=inf b_n
> and lim a_n=sup a_n
> lim b_n - lim a_n=lim(b_n - a_n)= lim L/2^(n-1)=0
> where L is the distance of I_0 and the length of I_n is L/2^(n-1).

I say this because if I_0 =(-L,L) and we bisect it repeatedly so that
I_1=(0,L) and so forth.

[rabbits77]

William Elliot wrote:
> On Mon, 6 Oct 2008, rabbits77 wrote:
>
>> Let (a_n,b_n) be a sequence of nested open intervals. Assume
>
> Of course, they are all not empty, ie no j with bj <= aj.
>
>> that intersection of all (a_n, b_n) is empty.
>> Show that Sup a_n=Inf b_n
>>
> a = sup_j aj <= b = inf b_j.
> If a < b, then
> nonnul (a,b) subset /\_j (aj, bj).

I am sorry(this may be a minor confusion) but I am not sure what you
mean by this?
If a < b then the open non-empty interval (a,b) is a subset of
the intersection of all (a_j,b_j) but since we know that this
is empty then we have a contradiction and we are done?

[William Elliot]

On Tue, 7 Oct 2008, rabbits77 wrote:
> William Elliot wrote:
> > On Mon, 6 Oct 2008, rabbits77 wrote:
> >
> >> Let (a_n,b_n) be a sequence of nested open intervals. Assume
> >
> > Of course, they are all not empty, ie no j with bj <= aj.
> >
> >> that intersection of all (a_n, b_n) is empty.
> >> Show that Sup a_n=Inf b_n
> >>
> > a = sup_j aj <= b = inf b_j.
> > If a < b, then
> > nonnul (a,b) subset /\_j (aj, bj).
>
> I am sorry(this may be a minor confusion) but I am not sure what you
> mean by this?

In this senctece you are speaking English, not math.
Thus a blank space is required to be between 'sorry' and '(".
So don't be confused about punctuation.

> If a < b then the open non-empty interval (a,b) is a subset of
> the intersection of all (a_j,b_j) but since we know that this
> is empty then we have a contradiction and we are done?
>

Yes, you got it; the easy way.

----

[Tim Smith]

In article <gcen6v$19c$1...@aioe.org>, rabbits77 <rabb...@my-deja.com>
wrote:

> Here is another attempt...
> Let (a_n,b_n) be a sequence of nested open intervals. Assume
> that intersection of all (a_n, b_n) is empty.
> Show that Sup a_n=Inf b_n

Here's an interesting supplemental problem that the above problem
suggests, that you might find interesting to have a go at:

Show that one of the sets {a_i}, {b_i} must be finite, and the
other one must be infinite.

If you are getting a good grasp of the material you are studying, this
should be a short and easy proof.

--
--Tim Smith