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I'm self-studying real analysis using Abbott's text "Understanding Analysis." I'm trying to think out/prove as much on my own as I can, so I am working on proving the Nested Interval Property (Theorem 1.4.1 in the book) using "just" the Axiom of Completeness. The author does prove it in the book, but as I say, I like to try to prove things myself before reading the author's proof.

So I have what I think is a convincing proof, but I would just like anyone who's willing to take a look and tell me if this is a convincing proof or not. Please don't give me a proof that works. It is best if you just tell me if you are not convinced and why.

Here is the theorem:

For each natural number n, assume we've been given a closed interval $I_n = [a_n, b_n] = \{x \in \mathbb R : a_n \le x \le b_n\}$. Assume also that each $I_n$ contains $I_{n+1}$. Then the resulting nested sequence of closed intervals has a nonempty intersection.

Here is my proof:

For any natural number n, consider the set $\{a_1, a_2, a_3, ...\}$. (I drew a diagram to work this out. Hopefully it is not unclear.) This set is bounded above, i.e. by any member of the set $\{b_1, b_2, b_3, ..., b_n\}$. By the axiom of completeness, therefore, the set has a supremum. Now suppose the intersection of the sequence of nested intervals is empty. Then for some natural number $n$, $[a_n, b_n]$ is empty. This means there can be no such number as $a_n$, for since $[a_n, b_n]$ is defined to include $a_n$, if $a_n$ existed, the set would not be empty. However, if there is no such number $a_n$, then the nonempty, bounded set $\{a_1, a_2, a_3, ...\}$ can have no supremum. This contradicts the axiom of completeness. Therefore the intersection cannot be empty. QED.

Right now I'm not so concerned with finer points of the proof, except insofar as they contribute to whether or not it is convincing.

Thank you.

EDIT: I believe I now have a convincing proof. To prove that the intersection of closed, nested intervals is non-empty, I need to show that for every natural number n, there is some real number x such that $a_n \le x \le b_n$ I broke this into two parts to make it easier for my simple brain. I need to show that there is some number x that is greater than or equal to a_n for every n, and I need to show that there is some number y that is less than or equal to b_n for every n. These need not be the same number.

So now I form the sets A = {$a_1, a_2, ..., a_n$} and B = {$b_1, b_2, ..., b_n$}. Since the intervals are nested, every member of B is an upper bound for A. Therefore, by the axiom of completeness, A has a supremum. Call it s. By definition of supremum $s \ge a$ for every a in A. Likewise, by definition of supremum and because every member of B is an upper bound of A, we have $s \le b$ for every b in B. This shows that for every a and for every b, there is a real number s such that $a \le s \le b$ This is enough to show that the intersection of the nested intervals always contains at least one real number, i.e that it is non-empty. QED.

My only concern with this is that the sets A and B have infinite number of members. But I don't think this matters, since they are both bounded. But it makes me less than comfortable with the "for every" quantifier, knowing that there are infinitely many. I considered induction but I don't see a remotely elegant way of doing that.

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  • $\begingroup$ To see that $a_n\le b_m$ for all $n,m\in \mathbb{N}$, simply consider the cases $m\le n$ and $n\le m$ separately. In the first case, it holds that $a_n \le b_n \le b_m$ and in the second case you have $a_n\le a_m\le b_m$. The rest of the proof looks fine. $\endgroup$ – Vincent Boelens Jul 7 '14 at 16:20
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    $\begingroup$ When you are self-studying you might not get to see just how picky mathematicians are. You write "So now I form the sets $A = \{a_1,a_2,...,a_n\}\ . . . $" Written this way suggests that you have some integer $n$ in mind [which you neglect to tell us about] and that your set $A$ contains only these few points. I know you want $A$ to contain the entire sequence $\{a_i\}$ but it is a requirement in a valid proof that you make every statement quite correct. $\endgroup$ – B. S. Thomson Nov 15 '15 at 1:58
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A few issues - but the main one is that the intersection of intervals being empty doesn't mean that you can find an $n$ such that $[a_n,b_n]$ is empty - it is here that you need to use the nested property.

Hint: Consider the "right" brackets and the "left" brackets. Can there be a "left" bracket on the "side" of the "right" brackets?

We certainly have $a_1<a_2<\ldots$ and $b_1>b_2>\ldots$. But this means that also $a_n < b_m$ $\forall n, m$.

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  • $\begingroup$ It isn't clear to me why the intersection of intervals being empty doesn't imply the emptiness of some [a_n, b_n] interval. The intersection is precisely the contents of the "skinniest interval", is it not? And as n goes to infinity, the contents of this interval is narrowed down to a single real number, i.e. no gaps. Right? Clearly there's something I misunderstand. $\endgroup$ – dustOfRumors Jul 7 '14 at 8:06
  • $\begingroup$ The issue here is that the result IS true. But it's only true because the intervals are nested, and to say that it is true without proving this implicitly assumes the result. For example $[1,2] [1.5,2]$ and $[1,1.25]$ have empty intersection but none of them are empty. They are of course not nested. $\endgroup$ – Mathmo123 Jul 7 '14 at 8:10
  • $\begingroup$ I guess I don't know what a "proof that the intervals are nested" would look like. Maybe this is where I misunderstand. It seems to me the intervals are assumed to be nested. But you are saying that I need to actually PROVE that they are nested? (Which would be the point of showing that no "left" bracket could appear on the "side" of the "right" brackets?) $\endgroup$ – dustOfRumors Jul 7 '14 at 8:14
  • $\begingroup$ Or are you saying that I have not used the fact that they are nested in my proof? $\endgroup$ – dustOfRumors Jul 7 '14 at 8:15
  • $\begingroup$ I'm saying the latter. That you need to explain why being nested gives you the property that you're claiming is true. $\endgroup$ – Mathmo123 Jul 7 '14 at 8:15
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In the part prior to your edit, you correctly state that $\sup_na_n\leq b_m$ for any $m.$ But after this, you state that $\phi=\cap_{n\in \mathbb N}[a_n,b_n] \implies \exists n\;( [a_n,b_n]=\phi)$ which does not immediately follow from what precedes it.

In the part after your edit, note that it is not assumed that $\{a_n\}_n$ or $\{b_n\}_n$ are infinite sets. It could be that $a_n=a_{42}$ for all $n\geq 42.$

For any non-empty subsets $A,B$ of $\mathbb R ,$ if $\;\forall a\in A \;\forall b \in B\;(a\leq b)\;$ then $\sup A\leq \inf B. $ You correctly showed this for the sets $A=\{a_n\}_n$ and $B=\{b_n\}_n.$

You should justify why any $s$ such that $\sup A\leq s\leq \inf B$ belongs to every $[a_n,b_n].$ Which is, that for any $n$ we have $a_n\leq \sup A\leq s\leq \inf B\leq b_n,$ which implies $a_n\leq s\leq b_n.$ This may be obvious, but when you write out all the details of a proof it is much easier for you to confirm your correctness, or to find errata.

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