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I came across the following problem on intervals and sequences:

Let $I$ be an interval, $(x_n)$ a sequence in $I$ that converges to a point $x$ of $I$. Prove that there exists a closed interval $J \subset I$ that contains $x$ and every $x_n$.

Here is my attempted proof:

Proof. We know that $x_n \to x$. Let us "add" $-x$ to both sides. Then $x_n \to 0$. Form a subsequence $x_{n_{k_{1}}}$ by deleting all the negative terms. Put $x_{n} := x_{n_{k_{1}}}$. Then we can look at the case where $x_n \geq 0$ for all $n$. Since $x_n \to 0$, $\exists N_1 \ni x_{n} \leq x_{N_{1}}$ ultimately. Form another subsequence $x_{n_{k_{2}}}$ by excluding all the positive terms. Put $x_n := x_{n_{k_{2}}}$. Then $x_n \to 0$ and it suffices to show that $(x_n)$ has a minimal element. In other words, there exists an $N_2$ such that $x_n \geq x_{N_{2}}$ ultimately.

Would this be correct?

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Your attempt is absolutely no good. You are required to find an interval that contains all terms of the sequence. You immediately go to a sub sequence and only work with the subsequence, all your conclusions are about the subsequence. Doesn't matter what you conclude about the terms of the subsequence, you haven't succeeded in proving anything about all terms of the original sequence, which is what you were asked to do in the first place. Worse, you don't even conclude anything about all terms of the subsequence, only for all terms from some point on. –  Arturo Magidin Jun 26 '11 at 20:02
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If he considers the subsequence of the negative terms and the subsequence of positive terms individually and unite both intervals he's okay. His idea wasn't that bad. –  Patrick Da Silva Jun 26 '11 at 20:06
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@Patrick: The problem asks for a closed interval that contains all terms; he produces a closed interval that eventually contains some terms (not even the entire subsequence chosen), and stops. I think that's not very good, regardless of how much extra work can be appended at the end to get this to be a good beginning. Doesn't contain all terms of the chosen subsequence, no indication that there is any concern about those missing terms from the subsequence, and absolutely no word or indication of what to do with the "negative" ones. –  Arturo Magidin Jun 26 '11 at 20:14
    
Your nomenclature is not very good; a subsequence is usually denoted $x_{n_k}$, and that means that the indices are $n_1$, $n_2$, $n_3$, etc. When you write $x_{n_{k_1}}$, you are specifying a single term, not a subsequence; likewise when you write $x_{n_{k_2}}$. Better would be to use $x_{n_k}$ for one subsequence, and $x_{m_k}$ for another, or perhaps $x_{n_{k,1}}$ and $x_{n_{k,2}}$. –  Arturo Magidin Jun 26 '11 at 22:57
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@Arturo : I said his IDEA wasn't that bad. I didn't say he was correct... and there is no point in being aggressive on someone trying to explain his point of view ; take it easy, brother. @Damien : Nice try for the editing, but you still lack some details. If you edit it one more time I'd be happy to comment. –  Patrick Da Silva Jun 27 '11 at 0:10

4 Answers 4

up vote 4 down vote accepted

Intuition. If $x$ is inside the interval $I$, then we can find a small closed interval $C$ around $x$ which is completely contained in $I$; because $x$ is the limit of the sequence, all except perhaps for finitely many terms of the sequence will be in $C$. Then just look at the missing terms; since there are only finitely many of them, we can enlarge the interval $C$ until it includes all the missing terms.

The idea works if $x$ is not inside the interval as well, though one has to be a bit careful then.

Proof. There is one degenerate case to consider; if singletons count as intervals, then we must consider the possibility that $I=\{p\}$; in that case, the limit point and all terms of the sequence must equal $p$, so the result holds with $J=I=\{p\}$. We therefore assume that $I$ has nonempty interior.

Consider first the case in which $x$ is in the interior of the interval $I$. Then there exists $\epsilon_1\gt 0$ such that $(x-\epsilon_1,x+\epsilon_1)\subseteq I$; if $\epsilon= \frac{1}{2}\epsilon_1$, then $[x-\epsilon,x+\epsilon]\subseteq I$.

Since the sequence converges to $x$, there exists $N\gt 0$ such that for all $n\geq N$, $x_n\in (x-\epsilon,x+\epsilon)$.

Let $b = \max \{ x+\epsilon, x_1, x_2,\ldots, x_N\}$. I claim first that $x_k\leq b$ for all $k$. Indeed, if $k\leq N$, then $x_k\leq \max\{x+\epsilon,x_1\ldots,x_N\}$, since $x_k$ is one of the elements of this finite set. And if $k\gt N$, then by the choice of $N$, $x_k\in (x-\epsilon,x+\epsilon)$, hence $x_k\lt x+\epsilon \leq b$.

I further claim that $b\in I$. Indeed, since $[x-\epsilon,x+\epsilon]\in I$, then $x+\epsilon\in I$. Since $(x_n)$ is a sequence in $I$, then $x_1,\ldots,x_N$ are each in $I$. Since this is a finite set of elements of $I$, the maximum is in $I$.

So $b\in I$, and $x_n\leq b$ for all $n\in\mathbb{N}$.

Now let $a = \min\{x-\epsilon, x_1,\ldots,x_N\}$. Then $x_k\geq a$ for all $k$: if $k\leq N$, because $x_k$ is in the set; and if $k\geq N$, because then $x_k\in (x-\epsilon,x+\epsilon)$. And $a\in I$, because each of $x_1,\ldots,x_N$ is in $I$, and $[x-\epsilon,x+\epsilon]\subseteq I$. So $a\in I$, and $a\leq x_n$ for all $n$. Also, clearly $a\lt b$, since $a\leq x-\epsilon\lt x+\epsilon\leq b$.

Since $I$ is an interval, $a\in I$, and $b\in I$, then $[a,b]\subseteq I$.

So we have the closed interval $J=[a,b]$, contained in $I$, and $a\leq x_n\leq b$ for all $n$, as desired.

If $x$ is not in the interior of $I$, then it is an endpoint of $I$. Suppose first that $x$ is the upper endpoint of $I$. Then there exists $\epsilon\gt 0$ such that $[x-\epsilon,x]\subseteq I$. Proceed then as above, working in the interval $(x-\epsilon,x+\epsilon)$, keeping in mind that all $x_i$ will necessarily lie in $(x-\epsilon,x]$. If $x$ is the lower endpoint of $I$ instead, then again we can find $\epsilon\gt 0$ such that $[x,x+\epsilon]\subseteq I$, and we can proceed the same way again.

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@Arturo: In my edited post, could I take $a = x_{N_{2}}$ and $b=x_{N_{1}}$? –  Damien Jun 26 '11 at 22:44
    
@Damien: No, because although $x_n\lt x_{N_1}$ ultimately, it is not necessarily true that $x_n\lt x_{N_1}$ for all $n$, which is the property you need for $b$. Likewise for $a$ and $x_{N_2}$. –  Arturo Magidin Jun 26 '11 at 23:01
    
@Damien: Correction: you only need $x_n\leq b$ for all $n$, not $x_n\lt b$ for all $n$. But the key is that it must happen for all $n$, not just "ultimately", because you want $J$ to contain $x$ "and every $x_n$". –  Arturo Magidin Jun 26 '11 at 23:21
    
@Damien: I think your approach would work, since every point of your sequence $x_n-x$ is in the nonnegative subsequence or in the nonpositive subsequence. But you have to make that explicit: for all $n \geq \max(N_1,N_2)$, if $x_n-x$ is positive it belongs in the nonnegative subsequence and falls in between $x_{N_1}$ and $x_{N_2}$ (by virtue of being nonnegative), and vice versa for $x_n-x$ being negative. (Arturo, feel free to correct me if I'm wrong.) –  Michael Chen Jun 27 '11 at 4:08
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@Michael: The rest is fairly easy to wrap up... if you're familiar with the usual techniques. For instance, the way in which one takes care of the outliers is precisely the way one shows that a convergent sequence is bounded. But when one is beginning, all the details should be taken care of carefully. –  Arturo Magidin Jun 27 '11 at 4:29

I guess it could, but there is a way to make it more general : since $E = \{ x_n \}_{n \in \mathbb N} \cup \{ x \} \subseteq I$, it is bounded, and since it contains all its limit points, it is also closed, hence compact. Therefore letting $a = \inf E$ and $b = \sup E$, you can notice that $E \subseteq [a,b] \subseteq I$ and letting $J = [a,b]$ completes the proof.

Hope that helps,

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I have looked with interest at the string of questions that you have asked recently, and have been impressed with your rapid progress.

I will describe my perception of a few weaknesses in your approaches, in the context of this specific question. Roughly speaking, I think you look too hard for standard tools that will solve a problem, and pay insufficient attention to the geometry. There is also, in my opinion, too much of an attempt to write very condensed proofs. A first proof should not be so condensed; later, if there is reason to condense, you can look for opportunities to do that.

In this problem, you subtracted the limit $x$, probably because you were fresh from work on null sequences. That's fine, although looking at the $n$ such that $x_n \ge x$, and separately at the $n$ such that $x_n<x$, would have been just as efficient. (The proof below does not take this path.)

The proof uses my style, not yours. Of course you need not, indeed should not, adopt my style, which has problems of its own. But do think in terms of producing first proofs that are not so condensed. Most importantly, this will give you greater confidence about the correctness of your arguments.

We start with an interval $I$. There are several kinds of interval. Let's settle on one kind, and write out the details. I will settle on $I=(a, \infty)$, because it illustrates all the issues. (Note that $[a, \infty)$, for example, is easier.) Trying for a single argument that will simultaneously take care of all the geometrically distinct situations is not a good idea.

Our sequence is convergent, so it is bounded. (Here I am using a "big" result, but a pretty basic one.) So if $M$ is an upper bound for $(x_n)$, and $c=M+1$, then $x$ and all the $x_n$ are $\le c$. (OK, they are $\le M$, but why bother?)

The other end is more interesting. First we show that there is a positive $\epsilon$ such that the interval $(a, a+\epsilon)$ contains no $x_n$. We know that $x_n>a$ for all $n$. If every neighbourhood of $a$ contained $x_n$ for some $n$, then some infinite subsequence of $(x_n)$ would converge to $a$, contradicting the fact that $(x_n)$ has a limit $x>a$.

It follows that $x\ge a+\epsilon$ (or, more obviously, $x\ge a+\epsilon/2$). Let $b=a+\epsilon/2$. Then $x$, and all the $x_n$, are in the closed interval $J=[b,c]$.

Now we could do the other cases, though it is clear from the above discussion that none of them will give trouble.

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Why not let $b = a+\epsilon$? –  Damien Jun 27 '11 at 4:43
    
@Damien: What I wrote in the solution, namely $x \ge a+\epsilon$, says we can. I was making the point that we should not sweat about the best choice for a bound if all we care about is showing that there is a bound. The $a+\epsilon/2$ is more trivial if we are pushed for details. But I should have been happy to make the point only once, with my choice of $c$! –  André Nicolas Jun 27 '11 at 6:14

Throwing in my two cents, your approach gets to the point where $x_{N_2} \leq x_n \leq x_{N_1}$, for all $n \geq M:=\max(N_1,N_2)$, but you now have to account for $n_1,n_2,\dotsc,n_{M-1}$, which may or may not lie in your interval $[x_{N_2},x_{N_1}]$. Since you are now dealing with a finite set of points which are all in the interval $I$, you can take the maximum and minimum of the points (together with $x_{N_1}$ and $x_{N_2}$). Those max/min points will be your closed interval for $x_n$.

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