# Subsequences and Convergence

I came across the following problems on subsequences during the course of my study of real analysis:

True or false: If $(a_n)$ is any sequence in $\mathbb{R}$, then the sequence $x_n = \frac{a_n}{1+|a_n|}$ has a convergent subseqeunce.

We know that $|x_n| < 1$ for all $n$. Hence by the Weierstrass-Bolzano Theorem $(x_n)$ has a convergent subsequence.

If $(a_n)$ is a sequence in $\mathbb{R}$ and $a \in \mathbb{R}$, the following conditions are equivalent. (a) $(\forall \epsilon >0) |a_n-a| < \epsilon$ frequently, (b) there exists a subsequence of $(a_n)$ converging to $a$. What happens if we change "frequently" to "ultimately."? How would (b) change?

So we know that $(\forall \epsilon >0)(\forall N) \ \exists n \geq N \ni |a_n-a| < \epsilon$. In other words, $(a_n-a)$ is a null sequence. Thus from a previous theorem $(a_{n_{k}}-a)$ is a null sequence where $(a_{n_{k}})$ is some arbitrary subsequence.

If we changed the wording to "ultimately" then we would have: $(\forall \epsilon >0) \ \exists N \ni n\geq N \Rightarrow |a_n-a| < \epsilon$. Then $(a_n)$ wouldn't have a subsequence converging to $a$ because $|a_n-a| \geq \epsilon$ frequently?

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Please, please stop putting multiple, distinct, unrelated questions into the same post. It means that they cannot have informative titles about the specifics, and it means that people searching for questions similar to the ones they might have will have an impossible time finding either question. –  Arturo Magidin Jun 25 '11 at 3:29
@Arturo: Actually, there is only one question (well disguised). The first three paragraphs are only motivation (if I understand this post correctly). –  t.b. Jun 25 '11 at 3:33
@Theo: How is the "True or False" question asked (and answered), related to the second quote box? It's not motivating in any way. Either it is irrelevant material, or (likely, given past history) it's a disguised question asking to validate the argument given. Either way, we have two unrelated questions in the post, even if the first one is being answered in the post itself. –  Arturo Magidin Jun 25 '11 at 3:40
@Arturo: Yes of course, the relation between the two things is beyond me, too. After all, both speak of sequences, no? :) I definitely agree with the thrust of your comment. Sorry about the noise. –  t.b. Jun 25 '11 at 3:49

For the first question, since every sequence contains either an increasing or a decreasing subsequence, every bounded sequence contains a convergent subsequence. So if you've shown it is bounded, you are done.

For the second question, I don't think you are correct in saying that $(a_n-a)$ is a null sequence; and you never said who $a_{n_k}$ was, so the paragraph is empty.. Now you are saying $a_{n_k}$ is an arbitrary subsequence; well, it doesn't follow then. Take $a_n = \frac{1}{n}$ if $n$ is even, $a_n = n$ if $n$ is odd, and take $a=0$. Then for every $\epsilon\gt 0$ we have that $|a_n - 0|\lt \epsilon$ frequently, but $(a_n-0)$ is not a null sequence, and there are plenty of subsequence $(a_{n_k})$ for which $(a_{n_k}-0)$ is not a null sequence either. For example, $n_k=2k+1$ gives the sequence $(2k+1)_{k=1}^{\infty}$, which is certainly not a null sequence.

Instead, for (a)$\Rightarrow$(b), construct the subsequence inductively; pick $n_1$ so that $|a_{n_1}-a|\lt 1$; then pick $n_2\gt n_1$ such that $|a_{n_2}-a|\lt \frac{1}{2}$; then pick $n_3\gt n_2$ such that $|a_{n_3}-a|\lt \frac{1}{4}$; etc. Of course, the condition you have in (a) is what you need to justify this is possible.

You never did anything about (b)$\Rightarrow$(a), however. If you have a subsequence $(a_{n_k})$ that converges to $a$, you need to prove that for every $\epsilon\gt 0$, $|a_n-a|\lt\epsilon$ frequently; it's easy, but you haven't done it.

Your attempted conclusion at the end does not follow; what makes you think that if $|a_n-a|\lt \epsilon$ from some point forward, then $|a_n-a|\geq \epsilon$ "frequently"? To occur frequently, it must first of all occur infinitely many times, but here it can only occur $N-1$ times at the most.

Think about the condition you wrote for the "ultimately" clause. It should look very familiar!

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@Arturo: For the "ultimately" clause, this just means that $a_n \to a$? Thus it has a convergent subsequence that converges to $a$? –  Damien Jun 25 '11 at 3:49
@Damien: "It has a subsequence that converges to $a$" is what you have to prove is equivalent to the "frequently" clause. If that was the right answeer here as well, you would be saying that the "frequently" and the "ultimately" statements are equivalent, which they most certainly are not. Remember, you are trying to find a statement that is equivalent to the "ultimately" clause, not just one that is implied by it; you seem to be forgetting the implication going back. How do you change (b) so that (a)-with-ultimately is equivalent to your new (b)? –  Arturo Magidin Jun 25 '11 at 3:52
@Arturo: You would also inductively create a subsequence. Pick $n_1 >N$ such that $|a_{n_{1}}- a| < 1$. Pick $n_2>n_1>N$ such that $|a_{n_{2}}-a| < \frac{1}{2}$ etc..So where we choose the indices of the subsequence changes. –  Damien Jun 25 '11 at 4:05
@Damien: You are missing the point. Something happens "ultimately" is a stronger condition than saying that something happens "frequently". The latter says it happens infinitely many times, but the former says it fails to happen only finitely many times (you can have something happen infinitely often, and also fail infinitely often). If you strengthen (a) by replacing the condition with a stronger condition, you are going to have to strengthen (b) in order to get something equivalent to your new (a). But you aren't even changing (b), you are trying to use the same (b). Won't work –  Arturo Magidin Jun 25 '11 at 4:08
@Arturo: $|a_n-a| \geq \epsilon$ for finitely many $n$ (in the "ultimately" clause). So there exists an increasing subsequence of $(a_n)$ that converges to $a$ but no decreasing ones? So we are restricting the type of subsequence. –  Damien Jun 25 '11 at 4:23