# Convergent subsequences

I'm studying for a real analysis exam tomorrow and I am curious about something:

Two true false questions: a) Every sequence in the interval (0,1) has a convergent subsequence. b) Every sequence in the interval (0,1) has a subsequence that converges to a point in (0,1)

This dude on Yahoo Answers says both are false and uses the counterexample {1/n}. I think he is wrong because the sequence {1/n} isn't even in my interval! (It fails for 1). In fact I think both are true, and I don't know the proper reasoning but all I know is that I am not convinced by this supposed "counterexample"

Secondly, my book describes the following proposition: "Let the sequence {An} converge to the limit a. Then every subsequence of {An} also converges to the same limit a." In feel like this confirms the validity of (b).

Any input on the truth of these two would be very much appreciated.

## 5 Answers

For $(a)$, you get sequences that are bounded. Use the Bolzano-Weierstrass theorem, and the Monotone Convergence theorem.

For $(b)$, consider the sequence $a_n=\frac 1 {n+1}$. Then $a_n\in(0,1)$ but $\lim a_n=0$.

ADD You say

Secondly, my book describes the following proposition: "Let the sequence $\{a_n\}$ converge to the limit $a$. Then every subsequence of $\{a_n\}$ also converges to the same limit $a$." In feel like this confirms the validity of $(b)$.

Note that that proposition talks about a convergent sequence, and the behaviour of its subsequences, while $(b)$ considers any sequence and it's subsequence, and the value it converges to. If you have an exam tomorrow, consider writing down the argument you "feel" confirms the validity of something, so that you know wether you're right or not, and reading propositions to understand what they are really stating. In this case, that proposition tells you nothing about $(b)$, unless you consider the convergent sequences in $(0,1)$ that converge to a point in $(0,1)$. But then you're missing a lot of other sequences that don't converge, but are bounded, so have a convergent subsequence, which may, or may not converge to a point in $(0,1)$. In particular, this is because $0$ and $1$ are limit points of $(0,1)$. If we were talking about $[0,1]$, the answer would be affirmative, because $[0,1]$ us a closed subset of $\mathbb R$.

The Bolzano–Weierstrass theorem states that every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence. Therefore, (a) is true.

For (b), $a_n = \dfrac{1}{n+1}$ is a counterexample.

• Dude, we wrote the same =P! – Pedro Tamaroff Sep 30 '12 at 22:28
• @PeterTamaroff When you see 'convergent subsequence', Bolzano-Weierstrass immediately comes to your mind. :) – Ayman Hourieh Sep 30 '12 at 22:30
• Yeah. It would be silly not to upvote this, so +1. – Pedro Tamaroff Sep 30 '12 at 22:35
• @PeterTamaroff +1 to you too, as it's the same answer. – Ayman Hourieh Sep 30 '12 at 22:42
• That was the point, =D – Pedro Tamaroff Sep 30 '12 at 22:46

The first example is true in the sense that every sequence taking values in $(0,1)$ has a convergent subsequence in $[0,1]$ and that's because $[0,1]$ is compact (or whatever reason your course might suggest, but I don't know which path you would take). Therefore if we don't "precise" where the convergent subsequence should converge, then 1) holds.

However 2) doesn't hold (if the beginning point $1/1$ annoys you then choose the sequence $1/(n+1)$ as a counter example). The idea here is that some sequences converge outside the interval $(0,1)$, therefore not admitting any subsequence in the interval $(0,1)$ because when a sequence is convergent every subsequence converges to the same limit.

Hope that helps,

Suppose that every sequence in $(0,1)$ has a convergent subsequence. $(0,1)$ is sequentially compact. Since $(0,1)$ is a metric space, compact equivalent to sequentially compact. So $(0,1)$ is compact. It is contradict. So there must be a sub sequence has no convergent subsequence.

$a_n=\{\frac{1}{2n}\}$ has convergent sequence in $\mathbb{R}$, but no in $(0,1)$.

The dude is almost right. Both are false. Take $a_n = \frac{1}{2n}$.

By the way, are you sure the question is stated with open intervals?

Edit: I have misunderstood your question, my answer applies only if your topological space is just $(0,1)$.

• Yes they are open intervals. If they were closed I could just immediately cite the theorem – Arthur Collé Oct 1 '12 at 1:27