Prove that a bounded sequence has two convergent subsequences.

Question

Let $a_1,a_2\dots$ be a bounded sequence in $\mathbb{R}$ that does not converge. Prove that the sequence has two subsequences that converge to different limits

Here is my proof: Let $s_n$ be a bounded sequence in $\mathbb{R}$ that does not converge. Then by the Bolzano Weierstrass theorem there exists a subsequence, $s_{n_k}$, that converges. Let $a$ be the limit of $s_{n_k}$. By definition $s_n$ does not converge so it can not converge to $a$. That is, there exists $\epsilon >0$ such that for all $N\in\mathbb{R}$, there exists $n>N$ such that $|s_n-a|\ge\epsilon$. So there exists a sequence $s_m$ that is a subsequence of $s_n$ with $m>N$. By its definition $s_m$ does not have a subsequence that converges to $a$. However, $s_m$ is bounded because $s_n$ is bounded. So by the Bolzano Weierstrass theorem there exists a subsequence, $s_{m_j}$, that converges. Clearly $s_{m_j}$ does not converge to $a$. $\square$

Am I correct in saying "By its definition $s_m$ does not have a subsequence that converges to $a$." This is what my proof relies on, but I'm not sure I have justified it correctly. Thanks in advance for any input!

Answer 1

Yeah, you're pretty much correct here. It might be more clear if you defined your $s_m$'s using a different letter, like $r_n$. For example:

$s_n$ does not converge to $a$. Therefore, there is some $\epsilon > 0$ such that for any $N>0$, we can find an index $M>N$ such that $|s_M - a| > \epsilon$. For each $N>0$, set $r_N$ equal to one such choice of $s_M$. Then $\{r_N\}_{N\in\mathbb{N}}$ is a subsequence of $\{s_n\}$ with the property that $|r_n - a| > \epsilon$ for every $n$. The sequence $\{r_N\}$ is bounded since it is a subsequence of the bounded sequence $\{s_n\}$. By Bolzano-Weierstrass, $\{r_n\}$ has a subsequence converging to some $b\in\mathbb{R}$. Then $b$ is a limit point of $\{s_n\}$ because a subsequence of $\{r_N\}$ is a subsequence of $\{s_n\}$, and moreover, $b \neq a$ because the sequence $\{r_N\}$ is bounded away from $a$. Therefore, the original sequence $\{s_n\}$ has two subsequences with different limit points.

You could also use double indices, and replace $r_N$ in the previous proof with $s_{n_N}$.

Answer 2

Here is my take. Le $a_n$ be a bounded sequence that doesn't converge. Then $\limsup a_n$ and $\liminf a_n$ exist and they are different.

Answer 3

Let $\{a_n\}_{n\geq 1}$ be a bounded real sequence, i.e. there exists $M>0$ such that $-M\leq a_n \leq M \quad \forall n\geq 1$, and doesn't converge.

Now, we define two sub-sequences:

$\alpha_k = \sup\{a_k,a_{k+1},\ldots\}=\sup\{a_n|n\geq k\} =\sup_{n\geq k} a_n$

and

$\beta_k = \inf\{a_k,a_{k+1},\ldots\}=\inf\{a_n|n\geq k\} =\inf_{n\geq k} a_n$.

Then,

1. $\{\alpha_k\}$ is monotonically decreasing and converging to $\limsup (a_n)$,
2. $\{\beta_k\}$ is monotonically increasing and converging to $\liminf (a_n)$.

Example: consider $a_n=(-1)^n, n\geq 1$, then $\alpha_k =1$ and $\beta_k =-1$.