If $ \infty $ is the only partial limit of $ a_n $, then $ a_n \to \infty $ I want to prove that $ \infty $ is the only partial limit of $ a_n $, then $ a_n \to \infty $, can you verify my proof?
Assume $ a_n \not \to \infty $.
If $ a_n $ is bounded, then it has a subsequence that approaches a real value according to BW, and therefore there is another partial limit - contradiction.
If $ a_n $ is not bounded, we will construct a subsequence, $ b_n$ that approaches $-\infty $. Choose $ b_1 = a_1 $, and for all $ n > 1 $, Choose $ b_n = a_k $ such that $ a_k \lt -n $ and the index of $ a_k $ is greater than the index of $ b_{n-1}$ in $ a_n $. Therefore, for all $ M $, there exists $ N $ such that for all $ n \gt N, b_n \lt -M $, and therefore there exists another partial limit - contradiction.
 A: No. If $\{a_n\}$ is not bounded, generally we can not extract a subsequence going to $-\infty$. For example: $\{a_n\}$ consists of $$1, 1, 2, 1, 3, 1, 4, 1, \cdots, n, 1, \cdots.$$
The right proof is as follows.
If $a_n\not\to\infty$, then by definition, there exists $M_0>0$ such that for all $N>0$, there exists $n>N$ such that $|a_n|\leq M_0$. Taking $N=1$ we can find $a_{n_1}$ such that $| a_{n_1}|\leq M_0$; taking $N=n_1$, we can find $n_2>n_1$ such that $|a_{n_2}|\leq M_0$; taking $N=n_2$ and $\cdots$. So we can find a subsequence $\{a_{n_k}\}$ such that $|a_{n_k}|\leq M_0$ for all $k$. By Bolzano-Weierstrass, this subsequence has a convergent subsequence, whose limit is not only finite, but also a partial limit of $a_n$, contradicting to the assumption that “$\infty$ is the only partial limit of $a_n$”.
A: As answered by Feng, your proof is not valid. Here’s a more general proof.
Recall that a space $S$ is sequentially compact when every sequence $a$ in $S$ has a convergent subsequence. Any metric space which is compact is sequentially compact.
For any sequentially compact space $S$, a sequence $a$ converges to $w$ if and only if $w$ is its only partial limit. For if $a$ did not converge to $w$, consider some open neighbourhood $V$ of $w$ such that there are arbitrarily large $n$ with $a_n \notin V$. This allows us to construct a subsequence $b$ of $a$ such that $\forall n (b_n \notin V)$. Since $S$ is compact, $b$ has a partial limit $y$. Since $b$ is a subsequence of $a$, $y$ is a partial limit of $a$ as well, and thus $y = w$. But $w$ clearly can’t be a partial limit of $b$ since $b_n \notin V$ for all $n$. $\square$
Then in particular, we see that $[-\infty, \infty]$ is a sequentially compact space, since it is homeomorphic to $[-\pi/2, \pi/2]$ using the $\arctan$ function. So if $\infty$ is the only partial limit of $a$, it is the only limit of $a$.
