Our professor showed us Bolzano-Weierstrass theorem for sets only (every infinite, bounded set $A\in\mathbb{R}$ has at least one accumulation point in $\mathbb{R}$) and left us to prove that every sequence of real numbers has at least one limit point and I would be grateful if anyone would tell me if my proof is correct:

  1. Sequence is infinite and bounded.

    Let $A=\{x_n|n \in\mathbb{N}\}.$ Since $A$ is both bounded and infinite existence of limit point comes directly from BW theorem for sets

  2. Sequence is infinite and unbounded.

    Let $G$ be some neighbourhood of $+\infty$ (same applies for $-\infty$). For any $M\in\mathbb{R}, \exists n\in\mathbb{N}$ such that $x_n\in(M,+\infty)$ $\forall n\geq$ some $n_0$ thus there is a subsequence of $x_n$ that converges to infinity and so we can say that $+\infty$ is limit point of $x_n$

  3. Sequence is finite and bounded

    There is certain real $a$ such that $x_n=a$ for finite $n$.$\implies \exists x_{n_k}=a; \forall k\in\mathbb{N}\implies lim_{k\to\infty} x_{n_k} = a$ thus there is subsequence of $x_n$ that converges to some point ($a$) which is its limit point.

  4. Sequence cannot be finite and unbounded in $\mathbb{R}$

I was looking also through lots of previous questions on this site that were somewhat similar but none of them have quite answered my question or I didn't understand the solution.

Please correct me if I am wrong somewhere I spent couple of hours to first find connections between BWT for sets and sequences and then prove this... Limits superior and inferior are introduced in later lectures so I am not allowed to use them.


1 Answer 1


You haven’t actually finished the first case. You know that the set $A$ has a limit point, say $p$, but you still have to show that the sequence has $p$ as a limit point (or as I would call it, a cluster point), i.e., that it has a subsequence converging to $p$. You can do this by recursively constructing the subsequence. Suppose that for $k=1,\ldots,m$ you’ve chosen $n_k\in\Bbb Z^+$ such that $n_1<\ldots<n_m$ and $|x_{n_k}-p|<\frac1k$; there are infinitely many $\ell\in\Bbb Z^+$ such that $|x_\ell-p|<\frac1{m+1}$, so let

$$n_{m+1}=\min\left\{\ell\in\Bbb Z^+:\ell>n_m\text{ and }|x_\ell-p|<\frac1{m+1}\right\}\;.$$

This allows the recursive construction to continue, and we get a subsequence $\langle x_{n_k}:k\in\Bbb Z^+\rangle$ of the original sequence that converges to $p$. This shows that $p$ really is a limit point of the original sequence.

In the second case your really ought to do something similar: you need to show that you can actually get a subsequence converging to $+\infty$. It would suffice to show that we can find $n_k\in\Bbb Z^+$ for $k\in\Bbb Z^+$ such that $n_1<n_2<\ldots$ and $x_{n_k}>k$ for each $k\in\Bbb Z^+$; this can be done by a recursive construction very similar to the one that I just did for the first case.

I think that you have a typo in your third case: I believe that you meant to say that there is an $a\in\Bbb R$ such that $x_n=a$ for infinitely many $n\in\Bbb Z^+$. In that case the subsequence $\langle x_n:x_n=a\rangle$ is a constant subsequence converging to $a$.

  • $\begingroup$ You were right, there was a typo in third case. Ty very much $\endgroup$ Jun 19, 2020 at 21:28
  • $\begingroup$ @BonkoZvogdan: You’re very welcome. $\endgroup$ Jun 19, 2020 at 21:33

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