1
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $a$.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.