2
$\begingroup$

I want to show given a sequentially compact subset $A \subseteq M \implies A$ is bounded. I read this Every sequentially compact set is closed and bounded. but the proof is poorly written and jumpy

Def: $A$ is sequentially compact if every sequence has a convergent subsequence.

  • By contradiction, assume $A$ is a sequentially compact subset of $M$ and $A$ is not bounded.

    Let $(x_n)$ be a sequence on $A$.

    Since $A$ is not bounded, then $\exists a \in A$ such that $\forall r > 0, \forall n \in \mathbb{N}, d(x_n, a) > r$.

    Let $(x_{n_k})$ be a subsequence of $(x_n)$, then $\exists x \in A$ such that $x_{n_k} \to x$ as $k \to \infty$

    Then $d(x_{n_k}, x) \leq d(x,a) + d(a, x_{n_k})$ (want to obtain some sort of contradiction)

How do I proceed from here? I know that $d(a, x_{n_k})> r$, but $d(x,a)$ is unknown....

$\endgroup$
  • $\begingroup$ I think you want to edit your definition of a sequentially compact set. It should be every sequence has a convergent sequence. Your statement is a tautology (i.e. every convergent sequence is the convergent sub-sequence). $\endgroup$ – Kevin Sheng Apr 29 '16 at 2:06
4
$\begingroup$

Assume by contradiction that $A$ is unbounded.

Let $a \in A$. Then for each $n$ there exists some $x_n$ such that $$d(x_n,a) >n$$

Now, $x_n$ has a converging subsequence $x_{k_n} \to b$.

Now, for $\epsilon=1$, since $x_{k_n} \to b$ there exists some $N$ so that for all $n >N$ we have $$d(x_{k_n}, b) <1$$

Then, for all $n >N$ we have $$d(a,b) \geq d(a, x_{k_n})-d(b,x_{k_n}) \geq k_n-1 \geq n-1$$

This leads to a contradiction, as $d(a,b)$ is a real number which is larger than any $n >N$.

$\endgroup$

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.