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.
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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….
Best Answer
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 $n >N$.