Sequential compactness and boundedness

Question

The definition of sequential compactness states that set $K$ is sequentially compact if $\forall$ sequences $a_n \in K$, $\exists a_{n_i}$, such that this subsequence converges to a point $p \in K$.

I'm thinking of the corollary to the open cover definition of compactness, that set $K$ is closed and bounded, and trying to relate this to the definition of sequential compactness. I can see how sequential compactness would imply that $K$ is closed, since every sequence contains a limit point in $K$ can follow the definition of sequential compactness

My Question How does the bounded property follow from the definition of sequential compactness? If we are given that set $K$ is sequentially compact then how could we show that $K$ is bounded? What restrictions or conditions apply to the definition of sequential compactness?

Thanks

Answer 1

Suppose it were not bounded. Then we see $X$ is not contained in any balls of finite radius. Pick one point in $X$ (this is possible as the empty set is bounded), say $x_0$ then choose a $x_{1} \notin B_{1}(x_0)$. This is possible as otherwise we would get a bounded $X$. Now choose a $x_{2} \notin B_{2}(x_1) \cup B_{1}(x_{1})$ as otherwise we could bound $X$. Keep repeating this and you'll get a sequence that cannot obtain a convergent subsequence, which is your desired contradiction.

Answer 2

Hint: suppose $K$ is not bounded. This means that there is a point $a_0\in K$, and for each $i$, a point $a_i\in K$ such that the distance between $a_0$ and $a_i$ is larger than $i$. Can $K$ be sequentially compact?