How can we prove that if $(X,d)$ is a metric space and if $K\subseteq X$ is pseudocompact, then $K$ is sequentially compact?
If not, we can find a sequence $(x_n)\subseteq K$ that it doesn't converge in $K$. Should we construct a continuous function $f:K\to\mathbb{R}$ not bounded?
Any hint? Thanks.
Suppose we have a sequence $(x_n)$ in $K$ such that no subsequence of the sequence converges in $K$ (this is the negation of sequential compactness).
Show that $S = \{x_n: n \in \mathbb{N}\}$ is a closed and discrete subset of $K$ (closed because a point in the closure would have a sequence converge to it, and discreteness is similar; we only need first countability of $X$), and so we can continuously extend the map $f(x_n) = n$ (which is continuous on $S$, having the discrete subspace topology) to $K$ and this is a non-bounded real-valued function on $K$.
