Show that every pseudocompact subset of $\Bbb R^n$ is sequentially compact.

Question

Let $X$ be a subset of $\mathbb{R}^n$ such that every continuous function $f: X \to \mathbb{R}$ is bounded. Show that X is sequentially compact

I'm currently self teaching metric spaces, and am not sure what to do. I'm thinking whether to maybe apply the Heine-Borel theorem? The definition of sequentially compact is that every sequence has a convergent subsequence.

Answer 1

Indeed, we can apply the Heine-Borel theorem. For metrizable topological spaces, compactness and sequential compactness are equivalent. Thus, in view of the Heine-Borel theorem, it suffices to show that $X$ is closed and bounded.

1. Consider the function $\psi: X \to \Bbb R$ defined by $\psi(x) = \|x\|$. $\psi$ is continuous and therefore bounded (by hypothesis.) Consequently, $X$ is a bounded set.

2. Suppose, for a contradiction, that $X$ is not closed. Then, there exists $x_0 \in \overline X\setminus X$. Consider the function $\varphi_{x_0}: X\to \Bbb R$ defined by $$\varphi_{x_0}(x) := \frac{1}{\|x-x_0\|}$$ The choice of $x_0$ helps us find a sequence $\{x_n\}$ in $X$, converging to $x_0$. Clearly, $\lim_{n\to\infty} \varphi_{x_0}(x_n) = \infty$, i.e., $\varphi_{x_0}$ is not bounded. Since $\varphi_{x_0}$ is continuous, this contradicts our hypothesis. Therefore, $X = \overline X$ and $X$ is a closed set.