Let $X \subset \mathbb{R^m}$. Show that if any continuous function of real value and defined in $X$ reaches a value maximum, then $X$ is compact.
I have seen several proofs of this exercise, but using more complex topics such as the Tietze Extension theorem and other answers I have seen prove the contrapositive. Can I make a simpler demonstration using only real analysis elements? And if that is possible, what would a demonstration be like for this exercise?
We will show it in standard way by contraposition, ie. for any non-compact $X \subseteq \mathbb R^m$ we construct an unbounded continuous function $f : X \to \mathbb R$.
We assume the following well known result: a subset of $\mathbb R^m$ is compact iff it is bounded and closed.
Let us suppose first $X$ is unbounded. Then it suffices to take simply $f(x):= \Vert x \Vert$, the Euclidean norm of $x \in X$.
Now, if $X$ is not closed then there is a point $y \in \mathbb R^m$ which belongs to the closure of $X$ but not to $X$ itself. Define now $f(x) := \frac1{\Vert x-y \Vert}$ for $x \in X$. This function is well-defined and continuous. You can show easily it is not bounded due to how we chose the point $y$.
