Why does the Weierstrass theorem fail if a set is not compact?

Question

By Weierstrass theorem I mean that if $f:\mathbb{R}^n \to \mathbb{R}$ is continuous and $C \subset \mathbb{R}^n$ is compact, then the theorem asserts that a solution

$x^*$ of $$ \text{min} _{x\in C}f(x)$$ exists.

How does this theorem come to fail if $C$ is either not closed, or it's not bounded, or $f$ isn't continuous?

Could someone show me with an example of each case?

For an unbounded set $C$ I would say if $C \in (-\infty,\infty)$ then $\text min f(x) = \emptyset$

But I can't think of anything for discontinuity or an open set?

Answer 1

For an unbounded set, $f(x)=x$ in $\mathbb{R}$.

For a non-closed set, $f(x)=1/x$ in $(0,1)$