Let $X$ be compact and $f:X\to\mathbb{R}$ s.t. each $x\in X$ has a nbh where $f$ attains its minimum. Show $f$ attains minimum on $X$.

Question

Consider a compact topological space $X$ and a map $f:X\to\mathbb{R}$ such that each $x\in X$ has a neighborhood where $f$ attains its minimum. Show that $f$ attains its minimum on $X$.

My attempt:

I was thinking of covering $X$ with all the neighborhoods where $f$ attains a minimum, so something like $X\subseteq \bigcup_{x\in X} U_x$. Then, by compactness, there would be a finite subcover. I don't see how I can conclude that a minimum is attained from this information: I could possibly go over all sets of the subcover and take the overall minimum that $f$ attains (which exists, as the subcover is finite), but am I guaranteed that this minimum is the minimum of $f$ on $X$?

Answer 1

For each $x\in X$ we have open neighborhood $U_x$ of $x$ such that there is $t_x\in U_x$ such that $f(t_x)\leq f(t) \ \forall t\in U_x$.

By compactness of $X$, we have finite subcover $\{U_{x_1},...,U_{x_n}\}$ of $X$.

Hence ,$f(t_{x_i})\leq f(t) \ \forall t\in U_{x_i}$ for $i=1,2,...,n.$

$\min\{f(t_{x_i})\}_{i=1}^n =f(t_0)$ is required minimum.

Answer 2

Yes, your approach is correct. Say the minimum over all the sets of the subcover is f(t). This guy is your candidate to be the minimum of f.

Suppose by way of contradiction, that there exists b ∈ X such that f(b) is strictly less than f(t). Then b is not an element of any set of the finite subcover. This would be a contradiction with the fact that every open covering of the space admits a finite subcover.