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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 $X$?

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  • $\begingroup$ I'm confused, isn't it true that every function $f: X \rightarrow \mathbb R$ where $X$ is compact has a global minimum? Why the condition on $f$? $\endgroup$ – Noel Lundström May 30 '20 at 0:48
  • $\begingroup$ Note that it's not given whether $f$ is continuous or not. $\endgroup$ – Zachary May 30 '20 at 8:27
  • $\begingroup$ Map means continuous function. In general it's better to write something like "given a possibly discontinuous function $f$" to make that fact clear. $\endgroup$ – Noel Lundström May 30 '20 at 12:08
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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.

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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.

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