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Let $f: \mathbb{R} \to \mathbb{R}$ be a function bounded below and lower semi-continuous. Show that for every compact $K \subset \mathbb{R}$, there exists $x_0 \in K$ such that $$ f(x_0)=\inf_{x\in K}f(x) $$

I've read a proof for a similar theorem here, but in that proof $f$ is a continuous coercive function. That makes the proof pretty straightforward.

user
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1 Answers1

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A characterization of lower semi continuity is that for any $a \in \mathbb{R}$, the set $f^{-1}((-\infty, a])$ is closed in $\mathbb{R}$ (I don't know what definition of lower semi continuity you're using becasue there are a few, but showing that the definition is equivalent to this is a good exercise, show it!)

Let $m = \inf_{x \in K} f(x)$ (finite by assumption). Let $A_{n} = \{x \in K: f(x) \leq m + \frac{1}{n}$}. By the characterization of lower semi continuity above, each $A_{n}$ is closed. Any closed subset of a compact set is compact, so each $A_{n}$ is compact. Also, clearly $A_{n} \subset A_{n+1}$ for each $n \in \mathbb{N}$ -- by definition of $A_n$. Thus, $$ A := \bigcap_{n \in \mathbb{N}} A_{n} \neq \emptyset $$ since nested sequences of compact sets in $\mathbb{R}$ (or any complete metric space, actually) have a non-empty intersection. Pick any $x_{0} \in A$. Check that $x_{0}$ is a minimizer like you want.

gtoques
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  • I just didn't get one part of your demonstration. Doesn't $A_n \subset A_{n+1}$ already imply $\bigcap_{n \in \mathbb{N}} A_{n} \neq \emptyset$? If so, then the latter is consequence of the former, and not the fact that you state right after (That nested sequences of compact sets in $\mathbb{R}$ have a non-empty intersection). – user Mar 04 '21 at 22:36
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    In general it's not true that any nested sequence of sets has a non empty intersection. Take $A_{n} = (0, \frac{1}{n})$. This is certainly a nested sequence, but $\bigcap_{n} A_{n} = \emptyset$. Nested sequences having a non-empty intersection is one reason why compact sets are so nice. – gtoques Mar 04 '21 at 22:44
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    If you want an example of a nested sequence of closed, but not compact, sets in $\mathbb{R}$ with an empty intersection, consider $A_{n} = [n, \infty)$. This hints at the fact that you want two things to ensure the property: closed and bounded (which in $\mathbb{R}^{n}$ is exactly the same as compact). – gtoques Mar 04 '21 at 22:46
  • Just to be sure, this is known as Cantor's Intersection Theorem? – user Mar 04 '21 at 22:52
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    Essentially, yes. Cantor's Intersection Theorem is a more general statement for any complete metric space (even general topological spaces, I think) and this is the same statement for $\mathbb{R}$. – gtoques Mar 04 '21 at 22:54