Htykuut

Let $M$ be a topological space, $Ksubset M$ compact, $fcolon Mtomathbb{R}cupleft{+inftyright}$ lower semi-continious. Show that $f$ takes its minimum on $K$.

Good day,

we defined lower semi-continuity as follows: $f$ is lower semi-continious on $M$ if for any $xin M$ and any $alpha < f(x)$ there is a neighbourhood $Usubset M$ of $x$ such that $alpha<f(y)$ for all $yin U$.

So my idea is the following.

$f$ is lower semi-continious on $K$. Take any $xin K$ and $alpha < f(x)$, then there is a neighbourhood $U_xsubset M$ of $x$ such that $alpha < f(y)~forall yin U$.

So it is
$$
Ksubsetbigcup_{xin K}U_x.
$$
Because of the compactness of $K$ there is a finite index set $I$ such that
$$
Ksubsetbigcup_{iin I}U_{x_i}.
$$
It is $f(y)>alpha~forall yin U_{x_i}, iin I$.

So for all $alpha <infty$ it is $f(x)>alpha$ for all $xin K$.

This means that it is $inf_{xin K}f(x)=-infty$.

But do not know what to do with this result, is it helpful at all??

As mentioned by Manuel, the first step of your reasoning is wrong because the $alpha$ depends on $x$.

However, your method of proof is quite correct, and with a minor adjustment it gives the statement you want. Here are the details.

Set $a:=inf{ f(x);; xin K}$. This is well defined, but possibly equal to $-infty$. It is enough to show that one can find $x$ such that $f(x)=a$. (In particular, this will show that $a>-infty$).

Assume this is not true. Then, for any $xin K$ you have $f(x)>alpha$; so you can choose a real number $alpha_x>a$ such that $f(x)>alpha_x$.

Then your above reasoning (together with Manuel's comments) gives that one can find $x_1,dots ,x_Nin K$ such that
$$forall xin K;: ; f(x)>alpha:=min (alpha_{x_1},dots ,alpha_{x_N}), . $$
It follows that $inf{ f(x);; xin K}geqalpha$, i.e. $ageqalpha$; but this is a contradiction since $alpha_{x_1},dots ,alpha_{x_N}$ are strictly greater than $a$ and hence $alpha>a$.