The infimum of a semi-continuous function on a compact set

Question

Define a funtion as follows: $$f:X \rightarrow \mathbb{R}_+$$ with the property that $f(x)=0$ if and only if $x\in S$, where $\mathbb{R}_+=[0, \infty)$ and $X$ and $S$ are subsets of $\mathbb{R}^n$ for some $n\in \mathbb{N}$.

And, consider two assumptions:

Assumption 1.

$X$ is a nonempty compact subset of $\mathbb{R}^n$ for $n\in \mathbb{N}$.

Assumption 2.

$f$ is a lower semi-continuous functuon.

Here, I want to show that

under Assumptions 1 and 2,

(i) $\inf_{x\in X}f(x) \geq 0$

(ii) $\inf_{x\in X} f(x) =0$ if and only if $S\neq \emptyset$

But, I failed to prove this. How does the statement above hold?

Answer 1

$\DeclareMathOperator{im}{im}$ Alternate proof of the second statement: clearly, if $S \neq \emptyset$ then the infimum is zero. Conversely, suppose $S = \emptyset$. Then for all $x \in X$, $f(x) > 0$. Now $\im_f(X)$ is compact and intervals of the form $(q, \infty)$, where $q \in \mathbb{Q}_+$, form an open cover of the image, since for all $x$, $f(x) > 0$. Then taking a finite subcover of intervals $(q_1, \infty), ..., (q_n, \infty)$, we see that $\min\limits_{1 \leq i \leq n} q_i$ is a lower bound of $\im_f(X)$.

Answer 2

The first statement already follows since $f$ is nonnegative by definition. For the second statement one only needs to verify the first implication, suppose $\inf_{x\in X} f(x)=0$ and construct a subsequence $\{x_n\}\subset X$ of points such that $f(x_n)$ converges to $0$ monotically, by compactness suppose a subsequence $x_{n_k}$ converges to some $x_0$, suppose $f(x_0)>0$ and use lower semicontinuity to get a contradiction.