Prove that $I(f) = \inf_{x \in X}f(x)$ is a continuous function

continuitymetric-spaces

Given an arbitrary set $X$, define the real functions $I,S: \mathcal{B}(X,\mathbb{R}) \to \mathbb{R}$ by $I(f) = \inf_{x \in X}f(x)$ e $S(f) = \sup_{x \in X}f(x)$. Prove that $I$ and $S$ are continuous.

Notation. $\mathcal{B}(X,\mathbb{R})$ denote the set of all real bounded functions.

With continuity of $I$ we can prove the continuity of $S$. I'm trying to use the definition of continuity, but I cannot manipulate $d(\inf f(X), \inf g(X))$ conveniently.

I suppose that just a hint for it will be enough. Thanks for de advance.

Best Answer

Let $f_n \to f$. For $n$ sufficiently large we have $f(x)-\epsilon <f_n(x) <f(x)+\epsilon$ for all $x$ so we can take infimum throughout and conclude that $\inf_x f(x)-\epsilon \leq f_n(x) \leq \inf_x f(x)+\epsilon$. This shows that infimum is continuous.

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