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

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.

• What's the topology on $\mathcal B(X,\mathbb R)$? – kimchi lover Mar 24 at 22:13
• The statement is false if $X$ is an infinite set and $\mathcal{B}(X, \mathbb{R})$ is equipped with the topology of pointwise-convergence. (For instance, choose $X = \{x_1, x_2, \dots \}$ as a countably infinite set and define $f_n = \mathbf{1}_{\{ x_1, \cdots, x_n\}}$. Then $f_n \to \mathbf{1}_X$ pointwise, but $I(f_n) = 0 \not\to 1 = I(\mathbf{1}_X)$.) As pointed out in the above comment, which topology are you working with? – Sangchul Lee Mar 24 at 22:16
• Is $\mathcal B(X,\Bbb R)$ meant to be endowed with the sup-norm $$\lVert f\rVert_\infty=\sup_{x\in X}\lvert f(x)\rvert\quad?$$ – Saucy O'Path Mar 24 at 22:20
• Oh, my bad! $\mathcal{B}(X,\mathbb{R})$ is equipped with sup-norm, as Saucy O'Path pointed out. – Lucas Corrêa Mar 24 at 22:26

Let $$f_n \to f$$. For $$n$$ sufficiently large we have $$f(x)-\epsilon for all $$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.