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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.

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    $\begingroup$ What's the topology on $\mathcal B(X,\mathbb R)$? $\endgroup$ – kimchi lover Mar 24 at 22:13
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    $\begingroup$ 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? $\endgroup$ – Sangchul Lee Mar 24 at 22:16
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    $\begingroup$ Is $\mathcal B(X,\Bbb R)$ meant to be endowed with the sup-norm $$\lVert f\rVert_\infty=\sup_{x\in X}\lvert f(x)\rvert\quad?$$ $\endgroup$ – Saucy O'Path Mar 24 at 22:20
  • $\begingroup$ Oh, my bad! $\mathcal{B}(X,\mathbb{R})$ is equipped with sup-norm, as Saucy O'Path pointed out. $\endgroup$ – Lucas Corrêa Mar 24 at 22:26
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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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  • $\begingroup$ This approach is more simple and elegant. However, I'm interested too in a proof using only the definition. If you have any hint for it, would be awesome. Btw, thank you very much! $\endgroup$ – Lucas Corrêa Mar 25 at 2:49

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