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.

  • 4
    $\begingroup$ What's the topology on $\mathcal B(X,\mathbb R)$? $\endgroup$ – kimchi lover Mar 24 at 22:13
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.