Let $X$ be any topological space and $\bar{\mathbb{R}} = [-\infty, \infty]$ with the standard topology. Is it true, in general, that the functionals $$\inf: C(X,\bar{\mathbb{R}}) \to \bar{\mathbb{R}}, f \mapsto \inf f$$ $$\sup: C(X,\bar{\mathbb{R}}) \to \bar{\mathbb{R}}, f \mapsto \sup f$$ are continuous, if $C(X,\bar{\mathbb{R}})$ is equipped with the compact-open topology?

If $X$ is compact, it should be true, if the following argument is correct (wlog. proof only for $\inf$ and $X \neq \emptyset$): Let $a \in \mathbb{R}$ arbitrary. Then $\inf^{-1}([-\infty,a)) = \bigcup_{x\in X} \langle x,[-\infty,a) \rangle$ is open. On the other hand $\inf^{-1}((a,\infty]) = \bigcup_{\varepsilon > 0}\langle X, (a+\varepsilon, \infty] \rangle$, which is open, because $X$ is compact. $[-\infty,a)$ and $(a,\infty]$ build a subbase of $[-\infty,\infty]$. Therefore $\inf$ is continuous.

What happens for arbitrary spaces $X$, or locally compact, etc.?

  • $\begingroup$ Hmm, in fact I think if $X$ is Tychonoff and not compact, then $\sup$ is never continuous: for any basic open set $\bigcap_{i=1}^n \langle K_i, U_i \rangle$ contaning the zero function, $\bigcup K_i$ is compact and therefore not the whole space. So choose $x \notin \bigcup K_i$; then there is a function which is 0 on $\bigcup K_i$ but 1 at $x$, so its sup is at least 1, but the function is still in $\bigcap_{i=1}^n \langle K_i, U_i \rangle$. (Not sure on the details here, though...) $\endgroup$ Commented Mar 16, 2023 at 18:23

1 Answer 1


I will present a counterexample and show that in the case $X = \mathbb{R}$, neither $\sup$ nor $\inf$ is continuous. To see this, let us consider the sequence of functions $f_n(x) := \arctan(x/n)$. I claim that in the compact-open topology, $f_n$ converges to the constant zero function. To see this, take some subbasic open neighborhood $V(K, U)$ of the zero function with $K$ nonempty. Then since the zero function is in the neighborhood, we must have $0\in U$; therefore, for some $\varepsilon > 0$, $(-\varepsilon, \varepsilon) \subseteq U$. Also, $K$ must be bounded, so take some $R \in \mathbb{R}^+$ such that $K \subseteq [-R, R]$. Now, we know that $\arctan(R/n) \to 0$ as $n\to \infty$, so for $n$ sufficiently large, we have $\arctan(R/n) < \varepsilon$. Therefore, for $n$ sufficiently large, for every $x\in K$, we have $|f_n(x)| \le \arctan(R/n) < \varepsilon$, implying $f_n(x) \in U$. This shows that for $n$ sufficiently large, $f_n \in V(K, U)$. Since this is true for every subbasic open neighborhood of the zero function, we have shown that $f_n \to 0$ in the compact-open topology.

On the other hand, $\sup(f_n) = \frac{\pi}{2}$ for each $n$, while $\sup(0) = 0$, showing that $\sup$ is not continuous. Similarly, $\inf(f_n) = -\frac{\pi}{2}$ for each $n$, while $\inf(0) = 0$, which implies that $\inf$ is not continuous.


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