Is the maximum of infinitely many functions continuous on a compact space?

[R136a1]

Hello everybody!

Given a topological space $X$ and two functions $f,g:X\rightarrow \mathbb{R}$, it is rather easy to prove that $x\rightarrow \max\{f(x),g(x)\}$ is continuous. I wonder if this also holds for infinitely many functions. Of course, the maximum doesn't need to exist, so we will at least need some compactness result to let the maximum exist.

The specific form I'm talking about is to let $C$ be compact and to give a function $\varphi:X\times C\rightarrow \mathbb{R}$ (perhaps continuous or something). Then we let
$$x\rightarrow \max_{c\in C} \varphi(x,c).$$ Is this continuous?

 

[WannabeNewton]

Yes. Let $\varphi:X\times C\rightarrow \mathbb{R}$ be a continuous map with $C$ compact, and define $\psi:X\rightarrow \mathbb{R}$ by $\psi(x) = \text{sup}\{\varphi(x,c):c\in C\}$; take $x_0\in X$ and $\epsilon > 0$. By continuity of $\varphi$, we have that for any $c\in C$, there exists a neighborhood $W_{c} = U_{c}\times V_{c}$ of $(x_0,c)$ (these are the basis sets in the product topology so no loss of generality here) such that for all $p\in W_{c}$, $\left | \varphi(x_0,c) - \varphi(p) \right | < \frac{\epsilon}{2}$. Now $(V_c)_{c\in C}$ is an open cover of $C$ so there exists a finite subcover of C given by $\{V_{c_1},...,V_{c_n}\}$. $\{U_{c_1},...,U_{c_n}\}$ is of course a finite collection of neighborhoods of $x_0$ so $\bigcap _{i}U_{c_{i}}, i\in \{1,...,n\}$ is also a neighborhood of $x_0$.

Let $y_0\in\bigcap _{i}U_{c_{i}}$ then, since $\bigcap _{i}U_{c_{i}}\times V_{c_{i}}\subseteq W_{c_{i}},\forall i\in \{1,...,n\}$, for any $c\in V_{c_{i}}$ we must have $\left | \varphi(x_0,c_i) - \varphi(y_0,c)\right |< \frac{\epsilon}{2}$. Now $\left | \varphi(x_0,c) - \varphi(y_0,c)\right |\leq \left | \varphi(x_0,c_i) - \varphi(x_0,c)\right | + \left | \varphi(x_0,c_i) - \varphi(y_0,c)\right |$ for all $c\in V_{c_i}$; consequently, $\left | \varphi(x_0,c_i) - \varphi(x_0,c)\right |< \frac{\epsilon}{2}$ since $(x_0,c)\in W_{c_i}$ so $\left | \varphi(x_0,c) - \varphi(y_0,c)\right | < \epsilon$. Since $\bigcup _{i}V_{c_{i}} = C$, the above holds for all $c\in C$.

Thus, $\varphi(x_0,c) < \varphi(y_0,c) + \epsilon , \varphi(y_0,c) < \varphi(x_0,c) + \epsilon$ for all $c\in C$ therefore $\text{sup}\{\varphi(x_0,c):c\in C\} < \varphi(y_0,c) + \epsilon \leq \text{sup}\{\varphi(y_0,c):c\in C\} + \epsilon$ and similarly $\text{sup}\{\varphi(y_0,c):c\in C\} < \text{sup}\{\varphi(x_0,c):c\in C\} + \epsilon$. Hence $\left | \text{sup}\{\varphi(x_0,c):c\in C\}- \text{sup}\{\varphi(y_0,c):c\in C\} \right | < \epsilon$ i.e. $\psi$ is continuous at $x_0$, as desired.

 

[R136a1]

Awesome, thanks a lot, miss!

I really like your signature by the way!

 

[WannabeNewton]

No problem! Are you a physics student then? :)

 

[R136a1]

No problem! Are you a physics student then? :)

Yes. I want to go into mathematical physics in grad school. I'm still a rising senior in high school though, but I greatly enjoy math and physics. Especially things to do with astronomy, as you can see from my username

 

[WannabeNewton]

Start getting into general relativity; it's the best subject in any field ever :)!

 

[R136a1]

Start getting into general relativity; it's the best subject in any field ever :)!

Oh yes, I'm very interested in general relativity! I'm actually trying to study it right now (hence my interest in topology). I might make a relativity thread later on!

 

[WannabeNewton]

I look forward to it ;)

 

[Bacle2]

I think if you want to define the max over an infinite collection of functions you could consider a map from a function space ℝ^S of real-valued functions into the reals. And in my experience, one often uses the compact-open topology. For a pair, or finite collection of functions, you can also use the pasting lemma.