# Continuity of the maximum

1. Jul 9, 2013

### 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?

2. Jul 9, 2013

### 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.

Last edited: Jul 9, 2013
3. Jul 14, 2013

### R136a1

Awesome, thanks a lot, miss!!

I really like your signature by the way!

4. Jul 14, 2013

### WannabeNewton

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

5. Jul 14, 2013

### R136a1

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

6. Jul 14, 2013

### WannabeNewton

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

7. Jul 14, 2013

### R136a1

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!

8. Jul 14, 2013

### WannabeNewton

I look forward to it ;)

9. Jul 15, 2013

### 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 ℝS 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.