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

In summary, the conversation covers the topic of continuity of the maximum of infinitely many functions on a topological space. It is proven that under certain conditions, this maximum function is continuous. The conversation also delves into the interest in general relativity and its connection to topology.
  • #1
R136a1
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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
[tex]x\rightarrow \max_{c\in C} \varphi(x,c).[/tex] Is this continuous?
 
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  • #2
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.
 
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  • #3
Awesome, thanks a lot, miss!

I really like your signature by the way!
 
  • #4
No problem! Are you a physics student then? :)
 
  • #5
WannabeNewton said:
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 :smile:
 
  • #6
Start getting into general relativity; it's the best subject in any field ever :)!
 
  • #7
WannabeNewton said:
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!
 
  • #8
I look forward to it ;)
 
  • #9
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.
 

1. What is "continuity of the maximum"?

"Continuity of the maximum" refers to the property of a function where the maximum value of the function remains unchanged as the input values approach a specific point or boundary.

2. How is continuity of the maximum related to the continuity of a function?

Continuity of the maximum is a specific case of continuity of a function. It is a more specific property that applies to functions with a maximum value, whereas continuity refers to a function being smooth and uninterrupted.

3. Can a function have a maximum value but not have continuity of the maximum?

Yes, it is possible for a function to have a maximum value but not have continuity of the maximum. This can occur when the function has a jump or discontinuity at the point where the maximum value is located.

4. How does continuity of the maximum affect the behavior of a function?

Continuity of the maximum ensures that the function is smooth and well-behaved near the point where the maximum value is located. This means that small changes in the input values will result in small changes in the output values, and the function will not have any sudden jumps or breaks.

5. Can continuity of the maximum be proven mathematically?

Yes, continuity of the maximum can be proven mathematically using the epsilon-delta definition of continuity. This involves showing that for any small positive value of epsilon, there exists a small positive value of delta such that the maximum value of the function remains within epsilon of the original maximum value when the input values are within delta of the point of interest.

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