Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X = 2^\mathbb{N}$ and $Y = \mathbb{R}^+$ (i.e. the non-negative numbers). Is there a topology in which functions similar to $f : X \to Y$, $$ f(A) = \begin{cases} \frac{1}{|A|}, & |A| < \infty \\ \\ \ \ 0, & |A| \not<\infty \end{cases}$$ or $$g(A) = f(A \cap \{2k \mid k \in \mathbb{N}\})$$ would be continuous, but $$ h(A) = \begin{cases} \frac{1}{|A|}, & |A| < \infty \\ \\ \ \ 1, & |A| \not<\infty \end{cases}$$ would not? (For $Y$ take the standard topology on $\mathbb{R}$.)

Is it possible for $X$ to be compact with such topology? The context is proving that some functions defined on $X$ attain the minimum/maximum value and I am wondering if it could be done via topology. The functions I am talking about are similar to $$F(A) = \sum_k [\text{if }B_k \subseteq A\text{ then }b_k\text{ else }0]$$

where $(B_k)$ is some countable family of finite sets, and we know that $F$ is bounded, i.e. there exists $M$ such that $|F(A)| < M$ for every $A$. I will appreciate comments on other approaches too.

Thanks in advance!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Presumably you'll want to define $f$ differently in the case $A = \phi$, but that shouldn't be much of an issue.

If any topology will work, then the smallest topology such that $f$ is continuous in particular will work. This is given by defining $U \subset 2^\mathbb{N}$ to be open iff $U = f^{-1}(V)$ for $V$ open in $Y$. A basis for this topology is $\{\{A : |A|=n \} : n \in \mathbb{N} \} \cup \{ \{A : |A| > n \} : n \in \mathbb{N} \}$.

In this topology, $f$ is indeed continuous, while $h$ is not since ${1}$ is open in the image of $h$ and $h^{-1}(1) = \{A : |A| =1$ or $|A| = \infty \}$ is not open. You can check for yourself that this topology is compact.

I haven't thought about the optimization part, but hopefully this gives you somewhere to start on that.

share|improve this answer
    
Thanks a lot, and sorry to break your nice 1111 reputation ;-) –  dtldarek Jun 19 '12 at 19:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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