MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a topological space $(X,\mathcal{O})$ can one assign a natural topology to $\mathcal{O}$ such that

1) The intersection of a compact set of open sets is again open,

2) The maps $\cap,\cup:\mathcal{O}^2\rightarrow \mathcal{O}$ are continuous,

3) For any continuous map $f: (X,\mathcal{O})\rightarrow (X',\mathcal{O}')$ is the induced map $f^{-1}: \mathcal{O}'\rightarrow \mathcal{O}$ also continuous?

Since these properties would allow taking the discrete topology and I am looking for a more interesting one let me also add:

4) For $X=\mathbb{R}$ the subset $\{]a,b[|a\le b\}$ of intervals has the topology coming from $\{(a,b)\in \mathbb{R}^2| a\le b\}/\Delta \mathbb{R}$.

share|cite|improve this question
I don't understand 4). Which sets of intervalls are open? – Martin Brandenburg Apr 2 '12 at 8:06
In any case, have you tried the compact-open topology on $\mathcal{O}(X) \cong \mathrm{Hom}(X,S)$, where $S$ is the Sierpinski space? – Martin Brandenburg Apr 2 '12 at 8:08
Have you seen… ? – Gjergji Zaimi Apr 2 '12 at 8:26
@Martin: in topology $\mathcal{O}(X)\cong\mathrm{Hom}(X,S)$ while in algebraic geometry we have $\mathcal{O}(X)\cong\mathrm{Hom}(X,\mathbb{A}^1)$... (just kidding :) ) – Qfwfq Apr 2 '12 at 8:28
4) seems to be a problem. By definition of the compact open topology the following system indexed over closed intervals $[a,b]$ is a subbasis: $\\{(a',b')|[a,b]\subset (a',b')\\}_{[a,b]}$. Especially for any open set $U\subset \mathcal{O}$ we can find a $r$ such that $(-R,R)\in U$ for $R>r$. Hence the compact open topology on $\mathcal{O}(\mathbb{R})$ is not Hausdorff and not the quotient topology as in 4). Concerning 2 & 3. Basically they would follow from usual properties of the compact open topology (for CGHaus-Spaces). But $S$ is not CGHaus. – HenrikRüping Apr 2 '12 at 13:12

Your Answer


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

Browse other questions tagged or ask your own question.