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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 $\{(a,b)\in \mathbb{R}^2| a\le b\}/\Delta \mathbb{R}$.

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I don't understand 4). Which sets of intervalls are open? –  Martin Brandenburg Apr 2 '12 at 8:06
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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
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Have you seen mathoverflow.net/questions/20738/… ? –  Gjergji Zaimi Apr 2 '12 at 8:26
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@Martin: ...so 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
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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

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