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Let $X$ and $Y$ be topological spaces and let $C(X,Y)$ denote the set of continuous maps from $X$ to $Y$. For any two subsets $A \subset X$ and $B \subset Y$ let $W(A,B) := \{ f \in C(X,Y) \mid f(A) \subset B\}$. The compact-open topology on $C(X,Y)$ is the topology with subbasis consisting of the sets $W(K,V)$ for all compact subsets $K\subset X$ and open subsets $V \subset Y$.

However, in Bourbaki the term compact means compact Hausdorff. Suppose we instead take the subbasis to consist of those sets $W(K,V)$ for all compact Hausdorff $K \subset X$ and open subsets $V\subset Y$. In general, does this give the same topology as the one above?

Victoria Flat
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1 Answers1

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There are many conflicts of notations/notions between French and English : $]0,1[/ (0,1)$, compact/compact Hausdorff, $\mathrm X / X$ or with the binomial coefficient $\mathrm C_n^p$.

As said by Wikipedia, usually the right definition of the compact-open topology is the French one.

Note that in Aglebraic Geometry, the Zariski topology is not Hausdorff and one say that sets are quasi-compact to mean that they satisfy Borel-Lebesgue Axiom.

So quasi-compact + Hausdorff = compact ; and compact should by used only for Hausdorff spaces.

And this give you the counter example you wanted. Take for $\rm X$, the Zariski spectrum of a ring. Then all the open subsets are quasi-compact but there are extremely few compact subsets.

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Damien L
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  • In the spectrum of a ring the only quasi-compact open subsets are finite unions of open basis-sets wrt. the standard basis. In general there are open subsets which are not quasi-compact. – Dune Mar 21 '13 at 12:21
  • Ok, so one has to make some assumptions on the ring, like noetherianity. – Damien L Mar 21 '13 at 14:09
  • Thanks for your answer. However, I don't see how the Zariski spectrum of a noetherian ring gives a counter-example. I'm interested in the compact-open topology on the set of continuous functions between two topological spaces. – Victoria Flat Mar 21 '13 at 22:44