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?
Best Answer
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.