Let X be in CGHaus and Y locally compact hausdorff. The usual product space XxY is CGHaus, so we dont need to apply that special functor to it (the one that takes a space to the space with same points and the strongest topology induced by its compact subsets). Is the right adjoint function space X^Y CGHaus? It would be convennient if this is true because I dont like applying that special functor.
1 Answer
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If you choose for Y the two-point discrete space, then $C(Y,X)=X^Y=X\times X$. So if the above would be true, then $X\times X$ would be k-space for any k-space X. This is not true. E.g., in Engelking, Example 3.3.29, an example of k-spaces $Y_{1,2}$, such that $Y_1\times Y_2$ is not k-space, is given. The space $X=Y_1\sqcup Y_2$ is a k-space, but $X\times X$ is not, as it contains $Y_1\times Y_2$ as a clopen subspace.
answered Mar 8, 2011 at 15:22
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$\begingroup$ "Clopen" is well-known terminology. $\endgroup$ Commented Mar 8, 2011 at 16:28
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