1

In the nlab, it is stated that a compactly generated space could equivalently be defined as a space $X$ such that there is a family of compact Hausdorff spaces $K_\alpha$ such that for any space $Y$, a set function $f:X \to Y$ is continuous iff for every continuous $u:K_\alpha \to X$ we have that $fu$ is continuous.

It is clear that this implies the usual definition, on the other way I was told the following argument:

For each non-open subset of $X$, choose a $C$ and a continuous $t: C \to X$, such that the preimage is not open. (If such a $C$ exists). Since $X$ has a set worth of subsets, this forms a set. Now assume $f: X \to Y$ isn't continuous. Then there exists a continuous function $t:C\to X$ and an open set $U$ in $Y$, such that $t^{-1}(f^{-1}(U))$ is not open. In particular $f^{-1}(U)$ satisfies our above assumptions, so we have a chosen $t$ and $C$ such that $t^{-1}(f^{-1}(U))$ is not open.

The problem with this argument is that the choice of $C$ is taken from a proper class of spaces. Is there anyway to bound this class (say, by using the cardinality of $X$) so that the only axiom of choice is needed, instead of global choice? Better yet, can this be proven without using the axiom of choice?

Fernando Chu
  • 2,511
  • 8
  • 22

1 Answers1

3

The problem with this argument is that the choice of $C$ is taken from a proper class of spaces. Is there anyway to bound this class (say, by using the cardinality of $X$) so that the only axiom of choice is needed, instead of global choice? Better yet, can this be proven without using the axiom of choice?

As Asaf Karagila notes in the comments, we use Scott's trick, which requires the axiom of regularity, but not choice. Explicitly, for each non-closed $A\subset X$, let $\alpha_A$ be the first ordinal $\alpha$ for which there is a continuous map $f\colon K\to X$ from some compact Hausdorff $K$ such that $f^{-1}(A)$ is not closed, and such that $f\in V_\alpha$.

Then let $\mathcal F_A$ denote the family of all such maps in $V_{\alpha_A}$, and let $$\mathcal F=\bigcup_{A\subset X\mid A \text{ is not closed} } \mathcal F_A.$$ Then the set $\mathcal K=\{\text{dom}(f)\mid f\in \mathcal F\}$ is the desired family of compact Hausdorff spaces. Note that since $V_{\alpha_A}$ is a set, the axiom of unions (and the axiom schema of replacement applied a few times) guarantees $\mathcal K$ is a set.

M W
  • 9,866