Weak Equivalence between an arbitrary space and a CW complex When working in the compactly generated spaces (weak Hausdorff $k$-spaces) it is true that any space is weakly equivalent to a CW complex. I'm interested in the converse: let $X$ be a CW complex and $Y$ be a topological space such that there is a weak equivalence $X \to Y$, then is it necessarily true that $Y$ is compactly generated? If that is not true in general, does it hold for homotopy equivalence? I figure the proof would be an abysmal exercise in point-set topology if either were true.
I wondered about this while trying to show that a certain space does not have the homotopy type${}^*$ of a CW complex, and that if one wanted to look for such a space one shouldn't waste any time looking in the category of compactly generated spaces. If the above were true I could prove no such weak equivalence exists if the space isn't compactly generated.
${}^*$It seems some authors take homotopy type to mean weak equivalence, which makes sense in the category of CW spaces, but in general is homotopy type taken to mean homotopy equivalence?
 A: Every space is weak equivalent to a CW-complex; in fact, I'm not familiar with any proof that only compactly generated spaces (as opposed to all spaces) are weak equivalent to CW-complexes.  For instance, the proof in Hatcher (Proposition 4.13) works for all spaces.
In any case, if you know the result for $k$-spaces, you can immediately deduce it for arbitrary spaces, since the map $kX\to X$ to any space from its $k$-ification is a weak equivalence (this is immediate from the fact that "weak equivalence" is defined in terms of maps from certain compact Hausdorff spaces to $X$).
For strong homotopy type, however, compact generation is neither necessary nor sufficient to be homotopy equivalent to a CW-complex.  For instance, if $X$ is any non-compactly generated space, the cone on $X$ is contractible and hence homotopy equivalent to a CW-complex, but not compactly generated.  On the other hand, plenty of compact Hausdorff spaces such as $\{0\}\cup\{1,1/2,1/3,\dots\}$ are not homotopy equivalent to CW-complexes.
