What is an example of an Eilenberg-MacLane space that is not homotopy equivalent to a CW complex?

Question

The proofs that any two CW complexes that are K(G,n) are homotopy equivalent has no obvious extension to the general case. I assume it is known to be false in general, but I can't find an example.

Answer 1

I think something like this might work: take a noncontractible space that has all homotopy groups zero (such a thing is necessarily not a CW complex, I think an example can be obtained from wedging two comb spaces) and wedge it with K(G,n). The resulting space should not be homotopy equivalent to a CW complex and (perhaps depending on the noncontractible space) be a K(G,n).

Answer 2

For a nice example consider the pseudocircle (https://en.wikipedia.org/wiki/Pseudocircle). This is a finite topological space which is weakly equivalent to the circle $S^1$. It follows that the psuedocircle is a $K(\mathbb{Z},1)$. However, any finite CW complex must be discrete, so it obviously cannot be CW.

Answer 3

The Hawaiian earring (which is a a compact path connected and locally path connected subset of the plane) is a $K(G,1)$. See https://en.wikipedia.org/wiki/Hawaiian_earring, https://mathoverflow.net/questions/163847/are-the-higher-homotopy-groups-of-the-hawaiian-earring-trivial. Its fundamental group $G$ is uncountable.

Another example is the Warsaw circle. It is $K(0,n)$ for all $n$. See How to show Warsaw circle is non-contractible? and Connor Malin's answer.