An infinite dimensional CW complex always has infinitely many non-trivial homology groups?

Question

Let $X$ denote an infinite dimensional CW complex, I wonder if $H_n(X)\ne 0$ for infinitely many $n$'s.

I think we may need to use cellular homology. The only thing I have got is that since $X$ is infinite dimensional, there are infinitely many $n$'s such that $H_n(X^{n}, X^{n-1}) \ne 0$.

Edit: This turns out to be a stupid question. But it seems that all the counterexamples below(great answers!) are contractible so far. What if I require the CW complex to be non-contractible?(I am working on $K(G,1)$ actually...)

Answer 1

A counterexample is $S^\infty$, the union of $S^n$ for all $n\geq 0$. This is a complex with two cells in each dimension, so it is infinite dimensional, but it is contractible.

Answer 2

A very simple counterexample is an infinite wedge sum like $\bigvee_{n=0}^\infty\Delta^n$. This is infinite-dimensional, but is contractible since each $\Delta^n$ is contractible.

Answer 3

Here is an example of a $K(G,1)$ which is not homotopy-equivalent to a finite-dimensional one but has zero integer homology (and cohomology) in all positive degrees, as asked by Dennis Sullivan.

Start with the Higman group $G_1$: It is a nontrivial acyclic group (all its homology and cohomology vanish in all positive degrees), see here. Aside: $G_1$ has a finite 2-dimensional aspherical presentation complex and, thus, has cohomological dimension 2. Let $X_1=K(G_1,1)$, for instance, one can take the 2-dimensional presentation complex of $G_1$. The complex $$ X=\prod_{i=1}^\infty X_1 $$ is the $K(G_,1)$ for the group $G$ which is the countably infinite direct product of copies of $G_1$. In particular, $X$ is acyclic, see for instance my answer here: This is just an application of the Kunneth formula.

In order to see that $X$ is not homotopy-equivalent to a finite-dimensional complex, one just needs to observe that the group $G$ has infinite cohomological dimension. (This is again an application of the Kunneth formula.)