Why a convergent succesion does not have the same homotopy type of a CW-Complex?

Question

The question is pretty much in the title;

If my space is $\{1/n\}_{n\in \mathbb{N}} \cup \{0\} $ why it isn't homotopically equivalent to a CW-Complex?

Answer 1

If your space $X$ were homotopy equivalent to a CW-complex, then this CW-complex would have homotopy groups $\pi_i = 0$ for $i>0$ with countably many connected components. Thus by Whitehead's theorem, this space would be homotopy equivalent to $\Bbb N$.

Now, given a homotopy equivalence $f: X \to Y$ where $Y$ is totally disconnected, $f$ must be a homeomorphism. For suppose that $g$ is its homotopy inverse; then $gf: X \to X$ is homotopic to the identity. Let $(gf)_t: X \times I \to X$ be such a homotopy. Then $h(t) = (gf)_t(x)$ is a continuous map $I \to X$ with $h(0) = (gf)(x)$ and $h(1) = x$; but $X$ is totally disconnected so this map must be constant, and $(gf)(x) = x$ for all $x$. Similarly for $fg$; so $f$ is a homeomorphism.

But $\Bbb N$ is not homeomorphic to your space $X$. This contradicts the above two paragraphs.

In general, if you want to see that some space isn't homotopy equivalent to a CW complex, 'it contradicts Whitehead's theorem' is a good reason for this to be so. Indeed, if a space isn't homotopy equivalent to a CW complex, it does contradict Whitehead's theorem: CW approximation says that there's a weak homotopy equivalence $Z \to X$, where $Z$ is a CW-complex and $X$ is your space; but if $X$ doesn't have the homotopy type of a CW complex this is of course not a homotopy equivalence, thus (with only a little more acrobatics) contradicting Whitehead.