Showing Hawaiian earrings are not CW complexes

Question

Let $S^1_{1/n}$ be circles with radius $1/n$ and origin $(1/n,0)$ in the plane. I want to show that the space $X = \cup_{n=1}^\infty S^1_{1/n} \subseteq R^2$, is not a CW complex.

I know that one way of proving this is that by showing the $X$ is not a weak topology. I want to know if there is another way of proving this namely by the fact that a CW complex X is compact if and only if it has finitely many cells.

Answer 1

Assume $X$ is a CW-complex. As you say, it must be a finite CW-complex because $X$ is compact. Let $e_1, \ldots, e_n$ be its open cells. Note that all $e_i$ are path connected.

1. If a space $Z$ is the union of $m$ path connected subsets, then it hast at most $m$ path components.

This is trivial.

2. For each $p \in X$, the subspace $X_ p = X \setminus \{ p \}$ has only finitely many path components.

There exists a unique $j$ such that $p \in e_j$. If $\dim e_j = 0$, then $X_p = \bigcup_{i \ne j} e_i$ which has at most $n-1$ path components. If $\dim e_j > 0$, then $X_p = \bigcup_{i \ne j} e_i \cup (e_j \setminus \{ p \})$. If $\dim e_j = 1$, then $e_j \setminus \{ p \}$ has two path compenents and $X_p$ has at most $n+1$ path components. If $\dim e_j > 1$, then $e_j \setminus \{ p \}$ is path connected and $X_p$ has at most $n$ path components.

3. Let $p = (0,0) \in X$. This is the point in which all circles $S^1_{1/n}$ meet. Then each $S^1_{1/n} \setminus \{p\}$ is a path component of $X_p$.

This is a contradiction: By 2. $X_p$ has only finitely many path components, by 3. it has infinitely many.

Answer 2

Here's another way to show it. As you say, if $X$ were a CW-complex, then it would have to be a finite CW-complex since it is compact. Since every finite CW-complex has finitely generated homology, it suffices to show the homology of $X$ is not finitely generated. To show this, just observe that for any $n$, $X$ has a retract which is a wedge of $n$ circles (namely, the union of $n$ of the circles that make up $X$; the retraction just maps all the other circles to the origin). The first homology group of a wedge of $n$ circles is $\mathbb{Z}^n$, which cannot be generated by fewer than $n$ elements. It follows that $H_1(X)$ cannot be generated by fewer than $n$ elements for any $n$, and thus cannot be finitely generated.