Topologists sine curve is not CW-decomposable

Question

I was going through the book "Topology" by Klaus Jaenich for some material on CW-complexes. After the definition of a CW-complex, the author gives a few first non-examples where one of the axioms is not satisfied. One of the non-examples is that of the topologist's sine curve (the author draws only a figure, but I did recognize it as the topologist's sine curve). In that example, he gives a particular cell-decomposition which does not satisfy the first axiom of CW-complexes and hence is not a CW-complex.

However, he then proceeds to write "This example cannot be 'fixed' by some other cell-decomposition, and hence the space is not CW-decompasble." I was looking for a proof of this statement. Is there an elementary proof which could be done at this stage or do we have to develop more theory around CW-complexes to get it?

Thanks in advance for any insights on the matter!

Answer 1

There are various approaches.

1. The topologist's sine curve $S$ is connected. If it were a CW-complex, then it would be path-connected which is not true.

2. $S$ is compact. If it were a CW-complex, then it would have only finitely many closed cells. Each closed cell must be contained in a path component of $S$. But $S$ has two path-components, one of which is not compact. This path-component must be the union of finitely many closed cells, i.e. must be compact. This is a contradiction.

Answer 2

Every CW complex $X$ is locally contractible, meaning that for each $x \in X$ there is a neighborhood basis of $x$ consisting of contractible open subsets (you can find a proof of this fact in an appendix of Hatcher's book "Algebraic Topology"). Of particular in for this problem is that for any $x \in X$ there is a neighborhood basis of $x$ consisting of path connected open subsets. Which is to say: every CW complex is locally path connected.

However, the topologists sine curve $S$ is not locally path connected: it has a point $s \in S$ which possesses an open neighborhood homeomorphic $U$ to the space $$V = \biggl(\bigl\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},...\bigr\} \cup \{0\} \biggr) \times (-1,1) $$ such that the homeomorphism $U \mapsto V$ takes $s \mapsto (0,0)$. Clearly no open neighborhood of $s$ that is a subset of $U$ is path connected. It follows that $s$ has no neighborhood basis of path connected open sets.

From this it follows that $S$ is not homeomorphic to any CW complex.