why is a set containing exactly 1 point closed in CW complex?

Question

Let $x \in X$, $X$ is a CW-complex. Why is the $\{x\}$ closed in $X$? Please do not answer with $X$ being Hausdorff because that is what I am trying to develop from this. I am new to the construction of CW complex and I am stuck on this probably trivial detail while trying to read Hatcher's Algebraic Topology. He just mentions it off-handedly without any proof.

Answer 1

Well, you should try to show that a CW complex is Hausdorff to begin with. Developing this from closed points might be an unnecessary or even pointless work. After all closed points do not imply Hausdorff in general.

Recall that a CW complex $X$ is constructed by successive glueing of disks into skeletons. So we start with a discrete space $X_0$ and then recursively construct $X_n$ by glueing disks $D^n$ to $X_{n-1}$ via continuous maps $S^{n+1}\to X_{n-1}$. Then $X$ is defined as union of $X_n$ with the weak topology.

Now $X_n$ is constructed from $X_{n-1}$ by glueing. This means that $X_n$ is the adjunction space of $X_{n-1}$ and a disjoint union of disks. And it is well known that an adjunction space is Hausdorff if both spaces are Hausdorff. Note that obviously a disjoint union of disks is Hausdorff.

Therefore we conclude that if $X_{n-1}$ is Hausdorff then so is $X_{n}$. Now since $X_0$ is discrete (and hence Hausdorff) then by simple induction we conclude that each $X_n$ is Hausdorff. Finally this implies that $X$ is Hausdorff as well, by the weak topology.

Answer 2

I'm assuming you want to start with the inductive definition of adding cells in order of dimension and giving the result the weak topology. Suppose $x$ lies in $X_n\setminus X_{n-1}$; then it must be contained in the interior of an $n$-dimensional cell, and the interiors are mapped homeomorphically into $X$ so $x$ is certainly closed in $X_n$. It is closed in any higher dimensional cell by induction on $m > n$: its preimage under any attaching map must be closed (because it is closed in $X_{m-1}$).