# Hausdorff condition for CW complexes

Consider the definition of CW complex from wikipedia. It is assumed that the space is Hausdorff.

Are there problems if we drop this assumption? What is an example of a space satisfying all the CW complex axioms except this condition?

• For one thing, it is really nice to have one's compact sets be closed! Commented Oct 5, 2011 at 17:58
• The right way to think about CW complexes would rather be the inductive definition, which is also mentioned in the wikipedia article. The spaces constructed in this way are Hausdorff. Commented Jan 3, 2012 at 19:39

It is unnecessary to add Hausdorff to the definition. Every CW complex is Hausdorff. A proof of this can be found in the appendix of Hatcher's Algebraic Topology which is available for free on his homepage: http://pi.math.cornell.edu/~hatcher/AT/ATpage.html

• The Smirnov metrization theorem (every paracompact locally metrizable space is metrizable) from Munkres Thm 42.1 actually requires Hausdorff in the hypotheses. It is straightforward to check that the line with two origins is not Hausdorff. But we cannot conclude that it is not paracompact. I am pretty sure it is paracompact in fact. Commented Apr 14, 2023 at 3:47
• You are absolutely right. I went to far afield there. I've updated my answer. Now that I think of it, the bug-eyed line may be paracompact. However, intuition says the long line should be paracompact, but it's not (otherwise it'd be metrizable). My intuition may be wrong about the bug-eyed line. Commented Sep 8, 2023 at 13:34
• The line with two origins is paracompact in fact. See math.stackexchange.com/questions/4756135. Commented Sep 9, 2023 at 6:27
• Neat! Thanks for sharing. Commented Sep 11, 2023 at 11:27

Here's a proof that the space given by Joe Johnson is not Hausdorff:

Let $R_1= R_2=\mathbb{R}$ and let $X$ denote the quotient of $R_1\dot{\cup}R_2$ by the relation given by above. Let $\pi\colon R_1\dot{\cup}R_2\rightarrow X$ be the projection map, and let $\pi_1,\pi_2$ be the restrictions to the $R_1,R_2$ (note that restricting a continuous function gives a continuous function).

Let $U_1,U_2$ be any open nhds of $0_1$ and $0_2$ respectively. Since $\pi_1^{-1}(U_1)\subset\mathbb{R}$ is open and contains $0$, $\exists \epsilon_1>0\ :\ (-\epsilon_1,\epsilon_1)\subset \pi_1^{-1}(U_1)$. Similarly, $\exists \epsilon_2>0\ :\ (-\epsilon_2,\epsilon_2)\subset \pi_2^{-1}(U_2)$. Let $\epsilon=$min{${\epsilon_1,\epsilon_2}$}. Then $[\epsilon/2]=\pi_1(\epsilon/2)=\pi_2(\epsilon/2)\in U_1\cap U_2$

Take two copies of $$\mathbb{R}$$, say $$R_1$$ and $$R_2$$. If $$x\neq 0$$ then identify $$x\in R_1$$ with $$x\in R_2$$. Then the two $$0$$'s will not have disjoint neighborhoods. This is also an example of a manifold that is not Hausdorff.

• One should show somehow that this is a non-Hausdorff CW-complex, though. Commented Oct 5, 2011 at 17:57

Consider the following example. Let $$X$$ be the set $$D^2$$, the closed disk not yet equipped with a topology. Define a characteristic function $$\varphi: D^2\to D^2$$ to be the identity on the interior of $$D^2$$, and define it on the boundary $$S^1$$ as follows: fix some distinct $$x,y$$ in $$S^1$$, and let $$\varphi$$ map $$y$$ to $$y$$ and the rest of $$S^1$$ to $$x$$. In addition, define two characteristic function $$D^1\to D^2$$, which specify paths along the boundary circle, first from $$x$$ to $$y$$ counterclockwise, and then from $$y$$ to $$x$$ counterclockwise. Finally, define two characteristic functions $$D^0\to D^2$$ that map to $$x$$ and $$y$$. Now endow $$X=D^2$$ with the coarsest topology making these characteristic maps continuous. Then $$X$$ satisfies all the conditions of a CW complex, except the Hausdorff property.

You can show that $$X$$ is not Hausdorff, for example by showing that every neighborhood of $$y$$ must also contain $$x$$.