# 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?

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For one thing, it is really nice to have one's compact sets be closed! – Mariano Suárez-Alvarez Oct 5 '11 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. – Alexander Thumm Jan 3 '12 at 19:39

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.
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 $[\epsilon/2]=\pi_1(\epsilon/2)=\pi_2(\epsilon/2)\in U_1\cap U_2$