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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
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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
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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.

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One should show somehow that this is a non-Hausdorff CW-complex, though. –  Mariano Suárez-Alvarez Oct 5 '11 at 17:57
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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 $[\epsilon/2]=\pi_1(\epsilon/2)=\pi_2(\epsilon/2)\in U_1\cap U_2$

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