Hausdorff condition for CW complexes

Question

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?

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

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 $x$.