Page 1 of 1

Not Hausdorff

Posted: Fri Jun 10, 2016 7:48 pm
by Tsakanikas Nickos
Find an example of a space locally homeomorphic to $\mathbb{R}$, but not satisfying the Hausdorff condition.

Re: Not Hausdorff

Posted: Sat Jun 11, 2016 5:42 pm
by Papapetros Vaggelis
Hi Nickos. Here is a possible answer.

Consider \(\displaystyle{M=\mathbb{R}\cup\,\left\{(0,1)\right\}}\) and the sets

\(\displaystyle{U_1=\left\{(t,0)\,,t\in\mathbb{R}\right\}\,\,,U_2=\left\{(t,0)\,,t\in\mathbb{R}-\left\{0\right\}\right\}\cup\left\{(0,1)\right\}}\)

Also, define the maps

\(\displaystyle{\phi_1(t,0)=t\,,(t,0)\in U_1}\)

\(\displaystyle{\phi_{2}(t,0)=t\,,t\in\mathbb{R}-\left\{0\right\}\,\,,\phi(0,1)=0}\).

We have that \(\displaystyle{M=U_1\bigcup U_2\,\,,U_1\bigcap U_2=\mathbb{R}-\left\{0\right\}}\)

and \(\displaystyle{\phi_1\,,\phi_2}\) are homeomorphisms.

If \(\displaystyle{p=(0,0)\in M\,\,,q=(0,1)\in M}\), then for each \(\displaystyle{\epsilon>0}\) holds

\(\displaystyle{(-\epsilon,\epsilon)\cap (-\epsilon,0)\cup (0,\epsilon)\cup\left\{0\right\}\neq \varnothing}\)

and \(\displaystyle{M}\) is not a \(\displaystyle{\rm{Hausdorff}}\) - space.