Not Hausdorff
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Not Hausdorff
Find an example of a space locally homeomorphic to $\mathbb{R}$, but not satisfying the Hausdorff condition.
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- Community Team
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Re: Not Hausdorff
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.
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.
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