# How to show that $X = \mathbb{R}\cup \{0'\}$ is homeomorphic to $\mathbb{R}$?

Actually, I need to show that $\forall x\in X,\exists U\subseteq X$ with $x\in U$ and $U$ is homeomorphic to an open set in $\mathbb{R}$. I have trouble showing the existence of neighbourhood of $0'$, and the homeomorphism.

Otherwise, for all $x\in\mathbb{R},$ any open set in $\mathbb{R}$ containing $x$, for example $(x-\epsilon,x + \epsilon), \forall\epsilon>0$ is homeomorphic to the same open set in the codomain $\mathbb{R}$, just by taking the identity map.

I have this requirement: $U\subseteq X$ is open if either $U\subseteq \mathbb{R}$ open or $0'\in X$ and $(X-\{0'\}\cup\{0\})$ is open in $\mathbb{R}$.

EDIT: The requirement should be: $U\subseteq X$ is open if either $U\subseteq \mathbb{R}$ open or $0'\in U$ and $(U-\{0'\}\cup\{0\})$ is open in $\mathbb{R}$.

• What's the topology on $X$? – Dan Rust Feb 23 '14 at 13:31

As you seem to be aware, the difficulty of this problem is in finding an open set around $0'$ which is homeomorphic to an open subset of $\mathbb{R}$. If we choose some open set around $0$ and then remove $0$ and insert $0'$ what happens? So, let $U'=U\setminus\{0\}\cup\{0'\}$ for some $U$ containing $0$. Can you see how this set will be open and, by definition, contains $0'$? Can you also then see that $U'$ is homeomorphic to $U$? (what would be a good choice of homeomorphism?)
• thank you. So the homeomorphism could be: $0'\mapsto 0, 0'\neq x \mapsto x$. Right? – ugstudent1243 Feb 23 '14 at 13:56
• From my definition of open set, I think it follows that every subset of $X$ is open, isn't? – ugstudent1243 Feb 23 '14 at 13:59
• by the second condition of open set. $(X-\{0'\})\cup\{0\}=\mathbb{R}\cup\{0\}=\mathbb{R}$ is open in $\mathbb{R}.$ Is this wrong idea? – ugstudent1243 Feb 23 '14 at 14:06
• @ugstudent1243 That is certainly the most natural choice of homeomorphism yes :). I'm not sure why you would think every subset of $X$ is open. As a note, I think you may have written the definition of the topology wrong as $U\ni 0'$ should be open if $U\setminus\{0'\}\cup\{0\}$ is open in $\mathbb{R}$ (you have $X$ instead of $U$ which doesn't make much sense). – Dan Rust Feb 23 '14 at 14:08
If you are talking about the line with two origins, then the topology of $X$ is such that $U=(a,0)\cup \{0'\}\cup (0,b)$ is open for all $a<0<b$. Then $U\approx (a,b)$ with the euclidean subspace topology, by definition, so you have your locally euclidean neighborhood of $0'$. Around any other point, choose $\mathbb{R}$ to be the locally euclidean neighborhood.