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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 $\mathbb{R}$.

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  • $\begingroup$ What's the topology on $X$? $\endgroup$ – Dan Rust Feb 23 '14 at 13:31
  • $\begingroup$ sorry, I have added more information. $\endgroup$ – ugstudent1243 Feb 23 '14 at 13:37
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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?)

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  • $\begingroup$ thank you. So the homeomorphism could be: $0'\mapsto 0, 0'\neq x \mapsto x$. Right? $\endgroup$ – ugstudent1243 Feb 23 '14 at 13:56
  • $\begingroup$ From my definition of open set, I think it follows that every subset of $X$ is open, isn't? $\endgroup$ – ugstudent1243 Feb 23 '14 at 13:59
  • $\begingroup$ @ugstudent1243 No, what makes you think so? $\endgroup$ – Hagen von Eitzen Feb 23 '14 at 14:03
  • $\begingroup$ 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? $\endgroup$ – ugstudent1243 Feb 23 '14 at 14:06
  • $\begingroup$ @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). $\endgroup$ – Dan Rust Feb 23 '14 at 14:08
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I am guessing you mean "locally homeomorphic" in your title.

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.

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