Locally Euclidean topology - but not Hausdorff We consider the set $X=\mathbb{R}\cup \{\star\}$, i.e. $X$ consists of $\mathbb{R}$ and an additional point $\star$.
We say that $U\subset X$ is open if:
(a) For each point $x\in U\cap \mathbb{R}$ there exists an $\epsilon>0$ such that $(x-\epsilon, x+\epsilon)\subset U$.
(b) If $\star \in U$ then there is an $\epsilon>0$ such that $(-\epsilon , 0)\cup (0, \epsilon)\subset U$.
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*Show that this defines a topology in $X$.


*Show that $X$ with this topology is locally Euclidean but not Hausdorff.
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For (1) we have to show that $X$ and $\emptyset$ are open, the union of two open sets is open and the intersection of two open sets is open.
First we show that $X$ is open :
(a) For each point $x\in X\cap \mathbb{R}=\mathbb{R}$ there exists an $\epsilon>0$ such that $(x-\epsilon, x+\epsilon)\subset X=\mathbb{R}\cup \{\star\}$. This is true since every neighboorhood of $x$ is contained.
(b) If $\star \in X$ then there is an $\epsilon>0$ such that $(-\epsilon , 0)\cup (0, \epsilon)\subset X=\mathbb{R}\cup \{\star\}$. Thisis true since the union of the intervals is a subspace of the real line.
Is that correct?
The emptyset is per definition open, or not? We don’t have to apply the given definition, do we?
Let $M_1$ and $M_2$ be two open sets. We consider the union $M_1\cup M_2$. For each point $x\in M_1\cup M_2$ it is either $x\in M_1$ or $x\in M_2$ (or both) so statement (a) follows from the fact that $M_1$ and/or $M_2$ are open. The same holds also for statement (b).
Let $M_1$ and $M_2$ be two open sets. We consider the intersection $M_1\cap M_2$. For each point $x\in M_1\cap M_2$ it is $x\in M_1$ and $x\in M_2$ so statement (a) follows from the fact that $M_1$ and $M_2$ are open. The same holds also for statement (b).
Therefore we get that the above defines a topology in $X$.
Is that correct and complete?
Could you give a hint for (2) ?
 A: Here is a proof of part (2): Using your definition of open sets in $X$, we denote with $\mathcal{O}_X$ the topology on $X$, i.e. the set of open sets in $X$. Let $\mathcal{O}_E$ denote the usual Euclidean topology on $\mathbb{R}$. To show that $X$ is locally Euclidean we'll want to show that $$\{U' \cap \mathbb{R} \mid U' \in \mathcal{O}_X \} = \mathcal{O}_E$$ (you can take a look at this page, if you want to verify that this is indeed what we need to show).
($\subseteq$) Let $U \in \{U' \cap \mathbb{R} \mid U \in \mathcal{O}_X \}$, i.e. $U = U' \cap \mathbb{R}$ with $U' \in \mathcal{O}_X$. If $U'$ consists only of numbers in $\mathbb{R}$, it is automatically contained in $\mathcal{O}_E$, so in particular $U \in \mathcal{O}_E$. If $U'$ is a neighborhood of $\star$, there exists $\varepsilon > 0$ such that $(-\varepsilon,0) \cup \{\star\} \cup (0,\varepsilon) \subseteq U'$. So we have $(-\varepsilon,0) \cup (0,\varepsilon) \subseteq U' \cap \mathbb{R} = U$, so $U\in \mathcal{O}_E$ (because again we have shown that every point in $U$ is contained within an open set).
($\supseteq$) Let $U \in \mathcal{O}_E$, i.e. for all $a \in U$ there exists $\varepsilon > 0$ such that $(a-\varepsilon,a+\varepsilon) \subseteq U$. So we have $U \in \{U' \cap \mathbb{R} \mid U' \in \mathcal{O}_X \}$.
Thus we have shown both inclusions, and equality follows. Hope this helps!
Edit: Sorry, forgot about the Hausdorff part. To show that $X$ is not Hausdorff, look at neighborhoods of $0$ and $\star$. For a neighborhood $U_0$ of $0$, there exists $\varepsilon > 0$ such that $(-\varepsilon,\varepsilon) \subseteq U_0$. For a neighborhood $U_\star$ of $\star$, there exists $\varepsilon' > 0$ such that $(-\varepsilon',0) \cup \{\star\} \cup (0,\varepsilon') \subseteq U_\star$. However, their intersection is non-empty, since
\begin{align}
(-\varepsilon,\varepsilon) \cap \big((-\varepsilon',0) \cup \{\star\} \cup (0,\varepsilon')\big) &= (-\varepsilon,\varepsilon) \cap \big((-\varepsilon',0) \cup (0,\varepsilon')\big)\\
&= \big((-\varepsilon,\varepsilon) \cap (-\varepsilon',0)\big) \cup \big((-\varepsilon,\varepsilon) \cap (0,\varepsilon')\big) \neq \emptyset,
\end{align}
so $X$ is not Hausdorff.
