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Let $X$ be a space that is the union of the subspaces $S_1,\dots,S_n$, where each $S_i$ is homeomorphic to the unit circle, and suppose that there is some $p \in X$ such that $S_i \cap S_j = \{p\}$ whenever $i \neq j$. I am trying to show that $X$ is Hausdorff if and only if each $S_i$ is closed in $X$, and also to construct an example of such a space $X$ which is not Hausdorff.

If we assume that $X$ is Hausdorff, then it's easy to show that each $S_i$ is closed in $X$, since the unit circle is compact and compact subspaces of Hausdorff spaces are closed. I'm having trouble with the converse, however. Let $x_1,x_2$ be distinct points of $X$, then $x_1 \in S_i, x_2 \in S_j$ for some $i,j$. Suppose that neither point is $p$. Then $ X - (S_i - \{p\}) = S_1 \cup \dots \cup S_{i-1} \cup S_{i+1} \cup \dots S_n $ is closed, so $S_i - \{p\}$ is open. If $i \neq j$, then $S_i - \{p\}, S_j - \{p\}$ are disjoint neighbourhoods of $x_1,x_2$.

Now if $i = j$, we have $x_1,x_2 \in S_i - \{p\}$, and I'm not seeing how we could find disjoint neighbourhoods of $x_1,x_2$, given that we don't know the topology of $S_i$. Do we have to assume that the unit circle, and hence each $S_i$, has the topology induced by the Euclidean metric?

The last case is when one of the points, say $x_1$, is $p$. Then $x_1 \in S_1$, which is closed, and $x_2 \notin S_1$, so we can find a neighbourhood $U_2$ of $x_2$ which does not intersect $S_1$, and therefore does not contain $x_1$. How could we produce a neighbourhood $U_1$ of $x_1$ not intersecting $U_2$, though?

Finally, any help on constructing a non-Hausdorff $X$ would be appreciated; I'm guessing what we proved means that we need the $S_i$'s to not all be closed in $X$, but I can't see how to use this fact to explicitly construct such an $X$.

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We do know the topology of $S_i$, since $S_i$ is homeomorphic to $S^1$. When the problem states that $S_i$ is homeomorphic to $S^1$, it can't mean anything but the normal topology on $S^1$ because $S^1$ is not just a set, it is a topology. In particular, since $S^1$ is a subset of $\mathbb R^2$ and the latter is Hausdorff, the former is, too. – Thomas Andrews Feb 19 '13 at 18:12
up vote 3 down vote accepted

Let $D=\{0,1\}$ be the two-point space with the indiscrete topology, and let $S^1$ be the unit circle. Fix a point $p\in S^1$. Let $Y=D\times S^1$, and let $X$ be the quotient of $Y$ obtained by identifying $\langle 0,p\rangle$ and $\langle 1,p\rangle$. This is a non-Hausdorff example with $n=2$, and you can easily generalize it to arbitrary $n\ge 2$.

For the theorem, suppose that each $S_k$ is closed in $X$. Note that for each $k\in\{1,\dots,n\}$ we must have $S_k\setminus\{p\}$ open in $X$, since its complement is the union of the $S_i$ with $i\ne k$, a union of finitely many closed sets. Suppose that $x,y\in X\setminus\{p\}$ with $x\ne y$. If there is a $k\in\{1,\dots,n\}$ such that $x,y\in S_k$, then $x,y\in S_k\setminus\{p\}$, which is an open Hausdorff subspace of $X$, so $x$ and $y$ can be separated by disjoint open sets in $X$. If $x\in S_i$ and $y\in S_k$ with $i\ne k$, then $S_i\setminus\{p\}$ and $S_k\setminus\{p\}$ are disjoint open sets containing $x$ and $y$ respectively. It only remains to separate $p$ and an arbitrary $x\in X\setminus\{p\}$.

The point $x$ has an open nbhd $U$ in $S_k$ whose closure in $S_k$ does not contain $p$. To finish the argument, show that $\operatorname{cl}U$ is closed in $X$ and hence that $X\setminus\operatorname{cl}U$ is open in $X$.

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