Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $X$.

share|improve this question
1  
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

1 Answer 1

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 $X$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.