# An open set in a space is homeomorphic to $S^1$. Is the set close?

Suppose I have a $1$-dimensional manifold $M$ and within it an open set $U$ such that $U$ is homeomorphic to $S^1$. Can I deduce from that that $U$ is also a closed set inside $M$?

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Is $M$ connected? Suppose that $M$ is connected. If $U$ is closed in $M$, then $U$ will be an open and closed set inside $M$, and therefore $M=U$. –  Hoedan Feb 29 '12 at 16:59
$M$ is indeed connected, and I do want to show that $M=U$. But the homeomorphism is between $U$ and $S^1$ as topological spaces, so of course $U$ will be closed relatively to itself. What I don't understand is why should it be closed with respect to $M$. –  Ahia Cohen Feb 29 '12 at 17:07
Couldn't it just be noted that topologically $M$ must be either $\mathbb{R}$ or $\mathbb{S}^1$, so that since $\mathbb{S}^1$ cannot be embedded inside of $\mathbb{R}$, we have $M$ homeomorphic to $\mathbb{S}^1$ and consequently $U=M$ (since no proper open subset of $\mathbb{S}^1$ is homeomorphic to $\mathbb{S}^1$)? –  youler Feb 29 '12 at 20:50

As $S^1$ is compact, $U$ is compact too (as it is homeomorphic) and assuming the manifold is Hausdorff (as is usual) then indeed $U$ is indeed closed. And as noted in the comments, if $M$ is connected this would imply $M = U$.