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 $M$?
An open set in a space is homeomorphic to $S^1$. Is the set close?
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general-topology
differential-geometry
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0Couldn'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 $\mathbb{S}^1$)? – 2012-02-29
1 Answers
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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 $M = U$.