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Suppose $(X, \mathcal{O})$ is a topological space and $A, U \subset X$ and $(A, \mathcal{O} _A), (U, \mathcal{O} _U)$ are the subspace topology of $(X, \mathcal{O})$. If $U$ is open in $(X, \mathcal{O})$ and $(U, \mathcal{O} _U), (A, \mathcal{O} _A)$ are homeomorphic, then $A$ is open in $(X, \mathcal{O})$.

Is this true or false? I think it is false but I can't make a counterexample. Please help me.

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    $\begingroup$ Sorry but what is $\mathcal{O}$? The topology on $X$? Then the assumption $A \in \mathcal{O}$ implies that $A$ is open. $\endgroup$
    – N.B.
    Oct 17 '18 at 14:13
  • $\begingroup$ Sorry, typo. I edited. $\endgroup$
    – moc79
    Oct 17 '18 at 14:27
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Without any other assumption the sentence seems to be false. Consider the topological space consisting of a line $l$ and one point $P$ external to the line. Then $P$ is open in the total space, however any other point in the line will not.

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  • $\begingroup$ I understand your explanation. Thank you very much! $\endgroup$
    – moc79
    Oct 17 '18 at 14:48
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There are many counterexamples. ALG gave one. But the interesting thing is that this does hold for some spaces, like $X= \mathbb{R}^n$ in the usual topology. This is called Invariance of domain (domain is an older name for "open set"; the openness is preserved as it were), so the intuition is understandable.

A simple minimalistic counterexample: $X = \{0,1\}$ with topology $\{\emptyset, \{0\}, X\}$ (Sierpinski space), then as subspaces $\{0\}$ and $\{1\}$ are homeomorphic (all one-point spaces are), the former is open, the latter is not.

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  • $\begingroup$ The theorem is interesting. Thank you. $\endgroup$
    – moc79
    Oct 17 '18 at 22:06

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