# $A$ and $U$ are homeomorphic; $U$ is open in $X$ $\Rightarrow$ $A$ is open in $X$?

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.

• Sorry but what is $\mathcal{O}$? The topology on $X$? Then the assumption $A \in \mathcal{O}$ implies that $A$ is open.
– N.B.
Oct 17 '18 at 14:13
• Sorry, typo. I edited. Oct 17 '18 at 14:27

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.
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.