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Let $X$ be a topological space and $A\subseteq X$ a subspace.

Let's say that $A$ has the property $(P)$ if for any nonempty open subset $U$ of $A$, there is a nonempty open subset $V$ of $X$ such that $V\subseteq U$.

Is this property actually known under a proper name ?

To simplify, let's take $X=\mathbb{R}^n$. Is there a way to characterize all the subspaces of $\mathbb{R}^n$ with the property $(P)$ ? It's clear that for instance a finite subset doesn't have the property $(P)$, while all open sets have. Could it be that $(P)$ is equivalent, or related, to having no isolated point ?

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$A$ has $(P)$ if and only if $\operatorname{int}_XA$ is dense in $A$.

Suppose first that $\operatorname{int}_XA$ is dense in $A$, and let $U$ be a non-empty relatively open subset of $A$. Then there is an open $V$ in $X$ such that $U=V\cap A$. Clearly $V\cap\operatorname{int}_XA\subseteq U$ is a non-empty open set in $X$.

Conversely, if $\operatorname{int}_XA$ is not dense in $A$, then $A\setminus\operatorname{cl}_X\operatorname{int}_XA$ is a non-empty relatively open subset of $A$ that contains no non-empty open subset of $X$.

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