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Let $\tau$ be a topology on a set $X$. Suppose a subset $A\subseteq X$ yields a subspace topology $\tau_{|A} := \{A \cap U \hskip4pt | \hskip4pt U \in \tau \}$ with the property that every open set of $\tau_{|A}$ is also an open set of $\tau$, meaning $\tau_{|A} \subseteq \tau$.

What's a minimal set of conditions that $A$ must satisfy in order for this to happen?

(For example, I think this is true if $A$ is open in $\tau$, because then every $V \in \tau_{|A}$ is an intersection of 2 open sets of $\tau$, so that $V$ is also open in $\tau$.)

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  • $\begingroup$ Hint: $A$ is open in $A$. $\endgroup$ – Batominovski Nov 29 '17 at 11:02
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For that it is necessary that $A\in\tau$ since $A\in\tau|_A$.

This condition is also sufficient since every set in $\tau|_A$ takes the form $A\cap U$ with $U\in\tau$.

Then $A\in\tau$ implies that also $A\cap U\in\tau$.

So we have:$$\tau|_A\subseteq\tau\iff A\in\tau$$

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