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Let $S$ be a closed subset of a topological space $X$. Show that the closed sets of $S$ in the induced topology are closed in $X$.

Any open set of $S$ with induced topology has the form $W\cap S$ where $W$ is open in $X$. So any closed set of $S$ with induced topology has the form $S-(W\cap S) = (S-W)\cup (S-S) = S- W$, so how is this closed in $X$?

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$S - W = S \cap W^c$, where $W^c$ is the complement of $W$ in $X$ and $W$ being open implies $W^c$ is closed in $X$. We know the intersection of two closed sets is closed. Therefore, it follows that $S - W$ is closed in $X$.

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