Munkres, Topology, Section 13 Exercises

Problem 1. Let $X$ be a topological space; let $A$ be a subset of $X$. Suppose that for each $x \in A$, there is an open set $U$ containing $x$ such that $U \subset A$. Show that $A$ is open in $X$.

Solution. For each $x \in A$, choose an open subset $U_x \subseteq A$ such that $x \in U_x$. I claim that $\bigcup\limits_{x \in A} U_x = A$. Indeed, we have $U_x \subseteq A$ for each $x \in A$ by definition, so $\bigcup\limits_{x \in A} U_x \subseteq A$. Furthermore, for each $x \in A$, one has $x \in U_x \subseteq \bigcup\limits_{x \in A} U_x$, which gives the opposite inclusion. As each $U_x$ is open and arbitrary unions of open sets are open, the union $\bigcup\limits_{x \in X} U_x$ is open in $X$ and hence $A$ is open in $X$.

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