Exercise 13.1

Proof. For each $x\in A$ we can choose an open set $U_x$ containing $x$ such that $U_x\subset A$. We then claim that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{x\in A}{U}_x=A$. So first consider any $y\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{x\in A}{U}_x$ so that there is an $x\in A$ such that $y\in U_x$. Then clearly also $y\in A$ since $U_x\subset A$. Hence ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{x\in A}{U}_x\subset A$ since $y$ was arbitrary. Now consider $y\in A$ so that clearly $y\in U_y$. Then obviously $y\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{x\in A}{U}_x$ so that $A\subset {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{x\in A}{U}_x$ since $y$ was arbitrary. Thus we have shown that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{x\in A}{U}_x=A$, and since each $U_x$ is open, it follows from the definition of a topology that the union ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{x\in A}{U}_x=A$ is open as well. □