Conversely, suppose that $A = \bigcup_a \in A B(a, r_a)$ for some $r_a > 0$. Let $x \in A$. Then, there exists $a \in A$ such that $x \in B(a, r_a)$. This implies that there exists an open ball around $x$ that is contained in $A$, and hence $A$ is open.
As the professor worked through the solution, Emma's eyes widened with understanding. "Oh, I see! I was overcomplicating things."
Conversely, suppose that $A = \bigcup_a \in A B(a, r_a)$ for some $r_a > 0$. Let $x \in A$. Then, there exists $a \in A$ such that $x \in B(a, r_a)$. This implies that there exists an open ball around $x$ that is contained in $A$, and hence $A$ is open.
As the professor worked through the solution, Emma's eyes widened with understanding. "Oh, I see! I was overcomplicating things." Introduction To Topology Mendelson Solutions
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