Let $A \subseteq X$. We need to show that $\overlineA$ is the smallest closed set containing $A$. First, we show that $\overlineA$ is closed. Let $x \in X \setminus \overlineA$. Then, there exists an open neighborhood $U$ of $x$ such that $U \cap A = \emptyset$. This implies that $U \subseteq X \setminus \overlineA$, and hence $X \setminus \overlineA$ is open. Therefore, $\overlineA$ is closed.
Prove that ( (0,1) ) in ℝ is connected. Introduction To Topology Mendelson Solutions
: Covers logic, set operations, and functions. Let $A \subseteq X$
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