fix: missed closing for math

html/topology/interior_and_closure.html

@@ -120,7 +120,7 @@
         Suppose that \( A \) is a subset of a topological space, then we have \[ \operatorname{Int}\left( A \right) \subseteq A \subseteq \bar{A} \]
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-        <p> Let A be a subset of the topological space \(X\). Let \(x\) be an arbitrary point in \(int(A)\). By definition, there exist an open ball \(B_\epsilon(p)) such that \(B_\epsilon(p)\subseteq A\). Since the open ball is entirely contained in \(A\), \(x\) must be an element of A. Since (\bar{A}=A^*\cup A\) by definition, since \(x\) is in A, \(x\in \bar{A}\). Since \(x\), which is an element of the interior of A, was chosen arbitrarily, the interior of A is a subset of A and is a subset of the closure.
+        <p> Let A be a subset of the topological space \(X\). Let \(x\) be an arbitrary point in \(int(A)\). By definition, there exist an open ball \(B_\epsilon(p)\) such that \(B_\epsilon(p)\subseteq A\). Since the open ball is entirely contained in \(A\), \(x\) must be an element of A. Since (\bar{A}=A^*\cup A\) by definition, since \(x\) is in A, \(x\in \bar{A}\). Since \(x\), which is an element of the interior of A, was chosen arbitrarily, the interior of A is a subset of A and is a subset of the closure.
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