Merge pull request #57 from Clement525/main

Edited proofs for interior and closure

html/topology/interior_and_closure.html

@@ -109,7 +109,7 @@
         Suppose that \( \operatorname{Int}\left( A \right) \) is the interior of \( A \) in a topological space, then given an open set \( U \subseteq A \) we have \( U \subseteq \operatorname{Int}\left( A \right) \)
     </div>
     <div class="proof">
-        TODO
+        <p> By definition, the interior of A is the union of all open sets contained in A. Let \(U_i\) be an arbitrary open subset of A. Since the interior is the largest open subset of the subset A, \(U_i\subseteq \bigcup_{i}^{}U_i \subseteq int(A)\) must be true. So, \(U_i \subseteq int(A)\). Since \(U_i\) was chosen arbitrary, any open subset is a subset of the interior.
     </div>
 </div>
 <div class="proposition" id="proposition-interior-is-smaller-closure-is-bigger">
@@ -120,7 +120,8 @@
         Suppose that \( A \) is a subset of a topological space, then we have \[ \operatorname{Int}\left( A \right) \subseteq A \subseteq \bar{A} \]
     </div>
     <div class="proof">
-        TODO
+        <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>
     </div>
 </div>
 <div class="proposition" id="proposition-open-sets-equal-their-interior">