Let \(A\), \(B\), and \(A_\alpha\) denote subsets of a space \(X\). Determine whether the following equations hold; if an equality fails, determine whether one of the inclusions \(\supset\) or \(\subset\) holds.
\(\overline{A \cap B} = \overline{A} \cap \overline{B}\).
\(\overline{\bigcap A_\alpha} = \bigcap \overline{A}_\alpha\).
\(\overline{A - B} = \overline{A} - \overline{B}\).
Solution
1. (a) \(\overline{A\cap B}\subseteq\overline A\cap\overline B\), and equality can fail.
Since \(A\cap B\subseteq A\) and \(A\cap B\subseteq B\), monotonicity of closure gives \(\overline{A\cap B}\subseteq\overline A\) and \(\overline{A\cap B}\subseteq\overline B\). In \(\mathbb R\) let \(A=(0,1)\) and \(B=(1,2)\). Then \(\overline{A\cap B}=\varnothing\), while \(\overline A\cap\overline B=[0,1]\cap[1,2]=\{1\}\).
2. (b) \(\overline{\bigcap A_\alpha}\subseteq\bigcap\overline{A_\alpha}\), and equality can fail.
For each index \(\beta\), \(\bigcap_\alpha A_\alpha\subseteq A_\beta\), so \(\overline{\bigcap_\alpha A_\alpha}\subseteq\overline{A_\beta}\); intersect over \(\beta\). In \(\mathbb R\) let \(A_n=(0,\frac1n)\) for \(n\in\mathbb Z_+\). Then \(\bigcap_nA_n=\varnothing\), while \(\bigcap_n\overline{A_n}=\bigcap_n[0,\frac1n]=\{0\}\).
3. (c) \(\overline A-\overline B\subseteq\overline{A-B}\), and equality can fail.
Let \(x\in\overline A-\overline B\). Since \(x\notin\overline B\), some open \(U\ni x\) satisfies \(U\cap B=\varnothing\). Let \(V\) be an open neighborhood of \(x\). Since \(x\in\overline A\), the neighborhood \(U\cap V\) contains a point \(y\in A\); then \(y\notin B\) because \(y\in U\), so \(y\in V\cap(A-B)\). Hence every neighborhood of \(x\) meets \(A-B\), and \(x\in\overline{A-B}\).
In \(\mathbb R\) let \(A=[0,1]\) and \(B=(0,1)\). Then \(\overline{A-B}=\{0,1\}\), while \(\overline A-\overline B=[0,1]-[0,1]=\varnothing\).
4. Q.E.D.
Steps 1, 2, and 3 decide (a), (b), and (c): in each, the inclusion stated in the step holds and the reverse inclusion fails.