A topological space \(X\) has the \(T_{2\frac12}\) property if for each pair \(a,b\) of distinct points of \(X\), there are open sets \(U,V\) of \(X\) such that \(a\in U\), \(b\in V\), and the closures satisfy \(\operatorname{Cl}_X(U)\cap\operatorname{Cl}_X(V)=\varnothing\). Show that if \(A\) is a subspace of a \(T_{2\frac12}\) space, then \(A\) is also \(T_{2\frac12}\).
Solution
Let \(a,b\in A\) be distinct.
1. There are open sets \(U,V\subseteq X\) such that\[ a\in U, \qquad b\in V, \qquad \operatorname{Cl}_X(U)\cap\operatorname{Cl}_X(V)=\varnothing. \]
The points \(a\) and \(b\) are also distinct points of \(X\). Apply the \(T_{2\frac12}\) property of \(X\).
2. For every subset \(S\subseteq A\),\[ \operatorname{Cl}_A(S)=A\cap\operatorname{Cl}_X(S). \]
Let \(x\in A\). By the definition of closure in the subspace topology, \[ x\in\operatorname{Cl}_A(S) \] if and only if every open neighborhood of \(x\) in \(A\) meets \(S\). Every such neighborhood has the form \(A\cap W\) for some open neighborhood \(W\) of \(x\) in \(X\). Since \(S\subseteq A\), \[ (A\cap W)\cap S=W\cap S. \] Thus every open neighborhood of \(x\) in \(A\) meets \(S\) if and only if every open neighborhood of \(x\) in \(X\) meets \(S\), which is equivalent to \[ x\in\operatorname{Cl}_X(S). \] Because \(x\in A\) throughout, this proves the formula.
3. Put\[ U_A=U\cap A, \qquad V_A=V\cap A. \]Then \(U_A,V_A\) are open in \(A\), contain \(a,b\) respectively, and satisfy\[ \operatorname{Cl}_A(U_A) \subseteq A\cap\operatorname{Cl}_X(U), \qquad \operatorname{Cl}_A(V_A) \subseteq A\cap\operatorname{Cl}_X(V). \]
1. By 2,\[ \operatorname{Cl}_A(U_A)=A\cap\operatorname{Cl}_X(U_A). \]Since \(U_A\subseteq U\), monotonicity of closure gives\[ \operatorname{Cl}_X(U_A)\subseteq\operatorname{Cl}_X(U), \]which proves the first inclusion.The second is identical.
4. The closures of \(U_A\) and \(V_A\) in \(A\) are disjoint.
Using 3 and then 1, \[ \begin{aligned} \operatorname{Cl}_A(U_A)\cap\operatorname{Cl}_A(V_A) &\subseteq A\cap\operatorname{Cl}_X(U)\cap\operatorname{Cl}_X(V)\\ &=\varnothing. \end{aligned} \]
5. The subspace \(A\) is \(T_{2\frac12}\).
For arbitrary distinct \(a,b\in A\), <1>3–<1>4 provide open neighborhoods in \(A\) whose closures in \(A\) are disjoint. This is exactly the defining \(T_{2\frac12}\) property.