Let \(X\) be a topological space. The following conditions are equivalent:
1. \(X\) is Hausdorff.
2. For each \(x\in X\) the intersection of all closed nbhds of \(x\) is \(\set x\).
3. The diagonal \(\del :=\set {\ang {x,x}\in X}\) is closed in \(X\tm X\).
Proof. Assume 1. We show that 2. holds. Let \(x\in X\) and let \(C\) be the intersection of all closed nbhds of \(x\). Suppose, for a contradiction, that \(C\neq \set x\). Then there is \(y\in C\sem \set x\). Let \(U,V\) be disjoint open sets with \(x\in U\) and \(y\in V\). Then \(X\sem V\) is a closed nbhd of \(x\) so \(C\sub X\sem V\). Since \(y\in C\cap V\), we have a contradiction.
Assume 2. We show that 3. holds. Let
\[ \ang {x,y}\in \cur {X\tm X}\sem \del . \]
Then \(x\neq y\) so there is a closed nbhd \(N\) of \(x\) with \(y\nin N\). Let \(U\) be open with \(x\in U\sub N\) and \(V:=X\sem N\). Then \(U\tm V\) is open in \(X\tm X\) with \(\ang {x,y}\in U\tm V\) and
\[ \cur {U\tm V}\cap \del =\emp . \]
Thus \(\cur {X\tm X}\sem \del \) is open in \(X\tm X\), which implies that \(\del \) is closed in \(X\tm X\).
Assume 3. We show that 1. holds. Let \(x,y\in X\) be distinct. Then \(\ang {x,y}\in \cur {X\tm X}\sem \del \) so there are open \(U\) and \(V\) in \(X\) with
\[ \ang {x,y}\in U\tm V\sub \cur {X\tm X}\sem \del . \]
Thus \(U\) and \(V\) are disjoint with \(x\in U\) and \(y\in V\). □
Assume that \(Y\) is a Hausdorff space.
1. If \(f:X\to Y\) is continuous for some topological space \(X\), then the graph of \(f\) is closed in \(X\tm Y\).
2. If \(f,g:X\to Y\) are continuous for some topological space \(X\), then
\[ \set {x\in X:f\of x=g\of x} \]
is closed in \(X\).
Proof. Assume that \(f:X\to Y\) is continuous for some topological space \(Y\). The graph of \(f\) is the set
\[ G:=\set {\ang {x,f\of x}:x\in X}. \]
Let \(1_{Y}:Y\to Y\) be the identity function and \(\del :=\set {\ang {y,y}\in Y}\). Since the function
\[ f\tm 1_{Y}:X\tm Y\to Y\tm Y \]
is continuous, \(\del \) is closed and \(G=\cur {f\tm 1_{Y}}^{-1}\bof {\del }\), it follows that \(G\) is closed.
Now assume that \(f,g:X\to Y\) are continuous for some topological space \(X\) and
\[ G:=\set {x\in X:f\of x=g\of x}. \]
Let \(h:X\to X\tm Y\) be defined by \(h\of x:=\ang {f\of x,g\of x}\). Since \(h\) is continuous and \(G=h^{-1}\bof {\del }\), it follows that \(G\) is closed. □
Any subspace of a Hausdorff space is Hausdorff.
If \(X_{\ga }\) is a Hausdorff space for every \(\ga \in A\) and \(X:=\prod _{\ga \in A}X_{\ga }\), then \(X\) is Hausdorff.
Proof. Let \(\cur {x_{\ga }}_{\ga \in A}\) and \(\cur {y_{\ga }}_{\ga \in A}\) be distinct elements of \(X\). There is \(\gb \in A\) with \(x_{\gb }\neq y_{\gb }\). Since \(X_{\gb }\) is Hausdorff, there are disjoint open \(U_{\gb }\) and \(V_{\gb }\) in \(X_{\gb }\) with \(x_{\gb }\in U_{\gb }\) and \(y_{\gb }\in V_{\gb }\). Let \(U_{\ga }:=V_{\ga }:=X_{\ga }\) for every \(\ga \in A\sem \set {\gb }\) and
\[ U:=\prod _{\ga \in A}U_{\ga }\qand V:=\prod _{\ga \in A}V_{\ga }. \]
Then \(U\) and \(V\) are disjoint open sets in \(X\) with \(x\in U\) and \(y\in V\). □
A \(T_{0}\) space is a topological space such that for any distinct \(x,y\in X\) there is an open set \(U\) with \(x\in U\), but \(y\nin U\). A \(T_{0}\) space is a topological space such that for any distinct \(x,y\in X\) there is an open set \(U\) with \(\set {x,y}\cap U\) having exactly one element.
Any Hausdorff space is \(T_{1}\) and any \(T_{1}\) space is \(T_{0}\).
The Sierpiński space is \(T_{0}\) but not \(T_{1}\). The space of natural numbers with the cofinite topology is \(T_{1}\) but is not Hausdorff.
Let \(X\) be a topological space. The following conditions are equivalent:
1. \(X\) is \(T_{1}\);
2. \(\set x\) is closed for every \(x\in X\);
3. the intersection of all nbhds of \(x\) is equal \(\set x\) for every \(x\in X\).
Proof. Assume \(X\) is \(T_{1}\). Let \(x\in X\). For every \(y\in X\sem \set x\) there is open \(U_{y}\) with \(y\in U_{y}\) and \(x\nin U_{y}\). Then
\[ U:=\bigcup _{y\in X\sem \set x}U_{y}=X\sem \set x \]
is open, so \(\set x\) is closed.
Assume that \(\set x\) is closed for every \(x\in X\). To prove 3. suppose for a contradiction that there is \(y\in X\sem \set x\) such that \(y\) belongs to every nbhd of \(x\). Since \(\set y\) is closed \(X\sem \set y\) is open and contains \(x\), which is a contradiction.
Now assume that 3. holds. If \(x,y\in X\) are distinct, then there is a nbhd \(U\) of \(x\) with so \(X\) is \(T_{1}\). □