K. Jänich, Topology. Chapter I: Fundamental Concepts. 7: The Hausdorff Separation Axiom

Definition 1 A topological space \(X\) is a Hausdorff space or \(T_2\) space iff, for all \(x,y \in X\) with \(x \neq y\), there exist disjoint open sets \(U\) and \(V\) with \(x \in U\) and \(y \in V\).

Definition 2 Every metric space is Hausdorff.

Proof Let \(X\) be a metric space. Let \(x,y \in X\) with \(x \neq y\). Let \(\epsilon = d(x,y) > 0\). Let \(U = B_{\epsilon / 2}(x)\) and \(V = B_{\epsilon / 2}(y)\). Then \(U\) and \(V\) are open sets with \(x \in U\) and \(y \in V\). We prove \(U\) and \(V\) are disjoint.

If \(z \in U \cap V\) then we have \[ \begin{align} \epsilon & = d(x,y) \\ & \leq d(x,z) + d(z,y) & (\text{Triangle Inequality}) \\ & < \epsilon / 2 + \epsilon / 2 \\ & = \epsilon \end{align} \] which is a contradiction. \(\Box\)

Proposition 3 Every discrete space is Hausdorff.

Proof Given distinct points \(x\) and \(y\), take \(U = \{x\}\) and \(V = \{y\}\). \(\Box\)

Proposition 4 If \(X\) has more than two elements, then the indiscrete topology on \(X\) is not Hausdorff.

Proof There is only one nonempty open set. \(\Box\)

Definition 5 Let \(X\) be a topological space. Let \((x_n)\) be a sequence of points in \(X\) and \(l \in X\). Then we say \((x_n)\) converges to \(l\), or \(l\) is the limit of \((x_n)\), and write \(x_n \rightarrow l\) as \(n \rightarrow \infty\), iff for every neighbourhood \(N\) of \(l\), there exists \(N\) such that \(\forall n \geq N. x_n \in N\).

Proposition 6 In a Hausdorff space, a sequence has at most one limit.

Proof Let \(X\) be a Hausdorff space. Assume for a contradiction \((x_n)\) converges to both \(l\) and \(m\) and \(l \neq m\). Pick disjoint open sets \(U\) and \(V\) with \(l \in U\) and \(m \in V\). Pick natural numbers \(M\) and \(N\) such that \(\forall n \geq M. x_m \in U\) and \(\forall n \geq N. x_n \in V\). Assume w.l.o.g. \(M \leq N\). Then \(x_N \in U \cap V\), contradicting the fact that \(U\) and \(V\) are disjoint. \(\Box\)

Proposition 7 In an indiscrete space, every sequence converges to every point.

Proof Let \(X\) be an indiscrete space. Let \((x_n)\) be any sequence and \(l\) any point. If \(N\) is a neighbourhood of \(l\) then \(N = X\) and so \(x_n \in N\) for all \(n\). Thus \(x_n \rightarrow l\) as \(n \rightarrow \infty\). \(\Box\)

Proposition 8 Every subspace of a Hausdorff space is Hausdorff.

Proof Let \(X\) be a Hausdorff space and \(Y\) a subspace of \(X\). Let \(x,y \in X\) with \(x \neq y\). Pick disjoint open sets \(U\) and \(V\) in \(X\) with \(x \in U\) and \(y \in V\). Then \(U \cap Y\) and \(V \cap Y\) are disjoint open sets in \(Y\) with \(x \in U \cap Y\) and \(y \in V \cap Y\). \(\Box\)

Proposition 9 Let \(X\) and \(Y\) be nonempty topological spaces. Then the following are equivalent.

  1. \(X\) and \(Y\) are Hausdorff.
  2. \(X + Y\) is Hausdorff.
  3. \(X \times Y\) is Hausdorff.

Proof

\(1 \Rightarrow 2\) Assume \(X\) and \(Y\) are Hausdorff. Let \(x,y \in X + Y\) with \(x \neq y\). We have the following cases:

If \(x,y \in X\) then pick disjoint open sets \(U\) and \(V\) in \(X\) with \(x \in U\) and \(y \in V\). Then \(U + \emptyset\) and \(V + \emptyset\) are disjoint open sets in \(X + Y\) with \(x \in U + \emptyset\) and \(y \in V + \emptyset\). The case \(x,y \in Y\) is similar.

If \(x \in X\) and \(y \in Y\), then \(X + \emptyset\) and \(\emptyset + Y\) are disjoint open sets with \(x \in X + \emptyset\) and \(y \in \emptyset + Y\).

\(2 \Rightarrow 1\) Assume \(X + Y\) is Hausdorff. We prove \(X\) is Hausdorff; the proof for \(Y\) is similar. Let \(x,y \in X\) with \(x \neq y\). Pick disjoint open sets \(U_1 + V_1\) and \(U_2 + V_2\) in \(X + Y\) with \(x \in U_1 + V_1\) and \(y \in U_2 + V_2\). Then \(U_1\) and \(U_2\) are disjoint open sets in \(X\) with \(x \in U_1\) and \(y \in U_2\).

\(1 \Rightarrow 3\) Assume \(X\) and \(Y\) are Hausdorff. Let \((x_1,y_1),(x_2,y_2) \in X \times Y\) with \((x_1,y_1) \neq (x_2,y_2)\). Assume w.l.o.g. \(x_1 \neq x_2\). Pick disjoint open sets \(U\) and \(V\) in \(X\) with \(x_1 \in U\) and \(x_2 \in V\). Then \(U \times Y\) and \(V \times Y\) are disjoint open sets in \(X \times Y\) with \((x_1,y_1) \in U \times Y\) and \((x_2,y_2) \in V \times Y\).

\(3 \Rightarrow 1\) Assume \(X \times Y\) is Hausdorff. We prove \(X\) is Hausdorff; the proof for \(Y\) is similar. Let \(x,x' \in X\) with \(x \neq x'\). Pick \(y \in Y\). Pick disjoint open sets \(W\),\(W'\) in \(X \times Y\) with \((x,y) \in W\) and \((x',y) in W'\). Pick open sets \(U\),\(U'\) in \(X\) and \(V\),\(V'\) in \(Y\) with \((x,y) \in U \times V \subseteq W\) and \((x',y) \in U' \times V' \subseteq W'\). We have \(U\) and \(U'\) disjoint (because in \(z \in U \cap U'\) then \((z,y) \in W \cap W'\)). So \(U\) and \(U'\) are disjoint open sets in \(X\) with \(x \in U\) and \(x' \in U'\). \(\Box\)

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