# A subspace of a Hausdorff space is Hausdorff

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## Statement

Suppose [ilmath](X,\mathcal{ J })[/ilmath] is a Hausdorff topological space; then for any [ilmath]A\in\mathcal{P}(X)[/ilmath] (so [ilmath]A[/ilmath] is an arbitrary subset of [ilmath]X[/ilmath]) considered as a topological subspace, [ilmath](A,\mathcal{J}_A)[/ilmath], of [ilmath](X,\mathcal{ J })[/ilmath] is also Hausdorff.

## Proof

• Let [ilmath]a,b\in A[/ilmath] be given. We wish to show there are neighbourhoods (with respect to the topological space [ilmath](A,\mathcal{J}_A)[/ilmath]) to [ilmath]a[/ilmath] and [ilmath]b[/ilmath] (which we shall call [ilmath]N_a[/ilmath] and [ilmath]N_b[/ilmath] respectively) such that [ilmath]N_a\cap N_b=\emptyset[/ilmath]
• As [ilmath](X,\mathcal{ J })[/ilmath] is Hausdorff, there exist neighbourhoods [ilmath]N_a'[/ilmath] and [ilmath]N_b'[/ilmath] neighbourhood to [ilmath]a[/ilmath] and [ilmath]b[/ilmath] respectively (with respect to the topological space [ilmath](X,\mathcal{ J })[/ilmath] in this case) such that [ilmath]N_a'\cap N_b'=\emptyset[/ilmath]
• Thus [ilmath]\exists U_a',U_b'\in\mathcal{J}[a\in U_a'\wedge b\in U_b'\wedge U_a'\cap U_b'=\emptyset][/ilmath] (by definition of neighbourhood, using [ilmath]U_a'\subseteq N_a'[/ilmath], [ilmath]U_b'\subseteq N_b'[/ilmath] and that [ilmath]N_a'\cap A_b'=\emptyset[/ilmath])
• By definition of the subspace topology, we see and define [ilmath]U_a:=U_a'\cap A\in\mathcal{J}_A[/ilmath] and [ilmath]U_b:=U_b'\cap A\in\mathcal{J}_A[/ilmath]
• Note that [ilmath]a\in U_a'[/ilmath], [ilmath]b\in U_b'[/ilmath] and [ilmath]a,b\in A[/ilmath], so [ilmath]a\in U_a[/ilmath] and [ilmath]b\in U_b[/ilmath]
• Furthermore notice [ilmath]U_a\cap U_b\subseteq U_a'\cap U_b'\subseteq N_a'\cap N_b'=\emptyset[/ilmath]
• So [ilmath]U_a\cap U_b=\emptyset[/ilmath]
• Since [ilmath]U_a[/ilmath] contains an open set (namely [ilmath]U_a[/ilmath]) containing [ilmath]a[/ilmath] it is a neighbourhood to [ilmath]a[/ilmath]
• Same for [ilmath]b[/ilmath] and [ilmath]U_b[/ilmath]
• Thus we have shown there exist disjoint neighbourhoods of [ilmath]a[/ilmath] and [ilmath]b[/ilmath] in the subspace.
• Since our choice of [ilmath]a[/ilmath] and [ilmath]b[/ilmath] was arbitrary we have shown this for all [ilmath]a,b\in A[/ilmath]