View source for A subspace of a Hausdorff space is Hausdorff

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==Statement==
Suppose {{Top.|X|J}} is a ''[[Hausdorff]]'' [[topological space]]; then for any {{M|A\in\mathcal{P}(X)}} (so {{M|A}} is an arbitrary [[subset of]] {{M|X}}) considered as a [[topological subspace]], {{M|(A,\mathcal{J}_A)}}, of {{Top.|X|J}} is also Hausdorff{{rITTMJML}}.
==Proof==
* Let {{M|a,b\in A}} be given. We wish to show there are {{plural|neighbourhood|s}} (with respect to the [[topological space]] {{M|(A,\mathcal{J}_A)}}) to {{M|a}} and {{M|b}} (which we shall call {{M|N_a}} and {{M|N_b}} respectively) such that {{M|1=N_a\cap N_b=\emptyset}}
** As {{Top.|X|J}} is [[Hausdorff]], there exist {{plural|neighbourhood|s}} {{M|N_a'}} and {{M|N_b'}} neighbourhood to {{M|a}} and {{M|b}} respectively (with respect to the topological space {{Top.|X|J}} in this case) such that {{M|1=N_a'\cap N_b'=\emptyset}}
*** Thus {{M|1=\exists U_a',U_b'\in\mathcal{J}[a\in U_a'\wedge b\in U_b'\wedge U_a'\cap U_b'=\emptyset]}} (by definition of neighbourhood, using {{M|U_a'\subseteq N_a'}}, {{M|U_b'\subseteq N_b'}} and that {{M|1=N_a'\cap A_b'=\emptyset}})
**** By definition of the [[subspace topology]], we see and define {{M|1=U_a:=U_a'\cap A\in\mathcal{J}_A}} and {{M|1=U_b:=U_b'\cap A\in\mathcal{J}_A}}
***** Note that {{M|a\in U_a'}}, {{M|b\in U_b'}} and {{M|a,b\in A}}, so {{M|a\in U_a}} and {{M|b\in U_b}}
***** Furthermore notice {{M|1=U_a\cap U_b\subseteq U_a'\cap U_b'\subseteq N_a'\cap N_b'=\emptyset}}
****** So {{M|1=U_a\cap U_b=\emptyset}}
***** Since {{M|U_a}} contains an [[open set]] (namely {{M|U_a}}) containing {{M|a}} it is a neighbourhood to {{M|a}}
***** Same for {{M|b}} and {{M|U_b}}
***** Thus we have shown there exist disjoint neighbourhoods of {{M|a}} and {{M|b}} in the subspace. 
* Since our choice of {{M|a}} and {{M|b}} was arbitrary we have shown this for all {{M|a,b\in A}}
==References==
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