A subspace of a Hausdorff space is Hausdorff

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Statement

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

Proof

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