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MATH31052 Topology

4 Hausdorff Spaces

  • 4.1 Definition .   Suppose that \((a_n)\) is a sequence of points in a topological space \(X\) and \(a\in X\). Then \(a_n\rightarrow a\) (as \(n\rightarrow \infty \)) when, for each open set \(U\) containing \(a\), there exists an integer \(N\) such that

    \[n\geq N \Rightarrow a_n\in U.\]

    In this case we say that the sequence \((a_n)\) converges to \(a\) or \(a\) is a limit of the sequence \((a_n)\).

  • 4.2 Example .  

    • (a) For \(X=\R ^n\) with the usual topology this is equivalent to the usual definition.

    • (b) For \(X\) a discrete space, \(a_n\rightarrow a\) means that there is an integer \(N\) such that \(n\geq N\Rightarrow a_n=a\) since \(\{a\}\) is an open set and \(a_n\in \{a\}\Rightarrow a_n=a\).

    • (c) For \(X\) an indiscrete space, every sequence converges to every point since the only open set containing \(a\in X\) is \(X\).

    • (d) For \(X\) with the Sierpinski topology \(\{\emptyset ,\{a\},X\}\), \(a_n\rightarrow a\) if and only if eventually \(a_n=a\) (as in (b)), but \(a_n\rightarrow b\) for all sequences (as in (c)).

  • 4.3 Definition .   The topological space \(X\) is Hausdorff (or \(T_2\)) if, for each distinct pair of points \(x\), \(y\in X\), there exist open sets \(U\) and \(V\) in \(X\) such that \(x\in U\), \(y\in V\) and \(U\cap V=\emptyset \).

  • 4.4 . Proposition   In a Hausdorff space a sequence can have at most one limit. So in this case we can refer to the limit of a convergent sequence and denote it by \(\lim _{n\to \infty }a_n\).

  • Proof. Exercise (a proof by contradiction).  □

  • 4.5 . Proposition   A subset \(X\subset \R ^n\) with the usual topology is Hausdorff.

  • Proof. Exercise.  □

  • 4.6 . Proposition   Points are closed in a Hausdorff space, i.e. given \(a\in X\), a Hausdorff space, the singleton subset \(\{a\}\) is a closed subset.

  • Proof. Exercise.  □

  • 4.7 Remark .   Hausdorff's original definition of a topological space (1914) was equivalent to our definition of a Hausdorff or \(T_2\) space. A topological space in which singleton subsets are closed is called a Fréchet space or a \(T_1\) space.

  • 4.8 . Proposition  

    • (a) A subspace of a Hausdorff space is Hausdorff.

    • (b) The disjoint union of two Hausdorff spaces is Hausdorff.

    • (c) The product of two Hausdorff spaces is Hausdorff.

  • Proof. (a) and (b) Exercises.

    (c) Suppose that \(X_1\) and \(X_2\) are Hausdorff spaces and that \((x_1,x_2)\), \((x_1',x_2')\in X_1\times X_2\) with \((x_1,x_2)\not =(x_1',x_2'\). Then \(x_1\not =x_1'\) or \(x_2\not =x_2'\).

    If \(x_1\not =x_1'\), since \(X_1\) is Hausdorff there are open subsets \(U\), \(V\subset X_1\) such that \(x_1\in U\), \(x_2\in V\) and \(U\cap V=\emptyset \). Then \(U\times X_2\) and \(V\times X_2\) are disjoint open subsets of \(X_1\times X_2\) as required.

    There is a similar argument if \(x_2\not = x_2'\).

    So in either case \((x_1,x_2)\) and \((x_1',x_2')\) lie in disjoint open subsets of \(X_1\times X_2\) as required to prove that \(X_1\times X_2\) is Hausdorff.  □

  • 4.9 Remark .   A quotient space of a Hausdorff space is not necessarily Hausdorff. For example, define an equivalence relation on the closed interval \([-1,1]\) with the usual topology (a Hausdorff space by Proposition 4.5) by \(t\sim \pm t\) for \(|t|<1\). Then the quotient space \([-1,1]/\!\!\sim \) is not Hausdorff.

  • Proof.
    Consider the points \(q(-1) = [-1] = \{-1\}\) and \(q(1) = [1]=\{1\}\in [-1,1]/\!\!\sim \), writing \(q\colon [-1,1] \to [-1,1]/\!\!\sim \) for the quotient map as usual.
    Suppose for contradiction that there are disjoint open subsets \(U\) and \(V\subset [-1,1]/\!\!\sim \) such that \([-1]\in U\) and \([1]\in V\).

    Since \([-1]\in U\), \(q^{-1}(U)\subset [-1,1]\) is an open set containing \(-1\) and so, by the definition of an open set in the usual topology, since \(-1 \in q^{-1}(U)\), there exists \(\e _1>0\) such that \(B_{\e _1}^{[-1,1]} = [-1,-1+\e _1)\subset q^{-1}(U)\). This means that \((1-\e _1,1)\subset q^{-1}(U)\) since each point is equivalent to a point of \([-1,-1+\e _1)\).

    Similarly, since \([1]\in V\), there exists \(\e _2>0\) such that \((1-\e _2,1]\subset q^{-1}(V)\). But then, for \(\e =\min (\e _1,\e _2)\), \(1 - \e /2 \in (1-\e ,1) = (1 - \e _1,1)\cap (1-\e _2,1] \subset q^{-1}(U)\cap q^{-1}(V) = q^{-1}(U\cap V)\) so that \([1-\e /2]\in U\cap V\) which are therefore not disjoint. So there do not exist disjoint open subsets of \([-1,1]/\!\!\sim \) containing \([-1]\) and \([1]\).  □

  • 4.10 Remark .   \(X=[-1,1]/\!\!\sim \) is covered by two open subsets, which are both homeomorphic to the unit interval \([0,1]\). Indeed, consider \(U^- =X\setminus \{[1]\}\). Then \(q^{-1}(U^-)=[-1,1)\) which is open in \([-1,1]\). On the other hand,

    \[\varphi \colon [0,1] \to U^-;\; t \mapsto [-t]\]

    gives a homeomorphism with inverse induced by

    \[f\colon [-1,1) \to [0,1];\; t \mapsto |t|.\]

    Indeed, one has \(f(t)=f(s)\) if and only if \(t \sim s\). Here, continuity follows by the continuity of \(t \mapsto -t\), \(q\), \(t \mapsto |t|\) and by the universal property of the quotient topology [Exercise]. Similarly \(U^+ =X\setminus \{[-1]\}\) is an open subset homeomorphic to \([0,1]\).

Exercises
  • 1.  Suppose that \(f\colon X\to Y\) is a continuous function of topological spaces. Prove that, if \(a_n\rightarrow a\) as \(n\rightarrow \infty \) in \(X\), then \(f(a_n)\rightarrow f(a)\) as \(n\rightarrow \infty \) in \(Y\).

  • 2.  Prove that in a Hausdorff space a sequence can have at most one limit (Proposition 4.4).

    [Hint: Give a proof by contradiction starting by supposing, for contradiction, that a sequence has two distinct limits.]

  • 3.  Prove that a subset \(X \subset \R ^n\) with the usual topology is Hausdorff (Proposition 4.5).

  • 4.  Prove that a set \(X\) with the discrete topology is Hausdorff. What about \(X\) with the indiscrete topology?

  • 5.  Prove that \(\R \) with each of the topologies of Problems 2, Question 2(b), Question 2(c), Question 2(e) and Question 2(f) is not Hausdorff.

  • 6.  Prove that, if \(X\) is a Hausdorff space and \(a\in X\), then the singleton subset \(\{a\}\) is a closed subset of \(X\) (Proposition 4.6).

    [Hint: Prove that \(X \setminus \{a\}\) is open by writing it as a union of open sets.]

  • 7.  Prove that a subspace of a Hausdorff space is Hausdorff (Proposition 4.8(a)).

  • 8.  Prove that the disjoint union of two Hausdorff spaces is Hausdorff (Proposition 4.8(b)).

  • 9.  Prove that if \(A\) is not a closed subset of \(X\) then \(X/A\) is not a Hausdorff space.

    [Hint: Use Proposition 4.6.]

  • 10.  Show that a topological space \(X\) is Hausdorff if and only if the diagonal \(\Delta _X = \{(x,y) \in X \times X \mid x=y\} \subset X \times X\) is closed (with respect to the product topology).