MATH31052 Topology
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 \([0,1]\).
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).