- Prove the following implications of separation axioms, and show that
they are strict:
Solution
Goal. Prove the implications \(T_4 \Rightarrow T_3 \Rightarrow T_2 \Rightarrow T_1 \Rightarrow T_0\) and show each is strict.
1. \(T_4 \Rightarrow T_3\).
1.1. \(T_4\) means normal and \(T_1\); \(T_3\) means regular and \(T_1\).
definitions.
1.2. A normal \(T_1\) space is regular.
given a closed set \(C\) and a point \(x \notin C\), since the space is \(T_1\) the singleton \(\theset{x}\) is closed, and normality separates the disjoint closed sets \(C\) and \(\theset{x}\) by disjoint open sets.
2. \(T_3 \Rightarrow T_2\).
2.1. A regular \(T_1\) space is Hausdorff.
given distinct \(x, y\), \(T_1\) makes \(\theset{y}\) closed and \(x \notin \theset{y}\); regularity separates \(x\) and \(\theset{y}\) by disjoint open sets.
3. \(T_2 \Rightarrow T_1\).
3.1. A Hausdorff space is \(T_1\).
given distinct \(x, y\), Hausdorff gives disjoint open \(U \ni x\), \(V \ni y\); then \(x \notin V\) and \(y \notin U\), so each singleton is closed.
4. \(T_1 \Rightarrow T_0\).
4.1. A \(T_1\) space is \(T_0\).
given distinct \(x, y\), \(T_1\) gives an open set containing \(x\) but not \(y\) (namely \(X \sm \theset{y}\)).
5. Each implication is strict.
5.1. \(T_0\) but not \(T_1\): the Sierpiński space \(\theset{0,1}\) with open sets \(\theset{\emptyset, \theset{1}, \theset{0,1}}\).
\(1\) has a neighborhood not containing \(0\), but \(\theset{0}\) is not closed.
5.2. \(T_1\) but not \(T_2\): the cofinite topology on an infinite set.
singletons are closed, but any two nonempty open sets intersect.
5.3. \(T_2\) but not \(T_3\): let \(K=\{1/n:n\ge1\}\) and give \(\mathbb R\) the \(K\)-topology generated by the basis\[ \{(a,b):a<b\}\;\cup\;\{(a,b)\setminus K:a<b\}. \]This space is Hausdorff but not regular.
The \(K\)-topology contains the usual topology, so it is Hausdorff. Also \(K\) is closed: for every \(x\notin K\), some basis set of the form \((a,b)\setminus K\) contains \(x\) and misses \(K\).
Suppose regularity separated the point \(0\) from the closed set \(K\) by disjoint open sets \(U\ni0\) and \(V\supseteq K\). Shrinking \(U\) to a basic neighborhood, we may choose \(\varepsilon>0\) with \[ (-\varepsilon,\varepsilon)\setminus K\subseteq U. \] Choose \(1/n<\varepsilon/2\). Since \(1/n\in V\), a basic neighborhood of \(1/n\) is contained in \(V\). Any basic neighborhood containing \(1/n\) must contain an ordinary interval \((c,d)\) about \(1/n\), because sets of the form \((a,b)\setminus K\) omit \(1/n\). This interval contains a point \[ y\in(-\varepsilon,\varepsilon)\setminus K, \] so \(y\in U\cap V\), a contradiction. Thus the space is not regular.
5.4. \(T_3\) but not \(T_4\): the Sorgenfrey plane \(S\times S\), where \(S\) is \(\mathbb R\) with basis \([a,b)\).
First, \(S\) has a basis of clopen sets: each \([a,b)\) is open, and its complement \[ (-\infty,a)\cup[b,\infty) \] is open in the Sorgenfrey topology. Hence \(S\) is regular and \(T_1\). Products of clopen basic sets form a clopen basis for \(S\times S\), so the Sorgenfrey plane is also regular and \(T_1\), hence \(T_3\).
It remains to prove that it is not normal. The countable set \(\mathbb Q\times\mathbb Q\) is dense, since every basic rectangle \([a,b)\times[c,d)\) contains a rational pair. On the other hand, the anti-diagonal \[ D=\{(x,-x):x\in\mathbb R\} \] is a closed discrete subspace of cardinality \(\mathfrak c\). It is discrete because \[ [x,x+\varepsilon)\times[-x,-x+\varepsilon) \] meets \(D\) only at \((x,-x)\). It is closed because, if \(a+b\ne0\), a sufficiently small basic rectangle about \((a,b)\) keeps the sign of the coordinate sum and therefore misses \(D\).
Now use the following Jones lemma. A separable normal \(T_1\) space cannot contain a closed discrete subspace of cardinality \(\mathfrak c\). Indeed, if \(D\) were such a subspace, then every subset \(E\subseteq D\) and its complement \(D\setminus E\) would be disjoint closed subsets of the ambient space. By Urysohn’s lemma there would be a continuous map \(f_E\) to \([0,1]\) taking value \(0\) on \(E\) and \(1\) on \(D\setminus E\). Distinct subsets give distinct continuous maps, so there would be at least \(2^{\mathfrak c}\) continuous maps to \([0,1]\). But on a separable space a continuous map to the Hausdorff space \([0,1]\) is determined by its values on a fixed countable dense subset, so there are at most \[ |[0,1]|^{\aleph_0}=\mathfrak c \] such maps, contradicting Cantor’s theorem \(2^{\mathfrak c}>\mathfrak c\).
Thus the separable Sorgenfrey plane is not normal, so it is \(T_3\) but not \(T_4\).
6. Q.E.D.
<1>1–<1>4 prove the implications; 5 shows each is strict.