Hereditary and Productive Behaviour of the Separation Axioms

Let $x,y\in S$ be distinct. If $X$ is $T_0$, choose an open $U\subseteq X$ containing exactly one of $x,y$; then $S\cap U$ does the same in $S$.

If $X$ is $T_1$, apply the preceding trace argument separately to the two open sets supplied by the $T_1$ condition, so each of $x,y$ has an open neighbourhood in $S$ missing the other.

If $X$ is Hausdorff, choose disjoint open $U,V\subseteq X$ containing $x,y$ respectively; $S\cap U$ and $S\cap V$ are disjoint open neighbourhoods in $S$.

Since $S$ was arbitrary, each of the three properties is hereditary.

Write $W=U\cap S$ for an open $U\subseteq X$ containing $x$.

Choose open $V\subseteq X$ with $x\in V\subseteq\overline V\subseteq U$.

The trace $V\cap S$ is open in $S$, contains $x$, and has $\operatorname{cl}_S(V\cap S)\subseteq\overline V\cap S\subseteq U\cap S=W$.

The closed-neighbourhood characterization now makes $S$ regular; as $S$ was arbitrary, regularity is hereditary.

Choose closed $C\subseteq X$ with $F=C\cap S$; since $x\in S\setminus F$, one has $x\notin C$.

Choose $f:X\to[0,1]$ continuous with $f(x)=1$ and $f[C]=\{0\}$.

The restriction $f|_S:S\to[0,1]$ is continuous, takes $x$ to $1$, and vanishes on $F\subseteq C$.

Thus $S$ is completely regular, and the arbitrariness of $S$ proves heredity.

The first three assertions are [L1], and regularity and complete regularity are [L2].

A subspace of a $T_3$ space is regular by [L2] and $T_1$ by [L1], hence is $T_3$.

A subspace of a Tychonoff space is completely regular by [L2] and $T_1$ by [L1], hence is Tychonoff.

These cover every property named in the statement.

If $I=\varnothing$, the product has one point and all three conditions hold vacuously.

Otherwise choose $i\in I$ with $x_i\ne y_i$. For a $T_0$ factor, the inverse image under $\pi_i$ of an open set distinguishing $x_i,y_i$ distinguishes $x,y$.

For a $T_1$ factor, pull back the two open sets separating $x_i$ from $y_i$ and $y_i$ from $x_i$.

For a Hausdorff factor, pull back disjoint open neighbourhoods of $x_i,y_i$; their inverse images remain disjoint.

Thus the product has the relevant property in every case.

Choose a basic neighbourhood $B=\bigcap_{i\in J}\pi_i^{-1}[U_i]$ of $x$ inside $W$, where $J$ is finite.

Since $J$ is finite, [L2] makes these factorwise choices simultaneously: take open $V_i$ with $x_i\in V_i\subseteq\overline{V_i}\subseteq U_i$ for every $i\in J$.

Put $V=\bigcap_{i\in J}\pi_i^{-1}[V_i]$. It is open, contains $x$, and its closure lies in $\bigcap_{i\in J}\pi_i^{-1}[\overline{V_i}]\subseteq B\subseteq W$.

The closed-neighbourhood characterization proves $P$ regular.

The map $m:[0,1]^2\to[0,1]$, $m(s,t)=\min(s,t)$ is continuous: for an interval $(a,b)$, $m^{-1}((a,b))=\{s>a,t>a\}\cap(\{s<b\}\cup\{t<b\})$, which is open in the product.

The empty minimum is constant one, hence continuous, and the one-term minimum is $f_0$.

Assume the minimum $g_n=\min_{i<n}f_i$ is continuous. Then $(g_n,f_n):X\to[0,1]^2$ is continuous and $g_{n+1}=m\circ(g_n,f_n)$ is continuous.

Induction gives the claim for every finite family.

Choose a finite-support basic neighbourhood $B=\bigcap_{i\in J}\pi_i^{-1}[U_i]$ of $x$ contained in $P\setminus C$.

The finite-choice result [L2] selects a map $h_i$ as in [F2] for every $i\in J$; put $h=\min_{i\in J}(h_i\circ\pi_i)$.

The map $h$ is continuous and $h(x)=1$. If $y\in C$, then $y\notin B$, so $y_i\notin U_i$ for some $i\in J$ and $h(y)=0$.

Thus $h$ separates $x$ from $C$ in the defining sense of complete regularity.

The assertions for $T_0$, $T_1$, $T_2$, regularity, and complete regularity are [L1].

A product of $T_3$ spaces is regular and $T_1$ by [L1], hence is $T_3$.

A product of Tychonoff spaces is completely regular and $T_1$ by [L1], hence is Tychonoff.

Therefore every property in the statement is productive.

Write $A=S\cap C$ and $B=S\cap D$ for closed $C,D\subseteq X$. Then $A$ and $B$ are closed in $X$, because $S$ is closed.

The sets $A,B$ are disjoint closed subsets of the normal space $X$, so choose disjoint ambient open sets $U,V$ with $A\subseteq U$ and $B\subseteq V$.

Their traces $U\cap S$ and $V\cap S$ are disjoint open subsets of $S$ containing $A,B$, so $S$ is normal.

Suppose $X$ is completely normal, let $S\subseteq X$, and let $A,B$ be disjoint closed subsets of $S$. By [F2] they are separated in $X$, so ambient disjoint open sets containing them trace to disjoint open sets of $S$. Thus $S$ is normal.

Conversely suppose every subspace of $X$ is normal, and let $A,B\subseteq X$ be separated. Put $Y=X\setminus((\overline A\setminus A)\cup(\overline B\setminus B))$; separation ensures that $A,B\subseteq Y$, and they are disjoint closed subsets of $Y$.

Normality of $Y$ gives disjoint open $U,V\subseteq Y$ containing $A,B$. Write $U=G\cap Y$ and $V=H\cap Y$ with $G,H$ open in $X$; then $G\cap H$ is contained in $(\overline A\setminus A)\cup(\overline B\setminus B)$.

The open sets $G\setminus\overline B$ and $H\setminus\overline A$ contain $A$ and $B$ respectively, and are disjoint: a point of their intersection would lie in $G\cap H$ but in neither of the two displayed closure differences.

Hence every separated pair in $X$ has disjoint open neighbourhoods, so $X$ is completely normal.

If $S\subseteq X$ and $X$ is completely normal, every subspace of $S$ is also a subspace of $X$ and hence is normal by [L1]. Therefore [L1] applied to $S$ makes $S$ completely normal.

If $X$ is $T_5$, step 1.1 gives complete normality of $S$ and [L2] gives $T_1$ of $S$, so $S$ is $T_5$.

By [L1], $X$ is completely normal. Every subspace of $S$ is then a subspace of $X$, hence normal by [L2]; applying [L2] to $S$ shows that $S$ is completely normal. The $T_1$ clause of [L2] also shows that $S$ is $T_1$ when $X$ is $T_6$.

Let $F$ be closed in $S$. Write $F=C\cap S$ with $C$ closed in $X$; perfect normality writes $C=\bigcap_{n\in\mathbb N}U_n$ with every $U_n$ open in $X$.

Then $F=\bigcap_{n\in\mathbb N}(U_n\cap S)$, a $G_\delta$ of $S$. Thus $S$ is perfectly normal, and with its inherited $T_1$ property it is $T_6$ when $X$ is $T_6$.

Let $\mathcal V$ be the family of all open $V\subseteq X$ satisfying $\overline V\subseteq X\setminus B$. For every $a\in A$, [L1] supplies a member of $\mathcal V$ containing $a$, so $\mathcal V\cup\{X\setminus A\}$ is an open cover of $X$.

By Lindelöfness, an at most countable subfamily $\mathcal W\subseteq\mathcal V$ covers $A$.

Put $U=\bigcup\mathcal W$ and $W=X\setminus\bigcup_{V\in\mathcal W}\overline V$. Then $U,W$ are open, $A\subseteq U$, $B\subseteq W$, and $U\cap W=\varnothing$.

Thus $X$ is normal.

Each $[a,b)$ is clopen: its complement is $(-\infty,a)\cup[b,\infty)$, a union of lower-limit basic intervals. Hence the line is regular by [L1].

Let $\mathcal U$ be an open cover and let $O$ be the union of the usual intervals $(a,b)$ for which $[a,b)$ lies in a member of $\mathcal U$. The rational-endpoint intervals $[p,q)$ contained in members of $\mathcal U$ cover $O$; they form an at most countable family by [L2].

Put $D=\mathbb R\setminus O$. For $x\in D$, some $[x,b_x)$ lies in a member of $\mathcal U$, and $[x,b_x)\cap D=\{x\}$. The first rational $q_x\in(x,b_x)$ in a fixed enumeration exists, and the intervals $(x,q_x)$ are pairwise disjoint; their first rationals $r_x$ are therefore distinct. Thus $x\mapsto r_x$ injects $D$ into $\mathbb Q$, so $D$ is at most countable.

By [A1], choose one member of $\mathcal U$ covering each point of the at most countable set $D$. Together with one covering member for each rational-endpoint interval used in step 1.2, these form an at most countable subcover of $\mathcal U$.

Therefore the lower-limit line is Lindelöf under countable choice.

Under the stated hypothesis, [L1] supplies a regular Lindelöf lower-limit line.

Applying [L2] gives its normality.

For every $A\subseteq D$, the sets $A$ and $D\setminus A$ are disjoint closed subsets of $X$. By normality there is an open $U_A$ containing $A$ and an open $V_A$ containing $D\setminus A$ with $U_A\cap V_A=\varnothing$.

Apply [A1] to choose one such pair $(U_A,V_A)$ for every $A\subseteq D$, and define $\Phi(A)=U_A\cap E\subseteq E$.

If $A\ne B$, take $d\in A\setminus B$ after interchanging them if necessary. Then $d\in U_A\cap V_B$, a nonempty open set meeting $E$; a point of $E\cap U_A\cap V_B$ lies in $\Phi(A)$ and not in $\Phi(B)$.

Thus $\Phi$ is injective. By [F3], this is the asserted cardinal inequality.

Every nonempty $[a,b)\times[c,d)$ contains a point of $\mathbb Q\times\mathbb Q$: choose rationals $p\in[a,b)$ and $q\in[c,d)$. Hence $\mathbb Q\times\mathbb Q$ is dense, and it is at most countable by [L1].

The map $x\mapsto(x,-x)$ is a bijection from $\mathbb R$ onto $D$, so $D$ has cardinality $|\mathbb R|$ and is uncountable.

For $(x,-x)\in D$, the rectangle $[x,x+1)\times[-x,-x+1)$ meets $D$ only at $(x,-x)$, so $D$ is discrete in its subspace topology.

If $(u,v)\notin D$ and $u+v>0$, every sufficiently small lower-limit rectangle at $(u,v)$ has positive coordinate sum; if $u+v<0$, choose its two right endpoints so that their total increment is less than $-(u+v)$. In either case the rectangle misses $D$, so the complement of $D$ is open.

Therefore $D$ is closed discrete, with the stated cardinality, and the plane has the stated countable dense subset.

By [A1] and [F1], $L$ is normal. Suppose, for a contradiction, that $L^2$ is normal.

Jones's lemma applied to the $D,E$ of [L2] injects $\mathcal P(D)$ into $\mathcal P(E)$.

The two injections of [L4], with the fixed bijection $\mathbb Q\approx\mathbb N$, give $\mathbb R\approx\mathcal P(\mathbb N)$ by Schröder-Bernstein. Therefore $D\approx\mathcal P(\mathbb N)$, while $E\preceq\mathbb N$.

Taking direct images under these injections turns step 1.2 into an injection $\mathcal P(\mathcal P(\mathbb N))\to\mathcal P(\mathbb N)$.

This contradicts Cantor's theorem in [L3]. Therefore $L^2$ is not normal, while $L$ is normal, proving nonproductivity.

The factors $\omega_1+1$ and $\omega+1$ are compact and $T_3$ by [L1], so [L2] makes $P$ compact, Hausdorff, regular, and normal.

Since $P$ is $T_1$, $\{p\}$ is closed; hence $T$ is open. Its regularity follows from the hereditary regularity conclusion in [L2].

Put $E=\{\omega_1\}\times\omega$ and $F=\omega_1\times\{\omega\}$, regarded as subsets of $T$. The clopen ordinal basis shows that they are disjoint closed subsets of $T$.