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 $x\in V\subseteq \bar{V}\subseteq U$.

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

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 $x\mathrm{\notin }C$.

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

The restriction $f{\mathrm{\mid }}_S:S\to [0,1]$ 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 $I=\mathrm{\varnothing }$, the product has one point and all three conditions hold vacuously.

Otherwise choose $i\in I$ with $x_i\mathrm{\ne }{y}_i$. 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 $y\mathrm{\notin }B$, so $y_i\mathrm{\notin }{U}_i$ 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 $Y=X\setminus ((\bar{A}\setminus A)\cup (\bar{B}\setminus B))$; 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 $(\bar{A}\setminus A)\cup (\bar{B}\setminus B)$.

The open sets $G\setminus \bar{B}$ and $H\setminus \bar{A}$ 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 $\bar{V}\subseteq X\setminus B$. 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 of $\mathcal{V}$ covers $A$. Interchanging $A$ and $B$ gives an at most countable family of open sets covering $B$, each with closure disjoint from $A$. List these families as $(U_n{)}_{n\ge 0}$ and $(V_n{)}_{n\ge 0}$, padding a finite family with empty sets. The countable-listing convention is part of [F1]. Only two subcovers and their listings are selected; no axiom of choice is needed.

Set $U={\bigcup }_{n\ge 0}(U_n\setminus {\bigcup }_{i\le n}\overline{V_i})$ and $W={\bigcup }_{n\ge 0}(V_n\setminus {\bigcup }_{i\le n}\overline{U_i})$. Each summand is open because it removes only finitely many closed sets from an open set. Every point of $A$ lies in some $U_n$ and in none of the $\overline{V_i}$, so $A\subseteq U$. Similarly $B\subseteq W$.

A point in the $n$th summand of $U$ and the $m$th summand of $W$ is impossible: if $m\le n$, the first summand excludes $\overline{V_m}$; if $n\le m$, the second excludes $\overline{U_n}$. Thus $U\cap W=\mathrm{\varnothing }$, and these open sets prove normality, including empty $A$ or $B$.

Each $[a,b)$ is clopen: its complement is $(-\mathrm{\infty },a)\cup [b,\mathrm{\infty })$, 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 $U_A\cap V_A=\mathrm{\varnothing }$.

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

If $A\mathrm{\ne }B$, 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 $\mathrm{\Phi }(A)$ and not in $\mathrm{\Phi }(B)$.

Thus $\mathrm{\Phi }$ 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 $\mathrm{\mid }\mathbb{R}\mathrm{\mid }$ 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 $(u,v)\mathrm{\notin }D$ 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 $E⪯\mathbb{N}$.

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 $T$.

Suppose, for a contradiction, that $T$ is normal. Choose disjoint open $U,V\subseteq T$ with $E\subseteq U$ and $F\subseteq V$.

For $n<\omega$, let $C_n=\{\xi <{\omega }_1:(\xi ,{\omega }_1]\times \{n\}\subseteq U\}$. By [F2] each $C_n$ is nonempty; [A1] chooses ${\alpha }_n\in C_n$ simultaneously. The countable set $\{{\alpha }_n:n<\omega \}$ is bounded by some $\alpha <{\omega }_1$.

Put $\beta =\alpha +1<{\omega }_1$. Since $(\beta ,\omega )\in V$, [F2] gives $\gamma <\beta$ and $m<\omega$ with $(\gamma ,\beta ]\times (m,\omega ]\subseteq V$.

The point $(\beta ,m+1)$ lies in $V$ by step 3.2 and in $U$ by step 3.1, because $\beta >\alpha \ge {\alpha }_{m+1}$. This contradicts $U\cap V=\mathrm{\varnothing }$, so $T$ is not normal; together with steps 1.1 and 1.2 this proves all the stated properties.

The parent $P$ in [L1] is normal, but its open regular subspace $T$ is not normal.

This one normal space and one nonnormal subspace refute the defining universal condition in [F1].

Let $T$ be the deleted Tychonoff plank. It is regular by [L1].

The same space $T$ is not normal by [L1], contradicting the displayed universal claim.