# What are the requirements for separability inheritance

Suppose we have an arbitrary separable topological space $X$. What are some (possibly nonequivalent) minimal requirements to put on $X$ to ensure that every subspace of $X$ is separable?

This is not true for arbitrary spaces, as witnessed by the the space $X$ which is uncountable, where open sets are precisely the sets containing some special point $x_0$. It is separable (because $\lbrace x_0\rbrace$ is dense), but $X\setminus \lbrace x_0\rbrace$ is uncountable and discrete, so not separable. However, this space is not even $T_1$ (though it is $T_0$).

It is clearly true for second-countable spaces (because weight is not smaller than density, and is inherited), but that is not necessary, as shown by an example similar to the above, but with $X$ countable.

Analogous question could be asked replacing $\aleph_0$ with arbitrary infinite cardinal $\kappa$: suppose we have a space $X$ with a dense subset of cardinality at most $\kappa$, what should we require of $X$ for this to be inherited? In this case we can perform the analysis which is exactly analogous to the above, but perhaps some more concrete results will be harder to arrive at with uncountable $\kappa$...

So I really have two somewhat related questions. In terms of cardinal invariants density $d$ and hereditary density $hd$, what requirements do we have to put on $X$ to have some of the following:

1. (Only for $X$ separable) $hd(X)\leq\aleph_0$
2. $d(X)= hd(X)$

By analysis similar to the above we know that 2. is not true in general, as well as that $d(X)=w(X)$ implies 2, but is not necessary.

I would appreciate some conditions which would imply either one, or some nice counterexamples which satisfy some stronger separation axioms than just $T_0$ (if there are any, I think there should be...), or a proof that there are none.

Edit: Sam L. suggested the example of Niemytzki plane, which shows that even somewhat strong separation axioms are not sufficient: it is completely regular and separable and of countable character, but has an uncountable discrete subspace. It is not hard to see that by taking a suitable subspace, we can strengthen it to $w(X)=\aleph_1$ (regardless of CH), and itself is a counterexample for all $\kappa<\mathfrak c$.

Edit 2: As per Arthur Fischer's suggestion, if we take an arbitrary nontrivial second-countable compact Hausdorff space $X$ (such as $2$ with discrete topology or $[0,1]$), $X^\mathfrak c$ will be separable (because product of at most $\mathfrak c$ separable spaces is separable), as well as Hausdorff and compact, and hence normal, but it has a discrete subspace of cardinality $\mathfrak c$. This shows that no usual separation axioms will suffice, not even augmented by compactness.

• @SamL.: Thanks! My mistake. I assumed that separability+countable character implies countable basis, which is of course not true (it only implies countable $\pi$-basis). Aug 9 '12 at 17:14
• Note, also, that the Cantor cube of weight $\mathfrak{c} = 2^{\aleph_0}$, $2^{\mathfrak{c}}$, is an example of a separable compact Hausdorff (hence normal) space of hereditary density $\mathfrak{c}$. (See, e.g., Example 11.8 (p.44) of Hodel, Cardinal Functions I in the Handbook of Set Theoretic Topology.) This space does have character $\mathfrak{c}$, however, so it is very far from satisfying any countability axioms. Aug 9 '12 at 21:50
• @ArthurFischer: I had thought about that example (well, specifically, I thought about $[0,1]^{\mathfrak c}$, but I don't suppose it's that much of a difference in this case), but wasn't sure about its hereditary density. Thanks. :) Aug 9 '12 at 22:23

The results in the comments are all negative. Here’s a positive result, albeit with a rather strong hypothesis:

Theorem: Every LOTS (space whose topology is generated by a linear order) has hereditary density equal to its density.

Proof: Let $$\langle X,\le\rangle$$ be a LOTS, and let $$Y$$ be a subspace of $$X$$. Let $$D$$ be a dense subset of $$X$$ of cardinality $$d(X)$$, and without loss of generality assume that $$D$$ contains the endpoint(s) of $$X$$, if any. For each $$a,b\in D$$ with $$a and $$(a,b)\cap Y\ne\varnothing$$ fix $$y(a,b)\in(a,b)\cap Y$$. Let $$I$$ be the set of isolated points of $$Y$$, and let $$E=D\cup\{y(a,b):a,b\in D\text{ and }a.

Claim: $$I\cup E$$ is dense in $$Y$$.

Proof: Let $$U$$ be any non-empty open subset of $$Y$$. If $$U\cap I\ne\varnothing$$, we’re done, so assume that $$U\cap I=\varnothing$$. Fix $$y\in U$$; there are $$a,b\in D$$ such that $$a and $$(a,b)\cap Y\subseteq U$$, and clearly this ensures that $$y(a,b)\in U$$. $$\dashv$$

Clearly $$|E|\le d(X)$$, so to show that $$d(Y)\le d(X)$$, it suffices to show that $$|I|\le d(X)$$. Fix $$y\in I$$. Suppose first that $$y$$ has no immediate predecessor or successor in $$X$$. Then there are $$a_y,b_y\in D$$ such that $$(a_y,b_y)\cap Y=\{y\}$$. Next, suppose that $$y$$ has both an immediate predecessor $$y^-$$ and an immediate successor $$y^+$$ in $$X$$; then $$y\in D\subseteq E$$.

Finally, suppose that $$y$$ has no immediate predecessor but does have an immediate successor $$y^+$$ in $$X$$. (The reverse case is entirely similar.) If $$y^+\in D$$, there is an $$a_y\in D$$ such that $$(a_y,y^+)\cap Y=\{y\}$$, and we set $$b_y=y^+$$. Note that this is necessarily the case if $$y^+$$ has an immediate successor in $$X$$, so we may assume that $$y^+\in\operatorname{cl}(y^+,\to)$$ and that $$y^+\notin D$$. Then there are $$a_y\in(\leftarrow,y)\cap D$$ and $$b_y\in(y^+,\to)\cap D$$ such that $$(a_y,b_y)\cap Y\subseteq\{y,y^+\}$$. The assignment $$y\mapsto(a_y,b_y)$$ is at most $$2$$-$$1$$, so $$|I|\le d(X)$$, and the proof is complete. $$\dashv$$

Added: The result can be strengthened a bit. A GO-space is a Hausdorff space $$\langle X,\tau\rangle$$ with a linear order $$\le$$ such that every point of $$X$$ has a local base of $$\le$$-intervals; these intervals may be open, closed, or half-open, and if they’re closed, they may be degenerate. It’s well-known $$X$$ is a GO-space iff $$X$$ is a subspace of a LOTS.

Corollary: If $$X$$ is a GO-space, then $$d(X)=hd(X)$$.

Proof: Let $$\langle X,\tau\rangle$$ be a GO-space with associated linear order $$\le$$, and let $$D$$ be a dense subset of $$X$$ of cardinality $$d(X)$$. Let $$\tau_\le$$ be the order topology on $$X$$ generated by $$\le$$; $$\tau_\le\subseteq\tau$$. Let $$L=\{x\in X:[x,\to)\in\tau\setminus\tau_\le\}$$ and $$R=\{x\in X:(\leftarrow,x]\in\tau\setminus\tau_\le\}\;.$$ Note that $$L\cap R\subseteq D$$.

Let $$X^+=\Big(X\times\{1\}\Big)\cup\Big(L\times\{0\}\Big)\cup\Big(R\times\{2\}\Big)\;,$$ and let $$\preceq$$ be the lexicographic order on $$X^+$$ determined by $$\le$$ on $$X$$ and the usual order on $$\{0,1,2\}$$, and let $$\tau_\preceq$$ be the order topology on $$X^+$$ induced by $$\preceq$$. If we identify $$X$$ with $$X\times\{1\}$$ in the obvious way, it’s a little tedious but not difficult to verify that the topology that $$X$$ inherits from $$\left\langle X^+,\tau_\preceq\right\rangle$$ coincides with $$\tau$$. Let $$D\,'=D\times\{1\}$$; I’ll show that $$D\,'$$ is dense in $$X^+$$.

Suppose that $$p=\langle x,m\rangle,q=\langle y,n\rangle\in X^+$$ with $$p\prec q$$ and $$(p,q)\ne\varnothing$$. If $$x=y$$, then $$m=0$$ and $$n=2$$, so $$x\in L\cap R\subseteq D$$, and $$\langle x,1\rangle\in(p,q)\cap D\,'$$. Assume now that $$x. If $$(x,y)$$ is non-empty in $$X$$, choose $$z\in(x,y)\cap D$$; then $$\langle z,1\rangle\in(p,q)\cap D\,'$$. Otherwise, $$y$$ is the immediate successor of $$x$$ in $$X$$. This implies that $$(\leftarrow,x],[y,\to)\in\tau$$, so $$x\notin R$$, and $$y\notin L$$. It follows that $$p=\langle x,0\rangle$$, $$q=\langle y,2\rangle$$, or both, as otherwise $$(p,q)=\varnothing$$. If $$p=\langle x,0\rangle$$, then $$x$$ is isolated in $$X$$, so $$x\in D$$, and $$\langle x,1\rangle\in(p,q)\cap D\,'$$. Similarly, if $$q=\langle y,2\rangle$$, then $$\langle y,1\rangle\in(p,q)\cap D\,'$$. In all cases $$(p,q)\cap D\,'\ne\varnothing$$, so $$D\,'$$ is indeed dense in $$X^+$$, and since $$X^+$$ is a LOTS we have $$hd(X^+)=d(X^+)\le|D|=d(X)$$. But $$X$$ is a subspace of $$X^+$$, so $$d(X)\le hd(X)\le hd(X^+)\le d(X)$$, and hence $$d(X)=hd(X)$$. $$\dashv$$

• @tomasz: It needn’t be. Suppose that $X=\Bbb R$, $D=\Bbb Q$, and $Y=[0,1]\cup\{\pi\}$; then $\pi\in I\setminus E$. Aug 12 '12 at 16:23
• Are you back, Prof.? Aug 12 '12 at 16:28
• @tomasz: Yes, of course: for some reason I was thinking that the point had to be in $D$. Change the example. Let $X=[0,1]\cup[2,3]$, let $D=\Bbb Q\cap\Big((0,1)\cup(2,3)\Big)$, and let $Y=\{1,2\}$. Then it’s possible that $y(a,b)=1$ whenever $a<1<2<b$ and $a,b\in D$. Aug 12 '12 at 16:28
• @Peter: If not, someone’s doing a pretty good impersonation! Aug 12 '12 at 16:30
• @tomasz: Oops! You’re right. Aug 12 '12 at 16:34