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The following is a problem a colleague has encountered. He would like to know whether the following conjecture is right, wrong, or neither:

Let $X$ be a topological space of countable tightness and $x\in X$ a point that has no countable local base. Then $x$ has a local base $W$ of neighborhoods, such that countably infinite intersections of elements of $W$ are never neighborhoods of $x$.

The tightness of a point $x$ in a topological space is the smallest cardinal number $\kappa$ such that for every set $S$ with $x\in\mathrm{cl}(S)$, there is $T\subseteq S$ with $|T|\leq\kappa$ and $x\in\mathrm{cl}(T)$. The tightness of a topological space is the supremum over the tightness of each point.

If there are other conditions that guarantee that a point that has no countable local base, has a local base $W$ of neighborhoods, such that countably infinite intersections of elements of $W$ are never neighborhoods of $x$, it would be great to know them too.

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  • $\begingroup$ By a "neighbourhood of $x$" do you just mean a set having $x$ as an interior point? or that the set itself is open and contains $x$? $\endgroup$ – user642796 Apr 30 '13 at 8:36
  • $\begingroup$ @ArthurFischer The French definition, a set that has $x$ as an interior point. $\endgroup$ – Michael Greinecker Apr 30 '13 at 8:38
  • $\begingroup$ Just to feel a flavor of the problem :-) I ask about not so trivial example: of a space $X$ and a element $x\in X$ which is not a $P$-point, but $x$ has no such local base $W$. $\endgroup$ – Alex Ravsky Apr 30 '13 at 18:46
  • $\begingroup$ @MichaelGreinecker When you have time, have a look whether Alex Ravsky's answer given here works as a counterexample for your conjecture. $\endgroup$ – Martin Sleziak May 1 '13 at 10:56
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    $\begingroup$ @MartinSleziak & MichaelGreinecker I propose to use the answer (see my edit below) as a reflection of the state of the work on the problem $\endgroup$ – Alex Ravsky May 2 '13 at 13:05
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Martin Sleziak has suggested a possible counterexample to the conjecture here. That it is indeed a counterexample was then shown by Alex Ravsky here.


AR: Started to compose a draft paper.

By a space in this paper we mean a $T_1$ topological space.

A point $x$ of a space $X$ is isolated iff the intersection of all open neighborhoods of $x$ is an open set. Based on this observation, for a non-isolated point $x$ of a $T_1$ space $X$ let $P_b(x)$ be the smallest non-zero cardinal $\lambda$ such that there exists a base $\mathcal B$ at $x$ such that $x\not\in\operatorname{int}\bigcap\mathcal B'$ for each family $\mathcal B'\subset\mathcal B$ of distinct sets with $|\mathcal B'|=\lambda$.

Clearly, $P_b(x)\ge\omega$ and $P_b(x)\ge\omega_1$ for each non-isolated $P$-point $x$. An upper bound for $P_b(x)$ is provided by the following

Proposition 1. For each space $X$ and each non-isolated point $x\in X$ holds $P_b(x)\le\chi(x)$.

Proof. Put $\kappa=\chi(x)$. Let $\mathcal B$ be an arbitrary base at the point $x$ of size $\kappa$. Therefore there exists a bijection $U:\kappa\to\mathcal B$. For the sake of simplicity we denote $U(\lambda)$ as $U_\lambda$. By transfinite induction we can define a transfinite sequence $\{U^*_\lambda:\lambda<\kappa\}$ of open neighborhoods of $x$ such that for each $\lambda<\kappa$ holds $U^*_\lambda\subset U_\lambda$ and $U^*_\lambda\not\supset U^*_\mu$ for each $\mu<\lambda$. This is possible because $\kappa$ is the minimal ordinal of size $\kappa$, therefore the family $\{U^*_\mu:\mu<\lambda\}$ has size less than $\kappa$ and cannot be a base at the point $x$.

Put $\mathcal B^*=\{U^*_\lambda:\lambda<\kappa\}$. Since the family $\mathcal B$ is a base at the point $x$, the family $\mathcal B^*$ is a base at the point $x$ too, by the construction. Suppose that there are a subfamily $\mathcal C\subset\mathcal B^*$ such that $|C|=\kappa$ and an open neighborhood $V$ of the point $x$ such that $U\supset V$ for each $U\in\mathcal C$. Since the family $\mathcal B^*$ is a base at the point $x$ there exists an index $\mu<\kappa$ such that $U^*_\mu\subset V$. Since $\kappa$ is the minimal ordinal of size $\kappa$ and $\mu<\kappa=|\mathcal C|$, there exists an index $\lambda>\mu$ such that $U^*_\lambda\in\mathcal C$. Then $U^*_\mu\subset V\subset U^*_\lambda$. But, by the construction of the family $\mathcal B^*$, $U^*_\mu\not\subset U^*_\lambda$, a contradiction. $\square$

We have the following generalization. Let $x$ be a point of a space $X$ and $\lambda\le\chi(x,X)$ be a cardinal. I say that a point $x$ is a weak $P_\lambda$-point, if every base $\cal B$ at $x$ contains a family $\cal C$ of distinct sets such that $|\cal C|=\lambda$ and $x\in\operatorname{int}\bigcap\cal C$.

General Problem. Generalizing the question of Michael Greinecker, investigate when a point $x$ of a space $X$ is a weak $P_\lambda$-point.

Our ideas are related with so-called small cardinals. Define on $\omega^\omega$ a partial order $\le^*$ as follows. Let $x,y\in \omega^\omega$. Put $x\le^* y$ if $x_n\le y_n$ for almost all $n\in \omega$ (that is, this inequality doesn’t hold only for the finite number of $n$). Let $\frak b$ be the smallest size of a $*$-unbounded subset of $\omega^\omega$ and $\frak d$ be the smallest size of a (*-)cofinal (also called dominating) subset of $\omega^\omega$.

Working problem 1. It seems that we showed that the center $c_0$ of the Fréchet-Urysohn fan $F$ is a weak $P_\omega$-point.

It seems that we can show that $c_0$ is a weak $P_\lambda$-point for each cardinal $\lambda<\frak b$ with $\operatorname{cf}(\lambda)>\omega$. Indeed, let $\cal B\subset\omega^\omega$ be a dominating family. Then $|\cal B|\ge\frak d$. Since $\frak d\ge\frak b>\lambda$, a family $\cal B$ contains a subfamily $\cal B'$ of cardinality $\lambda$. Since $\lambda<\frak b$, we see that the family $\cal B’$ is $\le^*$-bounded. That is there exists a sequence $b\in\omega^\omega$ such that $x\le^* b$ for each $x\in\cal B’$. For each $n\in\omega$ put $\cal B_n=\{x\in\cal B’:x\le b+n\chi^n\}$, where $\chi^n$ denotes a characteristic function of the set $\{0,...,n-1\}$. Since the sequence $b$ $\le^*$-bounds the set $\cal B’$, we see that $\cal B’=\bigcup\cal B_n$. Since $\operatorname{cf}(\lambda)>\omega$, there exist a number $n\in\omega$ such that $|\cal B_n|\ge \lambda$.

(ZFC) Is $c_0$ a weak $P_\lambda$-point for each cardinal $\lambda<\frak b$? A weak $P_{\omega_1}$-point? A weak $P_\frak b$-point? A weak $P_\frak d$-point?

Working problem 1’. Taras Banakh considered our poset result from a more topological point of view. It seems that we showed that for each cofinal subset $B$ in Baire space $\omega^\omega$ there is a compact $K\subset\omega^\omega$ such that $|K\cap B|\ge\omega$. He suggested the following idea. Let $B\subset \omega^\omega$, $|B|=\lambda<\frak b$ and $\operatorname{cf}(\lambda)>\omega$. Since $|B|=\lambda<\frak b$, we see that the set $B$ is $\le^*$-bounded. That is there exists a point $b\in\omega^\omega$ such that $x\le^* b$ for each $x\in\cal B$. For each $n\in\omega$ put $K_n=\{x\in \omega^\omega: x\le b+n\chi^n\}$. Tychonoff Theorem implies that $K_n$ is compact. Since the sequence $b$ $\le^*$-bounds the set $B$, we see that $B=\bigcup B\cap K_n$. Since $\operatorname{cf}(\lambda)>\omega$, there exist a number $n\in\omega$ such that $|B\cap K_n|\ge \lambda$. Conversely, each subset of $\omega^\omega$, contained in a compact set, should be bounded.

For this problem analogues of the questions from Working Problem 1 also hold.

Added: It seems that I can prove: if $\frak d\le\lambda<\frak c$ and $\operatorname{cf}(\lambda)=\omega$, then $c_0$ is not a weak $P_\lambda$-point. Added: Taras Banakh asked an other our student, Lyubomyr Zdomskyy (by the way, who is currently a postdoctoral researcher in Austria):

TB: Is it true that for any dominating subset $D\subset\omega^\omega$ and each cardinal $\lambda<\mathfrak b$ (or even $\lambda<\mathfrak d$) there is a compact subset $K\subset\omega^\omega$ such that $|K\cap D|\ge\lambda$. The answer is yes if $\lambda<\mathfrak b$ and $\operatorname{cf}(\lambda)>\omega$.

LZ: It is a good question. For the $\frak d$-case the answer is negative.

AR: As I understood Lyubomyr, he propose the following. If we add to a model of CH $\lambda$ many Cohen reals $\{c_i:i<\lambda\}$, then no real dominates uncountably many $c_i$'s. Now let $\lambda=\omega_{\omega+1}$. Then $\frak d=\frak c=\omega_{\omega+1}$ in the forcing extension. Let $D=\{d_i:i<\frak d\}$ be a dominating family such that $d_i(n)>c_i(n)$ for all $n$. Then $|D\cap K|=\omega$ for each compact subset $K$ of $\omega^\omega$.

This result should imply that in the extension $\frak d=\frak c=\omega_{\omega+1}$ and $c_0$ is not a weak $P_{\omega_1}$-point.

LZ: I do not know a simple answer to the $\frak b$-case. I shall think about it, but if I shall fail then I shall ask Goldstern. (Un)fortunately $\frak b=\omega_1$ in the Cohen model. The case $\frak b=\frak d$ smells as if there should be some PCF theory involved, but it is just my feeling, nothing more.

Added: I can formulate a new question from the colleague of Michael Greinecker as follows. Let $X$ be a compact space of countable tightness and $x\in X$ be a point with uncountable character. Is $x$ a weak $P_\omega$-point? I remark that this compact can not be Hausdorff dyadic, because each Hausdorff dyadic compact with countable tightness is metrizable (see, for instance, Ryszard Engelking, “General Topology” (Ex. 3.12.12 (h) in Russian edition)).


I expose our main results in terms of the cardinal invariant $P_b(x)$ (formerly called $P'(x)$) equal to the largest $\lambda$ such that for each $\kappa<\lambda$ every base $\mathcal B$ at $x$ contains a family $\mathcal C$ of distinct sets such that $|\mathcal C|=\kappa$ and $x\in\operatorname{int}\bigcap\mathcal C$. Since an intersection of a finite family of neighborhoods of the point $x$ is a neighborhood of the point $x$, we see that $P_b(x)\ge\omega$. Since there is a base at the point $x$ which containing no $\chi(x)^+$ many distinct sets, we should have that $P_b(x)\le\chi(x)^+$.

Concerning the case $x=c_o$. It seems that we showed that $P_b(c_0)\ge\omega_1$ and $P_b(c_0)\ge\sup \{\lambda^+:\lambda<\frak b$ and $\operatorname{cf}(\lambda)>\omega\}$. An old Lyubomyr's remark should imply that $P_b(c_0)=\omega_1$ under $\frak d=\frak c=\omega_{\omega+1}$.

Now I shall expose new Lyubomyr's results.

LZ: I read more thoroughly your construction, and understood that the case when $\mathfrak b$ is a successor of a cardinal with countable cofinality can be proved simialarly.

Proposition. Suppose that $(\kappa_k:k\in\omega)$ is an increasing sequence of uncountable regular cardinals below $\mathfrak b$. Then there is no family $B\subset\omega^\omega$ of size $\geq\mathfrak d$ with the following property: For every $x\in\omega^\omega$ there exists $k\in\omega$ such that $|\{b\in B:b\leq x\}|\leq \kappa_k$.

Proof. Suppose that such $B$ exists. Set $B^0=B$. Since $\kappa_0<\mathfrak b$ is regular, there exists $x_0\in\omega^\omega$ and $B_0\in [B^0]^{\kappa_0}$ such that (i) $b\leq x_0$ for all $b\in B_0$; (ii) $|\{b\in B^0: b(0)\leq x_0(0)\}|\geq\mathfrak d$. The first item can be achieved by the same argument as in the paragraph above starting with "It seems that we can show that $c_0$ is a weak $P_\lambda$-point...", and the second item we get simply by enlarging $x_0(0)$ (if necessary) and using that $cof(\mathfrak d)>\omega$.

Set $B^1=\{b\in B^0: b(0)\leq x_0(0)\}$. Since $\kappa_1<\mathfrak b$ is regular, there exists $x_1\in\omega^\omega$ such that $x_1(0)=x_0(0)$ and $B_1\in [B^1]^{\kappa_1}$ such that (i) $b\leq x_1$ for all $b\in B_1$; (ii) $|\{b\in B^1: b(1)\leq x_1(1)\}|\geq\mathfrak d$.

......

Set $B^{k+1}=\{b\in B^k: b(i)\leq x_k(i) for all i\leq k\}$. By the construction of $x_k$ we have $|B^k|\geq\mathfrak d$. Since $\kappa_{k+1}<\mathfrak b$ is regular, there exists $x_{k+1}\in\omega^\omega$ such that $x_{k+1}(i)=x_k(i)$ for all $i\leq k$ and $B_{k+1}\in [B^{k+1}]^{\kappa_{k+1}}$ such that (i) $b\leq x_{k+1}$ for all $b\in B_{k+1}$; (ii) $|\{b\in B^{k+1}: b(k+1)\leq x_{k+1}(k+1)\}|\geq\mathfrak d$.

and so on, by induction.

At the end it seems that, letting $x_\omega$ be the coordinatewise maximum of $\{ b_k:k\in\omega\}$ and $B_\omega=\bigcup_{k\in\omega}B_k$, we have that $b\leq x_\omega$ for all $b\in B_\omega$. It suffices to note that $|B_\omega|>\kappa_k$ for all $k$, a contradiction.$\square$

If the above ideas are right, then we should have $P_b(c_0)\ge\frak b$. Nevertheless, not necessarily $P_b(c_0)\ge\frak b^+$ (adding Cohen numbers) even provided $\omega_1=\frak b<\frak d$. It is interesting, are there models such that $P_b(c_0)\ge \frak b^+$ (in this case, obviously, we should have $\mathfrak b<\mathfrak d$). I have no ideas for this case.


It seems that if $\mathfrak u^+ <\mathfrak d$ (which is known to be consistent), then $P_b(c_0)\ge\sup \{\lambda^+:\lambda<\frak d$ and $\operatorname{cf}(\lambda)>\frak u\}$. What to do with small cofinalities I do not know.


LZ: I have an impression that in Miller's model $P_b(c_0)=\omega_1$. Since in this model $\mathfrak b=\mathfrak u=\omega_1<\mathfrak d=\omega_2$, we see that an existence of a cardinal between $\mathfrak u$ and $\mathfrak d$ is indeed essential for the following results. As soon I shall check this "impression", I shall wrote to you.

Proposition. ($\mathfrak u^+ < \mathfrak d $) Suppose that $\mathfrak u<\kappa<\mathfrak d $ is a cardinals of cofinality $>\mathfrak u$. Then for every family $B\subset\omega^\omega$ of size $\geq\mathfrak \kappa$ there exists $x\in\omega^\omega$ such that $|\{b\in B:b\leq x\}|\geq \kappa$.

In particular, $P_b(c_0)>\kappa$.

Proof For each $b\in B$ define a non-decreasing function $b^*\in\omega^\omega$ by putting $b^*(n)=\max\{b(k):k\le n\}$. Let $\mathcal U$ be an ultrafilter on $\omega$ with a base $\{U_\alpha:\alpha<\mathfrak u\}$. Consider the linear preorder $\leq_\mathcal U$ on $\omega^\omega$ defined as follows: $x\leq_{\mathcal U} y$ if $[x\leq y]\in \mathcal U$, where $[x R y]$ denotes the set $\{n\in\omega:x(n)\: R\: y(n)\}$ for any relation $R$ on $\omega$. It is well known [Theorem~12, BlaMil] that any subset $A$ of $\omega^\omega$ of size $|A|<\mathfrak d$ is bounded with respect to $\leq_\mathcal U$. In particular, there exists an increasing $x\in\omega^\omega$ such that $b^*\leq_{\mathcal U} x$ for all $b\in B$. Therefore for every $b\in B$ there exists $\alpha<\mathfrak u$ such that $U_\alpha\subset [b^*\leq x]$. Since $\mathrm{cof}(\kappa)>\mathfrak u$ there exists $\alpha$ such that $|\{b\in B: U_\alpha\subset [b^*\leq x]\}|\geq\kappa$. Set $B_0=\{b\in B: U_\alpha\subset [b^*\leq x]\}$ and let $y\in\omega^\omega$ be such that $y(n)=x(\min(U_\alpha\setminus n))$. We claim that $b\leq y$ for all $b\in B_0$. Indeed, fix $b\in B_0$, $k\in\omega$, and let $n_0<n_1$ be two consecutive elements of $U_\alpha$ such that $n_0<k\leq n_1$. Then $b(k)\leq b^*(n_1)\leq x(n_1)=y(k)$.$\square$

Assume that $V$ is a model of GCH and $\nu<\delta$ are regular uncountable cardinals. Then there exists a c.c.c. poset $\mathbb{P}$ such that in the forcing extension $V^\mathbb{P}$ we have $\mathfrak b=\mathfrak u=\nu$ and $\mathfrak d=\delta$, see [BlaShe]. In particular, the premises of the above proposition are consistent.

Using the above proposition, the following statement can be proved in exactly the same way as the fact that $c_0$ is a weak $P_\lambda$-point for $\lambda<\mathfrak b$ of countable cofinality, see my ideas above.

Corollary. ($\mathfrak u^+ < \mathfrak d $) Let $\langle\kappa_k:k\in\omega\rangle$ be an increasing sequence of uncountable regular cardinals below $\mathfrak d$. Then for any family $B\subset\omega^\omega$ of size $\geq (\sup_{k\in\omega}\kappa_k)^+$ there exists $x\in\omega^\omega$ such that $|\{b\in B:b\leq x\}|\geq (\sup_{k\in\omega}\kappa_k)^+$.

In particular, $c_0$ is a weak $P_\kappa$-point in the Fr'echet-Urysohn fan $S_\omega$ for all $\kappa<\mathfrak d$ of countable cofinality. That is $P_b(c_0)\ge\sup \{\kappa^+:\kappa<\frak d$ and $\operatorname{cf}(\kappa)=\omega\}$.

The following question remains open.

Question. Suppose that $\mathfrak u<\kappa<\mathfrak d=\kappa^+$ and $\omega<\mathrm{cof}(\kappa)\leq\mathfrak u$. Must $c_0$ be a weak $P_\kappa$-point in the Frechet-Urysohn fan $S_\omega$ (that is, is $P_b(c_0)>\kappa$)? What about the model constructed in [BlaShe] for $\nu=\omega_1$ and $\delta=\omega_{\omega_1+1}$?

References

[BlaMil] Blass, A.; Mildenberger, H., On the cofinality of ultrapowers, J. Symbolic Logic 64 (1999), 727--736.

[BlaShe] Blass, A.; Shelah, S., Ultrafilters with small generating sets, Israel J. Math. 65 (1989), 259--271.

AR: I have a question too. Can anybody evaluate these $\sup\{\lambda+:\dots\}$? Maybe they are known or easily obtainable and then we shall have some more clear results.

Added: It seems that the answer to this question is not hard obtainable. For each cardinal $\mu$ as $\lfloor\mu\rfloor$ I denote the largest limit cardinal not greater than $\mu$. Then $\mu=\lfloor\mu\rfloor^{+\dots+}$ and $\sup\{\lambda^+:\lambda<\mu$ and $\operatorname{cf}(\lambda)=\omega\}= \sup\{\lambda:\lambda<\mu$ and $\operatorname{cf}(\lambda)=\omega\}=\lfloor\mu\rfloor$. Indeed, if $\lambda<\lfloor\mu\rfloor$ then $\lambda^+<\lfloor\mu\rfloor$. Therefore $\lambda^{++}<\lfloor\mu\rfloor$ and so on. Thus $\lambda^\infty=\lambda\cup\lambda^+\cup\lambda^{++}\cup\dots$ is a limit cardinal of countable cofinality and $\lambda^\infty\le\lambda<\lfloor\mu\rfloor$. Since this inequality holds for each cardinal $\lambda<\lfloor\mu\rfloor$, we see that $\sup\{\lambda^+:\lambda<\mu$ and $\operatorname{cf}(\lambda)=\omega\}= \sup\{\lambda:\lambda<\mu$ and $\operatorname{cf}(\lambda)=\omega\}=\lfloor\mu\rfloor$.

By Proposition 1, $P_b(c_0)\le\frak d$.

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