# Maximum (or upper bound on the) weight of a topology on a set X.

Let $X$ be a set and $\tau$ a topology on $X$. Let's define the weight of $\tau$, $w(\tau)$, as the minimum cardinality of a basis for $\tau$.

1) What is the supremum of the weights of all topologies on $X$?

2) Is this supremum a maximum?

3) Is this supremum greater than the cardinality of $X$?

Let $s=\sup\{w(\tau) : \tau \text{ is a topology on } X\}$. This is what I deduced:

Clearly $s\geq |X|$, since $w(\tau)=|X|$ when $\tau$ denotes the discrete topology.

If $X$ is finite then $s=|X|$ is a maximum. So the answer to 3) in this case is no.

• I interpret "can be" as "Is there an $X$ for which the supremum is greater than $|X|$?" Apr 12, 2017 at 13:39
• @MeesdeVries that's correct. I made an edit to make it more clear. Apr 12, 2017 at 13:45
• I significantly expanded my answer, check it out Apr 16, 2017 at 9:38
• Finished with my 3 types of examples, 2 from large ultrafilters, 1 from large products. Hopefully this answer can be used as a reference for other answers as well. Apr 17, 2017 at 16:14
• If you allow finite values for $w(\tau)$ (this is unusual) any topology on a set of size $n$ has a base of size $\le n$: the minimal neighbourhoods of all points. So clear is $s \le |X|$. But the max is assumed for the discrete topology on $X$, so we have the $s = |X|$ is a maximum for finite spaces indeed. Apr 17, 2017 at 16:32

There is a trivial bound for all spaces $$(X,\tau)$$: $$w(X) \le |\tau| \le 2^{|X|}$$. (A base is a subset of the topology and the topology is a subset of the powerset)

As an extra: for $$T_0$$ spaces we also have $$|X| \le 2^{w(X)}$$ (as $$x \rightarrow \{B \in \mathscr{B}: x \in B\}$$ is 1-1 for any base $$\mathscr{B}$$ then take one of size $$w(X)$$)

and $$|X| \le |\tau|$$ (as $$x \rightarrow X\setminus\overline{\{x\}}$$ is 1-1 for $$T_0$$ spaces). So for lots of spaces there are bounds on $$|X|$$ from $$w(X)$$ and vice versa.

And for any set $$X$$ of some infinite size, an ultrafilter space on $$X$$ will have weight $$2^{|X|}$$, so the maximum is achieved.

So the answer is : the bound on all weights is $$2^{|X|} > |X|$$ for infinite $$X$$ and it is a maximum.

I'll clarify in this remark with some constructions of such spaces:

## Large ultrafilters

I'll assume in the rest all basic theory on ultrafilters. In this answer Brian Scott shows how to construct for a set $$X$$ of (infinite) size $$\kappa$$ an independent family $$\mathscr{F}$$ of size $$2^\kappa$$. This means that for any two finite disjoint subfamilies $$\mathscr{A}$$ and $$\mathscr{B}$$ of $$\mathscr{F}$$, we have that

$$\left( \bigcap_{A \in \mathscr{A}} A\right) \cap \left(\bigcap_{B \in \mathscr{B}} (X\setminus B)\right) \neq \emptyset$$

Assuming this fact (due to Hausdorff already) we will "construct" an ultrafilter as follows: define the family $$\mathscr{G} = \mathscr{F} \cup \{X \setminus \bigcap_{A \in \mathscr{F'}} A \text{, where } \mathscr{F'} \subseteq \mathscr{F} \text{ is infinite }\}$$

It's not too hard to see that $$\mathscr{G}$$ has the finite intersection property:

Finitely many sets from only $$\mathscr{F}$$ already intersect (from independence) and if we have only finitely many sets of the "complement type" for families $$\mathscr{F_1},\ldots,\mathscr{F_n}\subseteq \mathscr{F}$$, then for any choices $$F_1 \in \mathscr{F_1},\ldots, F_n \in \mathscr{F}_n$$, a point in $$\cap_{i=1}^n (X\setminus F_i)$$ (which again exists by independence) is in the intersection of the $$X \setminus \bigcap_{A \in \mathscr{F}_i} A, i=1,\ldots n$$. If we have a mixed finite subset with some $$F_1,\ldots,F_n$$ plus some complement type $$X \setminus \bigcap_{A \in \mathscr{F}_{n+1}} A, \ldots, X \setminus \bigcap_{A \in \mathscr{F}_{n+m}} A$$, then we pick $$F_{n+1} \in \mathscr{F}_{n+1},F_{n+m} \in \mathscr{F}_{n+m}$$ such that $$\{F_{n+1},\ldots, F_{n+m}\}$$ is disjoint from $$\{F_1,\ldots,F_n\}$$, which can be done as all families $$\mathscr{F}_{n+i}$$ are infinite. The intersection of all $$F_i$$ ($$i\le n)$$ and $$X\setminus F_{n+i}, i =1,\ldots m$$ is non-empty by independence ,and is a subset of the intersection of the sets we started with.

We can extend $$\mathscr{G}$$ to an ultrafilter $$\mathscr{U}$$ on $$X$$.

This has the property that there is no small generating set for it:

(*) If $$\mathscr{B} \subseteq \mathscr{U}$$ has the (generating set) property that $$\forall U \in \mathscr{U}, \exists B \in \mathscr{B}: B \subseteq U$$, then $$|\mathscr{B}| =2^\kappa$$

Proof: suppose that such we have such a family $$\mathscr{B}=\{B_\alpha: \alpha < \lambda\}\subseteq \mathscr{U}$$ of size $$\lambda < 2^\kappa$$. Then for each $$F \in \mathscr{F}(\subseteq \mathscr{G} \subseteq \mathscr{U})$$ we have $$f(F) < \lambda$$ such that $$B_{f(F)} \subseteq F$$. As $$\lambda < 2^\kappa = |\mathscr{F}|$$, there is some $$B_\alpha \in \mathscr{B}$$ such that $$\mathscr{F}' = \{F \in \mathscr{F}: B_\alpha \subseteq F\}$$ is infinite. But then $$B_\alpha$$ is disjoint from $$X \setminus \bigcap_{A \in \mathscr{F}'} A \in \mathscr{G} \subseteq \mathscr{U}$$ This is a contradiction and the statement (*) has been shown. This idea I learnt from this interesting paper by Blass and Rupprecht.

## Spaces from an ultrafilter

The above set theory results allow us to construct the spaces we need. I know of 2 basic constructions that start with an ultrafilter $$\mathscr{F}$$ on $$X$$ of size $$\kappa$$.

Firstly, add a point at infinity, so define the space $$X(\mathscr{F}) = X \cup \{\infty\}$$ where all points of $$X$$ are isolated (so have a local base $$\{\{x\}\}$$ at $$x$$) and a neighbourhood of $$\infty$$ is of the form $$\{\infty\} \cup A$$, where $$A \in \mathscr{F}$$. This space $$X(\mathscr{F})$$ is always $$T_2$$ and normal (as there is only one non-isolated point) and even though $$\infty \in \overline{X}$$, no sequence from $$X$$ can converge to $$\infty$$. And if $$\{\infty\} \cup A_\alpha, \alpha < \lambda$$ is a neighbourhood base for $$\infty$$, then $$A_\alpha$$ is a generating set for $$\mathscr{F}$$, almost by definition.

So $$X(\mathscr{U})$$ for the "large" ultrafilter constructed above, every local base of $$\infty$$ has size $$2^\kappa$$, which implies that $$w(X) = 2^\kappa$$, as required.

So the countable version is (hereditarily) Lindelöf ,(hereditarily) separable, all points are $$G_\delta$$, but the weight is $$\mathfrak{c}$$, and the space is not sequential.

Secondly we can just use $$X$$ as the topological space and use $$\mathscr{F}\cup \{\emptyset\}$$ as the topology (one readily checks that this is a topology: closed under unions follows from closedness under larger sets, closed under finite intersections is also clear, $$\emptyset$$ is a special case here)

This construction yields weirder spaces still: $$T_1$$ but anti-Hausdorff (any two non-empty open sets intersect, so it is connected) and a so-called "door space" (every subset of $$X$$ is closed or open or both). We can completely characterise such spaces as anti-Hausdorff door spaces. It's also clear that a base for this topology is in fact a generating set for $$\mathscr{F}$$, so if we use the "large" $$\mathscr{U}$$ from above, we again get a space of size $$\kappa$$ and weight $$2^\kappa$$.

There are other constructions from large products which I will write down later, For which I'll need:

# Some preliminaries on bases and weight.

A useful theorem (I call it the "thinning out lemma") is the following:

Theorem Let $$X$$ be a space and $$w(X) =\kappa$$, some infinite cardinal. If $$\mathscr{B}$$ is any base for the topology of $$X$$, then there exists a subfamily $$\mathscr{B}' \subseteq \mathscr{B}$$ such that $$\left|\mathscr{B}'\right| = \kappa$$ and $$\mathscr{B}'$$ is still a base for the space $$X$$.

So for example, if $$X$$ has a countable base (so $$w(X) =\aleph_0$$), all other bases of $$X$$ can be "thinned out" to a countable base, by possibly throwing some sets away.

Proof: we start by fixing a minimal base promised by $$w(X) = \kappa$$: find $$\mathscr{M}$$ a (minimally sized) base for $$X$$ such that $$\left|\mathscr{M}\right| = \kappa$$.

Now let $$\mathscr{B}$$ be any base for $$X$$.

Then define $$I = \{(M_1, M_2) \in \mathscr{M} \times \mathscr{M}: \exists B \in \mathscr{B}: M_1 \subseteq B \subseteq M_2 \}$$

Note that $$\left|I\right| \le \kappa^2 = \kappa$$, and we apply the Axiom of Choice to pick for each $$i \in I$$ (where $$i = (M^i_1, M^i_2)$$), some $$B_i \in \mathscr{B}$$ such that $$M^i_1 \subseteq B_i \subseteq M^i_2$$.

We claim that $$\mathscr{B}' := \{B_i: i \in I\} \subseteq \mathscr{B}$$ is also a base for $$X$$, and clearly $$\left| \mathscr{B'}\right| \le |I| \le \kappa$$ and so we would be finished (as all bases, hence also $$\mathscr{B}'$$, are of size $$\ge \kappa$$ (by $$w(X) = \kappa$$) and we'd have equality of sizes).

To see it is a base: let $$O$$ be open in $$X$$ and $$x \in O$$. We have to find some $$B_i$$ that sits between them.

First use that $$\mathscr{M}$$ is a base and find $$M_2 \in \mathscr{M}$$ such that

$$x \in M_2 \subseteq O$$

Then use that $$\mathscr{B}$$ is a base (applied to $$x$$ and $$M_2$$) and find $$B \in \mathscr{B}$$ such that

$$x \in B \subseteq M_2 \subseteq O$$

Again apply that $$\mathscr{M}$$ is a base (to $$x$$ and $$B$$) and find $$M_1 \in \mathscr{M}$$ such that

$$x \in M_1 \subseteq B \subseteq M_2 \subseteq O$$

Aha! We have that $$i:= (M_1, M_2) \in I$$ (we've forced it that way using the base property ) So we have already picked some $$B_i = B_{(M_1, M_2)}\in \mathscr{B}'$$ (it's probably some other member not necessarily our $$B$$ from above), such that

$$x \in M_1 =M^i_1 \subseteq B_i \subseteq M^i_2 = M_2 \subseteq O$$

And we have found the required member of $$\mathscr{B}'$$ between $$x$$ and $$O$$. This finishes the proof.

# Powers of discrete spaces

If $$D$$ is a set in the discrete topology and $$I$$ an index set, we give the set $$D^I = \{f: I \rightarrow D: f \text{ a function } \}$$ the product topology., which is the smallest topology such that all projections $$p_i : D^I \rightarrow D, p_i(f) = f(i)$$ are continuous.

It's then easily checked that the following collection is a base for this topology on $$D^I$$:

$$\mathscr{B}(D,I) = \{\langle i_1,\ldots i_n; d_1, \ldots d_n \rangle: n \in \mathbb{N}, i_1,\ldots i_n \in I, d_1,\ldots d_n \in D \}$$

Where

$$\langle i_1,\ldots i_n; d_1, \ldots d_n \rangle = \{f \in D^I: \forall j \in \{1,\ldots,n\}: f(i_j) = d_j\}= \bigcap_{j=1}^n p_{i_j}^{-1}[\{d_j\}] \text{.}$$

It's clear that $$\left|\mathscr{B}(D,I)\right| \le \sum_{n \in \mathbb{N}} |I|^n |D|^n$$ and also that all such powers $$D^I$$ are Hausdorff : if $$f \neq g$$ in $$D^I$$ then there is some $$i \in I$$ with $$f(i) \neq f(j)$$, and then $$\langle i;f(i)\rangle$$ and $$\langle i; g(i)\rangle$$ are disjoint open neighbourhoods of $$f$$ resp. $$g$$. More general theory on product spaces tells us that $$D^I$$ (and all of its subspaces) are Tychonoff (completely regular), which we will not need.

And the two special cases I will use later: $$D = 2:=\{0,1\}, I = \kappa$$ an infinite cardinal, where $$|\mathscr{B}(2,\kappa)| = \kappa$$ and $$D = \kappa$$ and $$I = 2^\kappa$$ where $$\left|\mathscr{B}(\kappa, 2^\kappa)\right| = 2^\kappa$$

The last case is the one I'm interested in for the purpose of finding a space with large weight: there is a dense set $$D$$ of size $$\kappa$$ in for the case $$D = \kappa$$ and $$I = 2^\kappa = \{0,1\}^\kappa$$. This nicely uses that $$2^\kappa$$ is a Hausdorff space of weight $$\kappa$$, as witnessed by $$\mathscr{B}(2,\kappa)$$,

Define $$D(\kappa) \subset \kappa^{(2^\kappa)}$$ as follows:

$$D(\kappa) = \{f: 2^\kappa \rightarrow \kappa: \exists n \in \mathbb{N} \exists B_1, \ldots, B_n \in \mathscr{B}(2,\kappa): \exists \alpha_1,\ldots, \alpha_n \in \kappa: \forall j \in \{1,\ldots, n\}: f|_{B_j} \equiv \alpha_j \text{and} f|_{2^\kappa \setminus \cup_{j=1}^n B_j} \equiv 0\}$$

So, all functions that are constant on finitely many base elements and $$0$$ outside them. Any $$f \in D(\kappa)$$ is determined by picking some finite number $$n$$ and $$n$$ many base elements from among the $$\kappa$$ many (in $$\kappa^n = \kappa$$ ways) and finally picking $$n$$ values from $$\kappa$$ in $$\kappa^n = \kappa$$ ways. So the size of $$D(\kappa)$$ is $$\aleph_0 \kappa = \kappa$$.

The only thing left to verify is that $$D(\kappa)$$ is dense in $$\kappa^{2^\kappa}$$, i.e. It must intersect any base element from $$\mathscr{B}(\kappa, 2^\kappa)$$. So let $$\langle f_1,\ldots,f_n; \alpha_1,\ldots, \alpha_n \rangle$$ be such a basic element with $$f_i \in 2^\kappa$$, $$\alpha_i \in \kappa$$. As the $$f_i \in \{0,1\}^\kappa$$ live in a Hausdorff space and $$\mathscr{B}(2,\kappa)$$ form a base for the space, we can find pairwise disjoint basic sets $$B_1, \ldots, B_n \in \mathscr{B}(2, \kappa)$$ such that $$f_i \in B_i$$ for all $$i = 1,\ldots n$$. Now define $$f: 2^\kappa \rightarrow \kappa$$ as follows: if $$x \in B_j$$ for some $$j$$: f$$(x) = \alpha_j$$, and if $$x$$ is in none of the $$B_i$$, define $$f(x) = 0$$. By construction now $$f \in D(\kappa)$$, and $$f_i \in B_i$$ then ensures that indeed $$f \in \langle f_1,\ldots,f_n; \alpha_1,\ldots, \alpha_n \rangle$$ as well. So $$D(\kappa)$$ intersects every basic element, so is indeed dense.

Corollary (Hewitt-Marczewski-Pondiczery theorem) if $$X_i, i \in I$$ is any family of spaces and all $$X_i$$ have a dense set of size at most $$\kappa$$ and $$|I| \le 2^\kappa$$ then $$\prod_i X_i$$ also has a dense set of size $$\kappa$$.

Proof sketch : find maps $$f_i$$ from discrete $$\kappa$$ onto the different dense sets, and WLOG we can replace $$I$$ by $$2^\kappa$$. Then the image under the product map of the above $$D(\kappa)$$ is as required.

# The point: $$w(D(\kappa))= 2^\kappa$$

proof: we already know a base for $$D(\kappa)$$, namely all sets $$\langle f_1,\ldots f_n; \alpha_i,\ldots \alpha_n \rangle \cap D(\kappa)$$, the relativised version of the standard base $$\mathscr{B}(\kappa, 2^\kappa)$$.

Suppose that $$w(D(\kappa)) = \lambda < 2^\kappa$$. Then the thinning out lemma allows us to reduce the standard relativised base to a subfamily of size $$\lambda$$: for every $$\beta < \lambda$$ we have some $$n = n(\beta) \in \mathbb{N}$$ and some sequence $$f^\beta_1, \ldots, f^\beta_n$$ all in $$2^\kappa$$ and a sequence $$\alpha^\beta_1,\ldots \alpha^\beta_n \in \kappa$$ such that

$$\mathscr{B}' = \{\langle f^\beta_1, \ldots, f^\beta_{n(\beta)}; \alpha^\beta_1,\ldots \alpha^\beta_{n(\beta)}\rangle \cap D: \beta < \lambda \}$$ is a base for $$D(\kappa)$$.

Now consider the subset $$I' = \{f^\beta_i: 1 \le i \le n(\beta), \beta < \lambda\}$$ which is the set of used coordinates in this base. $$I'$$ has size at most $$\aleph_0 \lambda = \lambda < 2^\kappa$$, so there is some $$g \in 2^\kappa \setminus I'$$.

Now $$\langle g; 0\rangle$$ is open and non-empty, so $$\langle g;0\rangle \cap D(\kappa)$$ is non-empty and so must contain some member of $$\mathscr{B}'$$ as the is a base. So there exists some $$\beta_0 < \lambda$$ such that $$\text{(*)} \langle f^{\beta_0}_1, \ldots, f^{\beta_0}_{n(\beta_0)}; \alpha^{\beta_0}_1,\ldots \alpha^{\beta_0}_{n(\beta_0)}\rangle \cap D(\kappa) \subseteq \langle g; 0\rangle \cap D(\kappa)$$

But we can pick $$h \in \langle f^{\beta_0}_1,\ldots,f^{\beta_0}_{n(\beta_0)},g; \alpha^{\beta_0}_1, \ldots ,\alpha^{\beta_0}_{n(\beta_0)}, 1\rangle \cap D(\kappa)$$ which is possible as we have a basic open set of $$\kappa^{2^\kappa}$$ so this has non-empty intersection with $$D(\kappa)$$, as $$D(\kappa)$$ is dense.

But this $$h$$ contradicts $$\text{(*)}$$ as $$h$$ is in the left hand side but $$h(g) = 1$$ ensures it is not in the right hand side. This contradiction shows that $$w(D(\kappa)) < 2^\kappa$$ is false, so $$w(D(\kappa)) = 2^\kappa$$, and $$D(\kappa)$$ is thus a Tychonoff space of size $$\kappa$$ (without isolated points, BTW, check this) with maximal weight. $$D(\omega)$$ is a nice counter example to a lot of hypotheses. It's hereditarily Lindelöf (normal) and hereditarily separable and all points are $$G_\delta$$ but has weight equal $$\mathfrak{c}$$, and it's nowhere first countable.

• Do you know an example of a topological space $X$ with weight $2^{|X|}$? Apr 13, 2017 at 13:28
• A homogeneous ultrafilter on $X$ with \emptyset$added is such a topology. @Anguepa Apr 13, 2017 at 20:51 • I can't find the definition of homogeneous ultrafilter Apr 14, 2017 at 0:13 • If homogeneous we get at least$w(X) > |X|$. We can get one with$w(X) = 2^{|X|}$using an independent family of size$2^{|X|}$. I'll maybe add details later ,but it's a pure set theory result. Apr 14, 2017 at 5:12 • You could get another example using Hewitt-Marczewski-Pondiczery: a dense set of size$X$of$D(X)^I$, where$I$has size$2^{|X|}$, will also do I think, and$D(X)$is the discrete topological space on the set$X\$. Apr 14, 2017 at 5:16