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.

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    $\begingroup$ I interpret "can be" as "Is there an $X$ for which the supremum is greater than $|X|$?" $\endgroup$ Apr 12, 2017 at 13:39
  • $\begingroup$ @MeesdeVries that's correct. I made an edit to make it more clear. $\endgroup$
    – Anguepa
    Apr 12, 2017 at 13:45
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    $\begingroup$ I significantly expanded my answer, check it out $\endgroup$ Apr 16, 2017 at 9:38
  • $\begingroup$ 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. $\endgroup$ Apr 17, 2017 at 16:14
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    $\begingroup$ 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. $\endgroup$ Apr 17, 2017 at 16:32

1 Answer 1


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 $D^I$:

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


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

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.

  • $\begingroup$ Do you know an example of a topological space $X$ with weight $2^{|X|}$? $\endgroup$
    – Anguepa
    Apr 13, 2017 at 13:28
  • $\begingroup$ A homogeneous ultrafilter on $X$ with \emptyset$ added is such a topology. @Anguepa $\endgroup$ Apr 13, 2017 at 20:51
  • $\begingroup$ I can't find the definition of homogeneous ultrafilter $\endgroup$
    – Anguepa
    Apr 14, 2017 at 0:13
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    $\begingroup$ 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. $\endgroup$ Apr 14, 2017 at 5:12
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    $\begingroup$ 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 $X$. $\endgroup$ Apr 14, 2017 at 5:16

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