1 Introduction

The collection $\Sigma \left(X\right)$ of all topologies on a given set $X$ forms a lattice under inclusion, in which the meet of two topologies is their intersection, while the join is the topology with their union as a sub-basis. This lattice has been an object of study since it was first defined by Birkhoff [1]: see the survey article by Larsen and Andima, [4], for more details.

In this paper we are concerned with the local structure of this lattice. Specifically, we are looking at the problem of determining the finite lattices $L$ such that there exist some set $X$ and some $\sigma ,\tau \in \Sigma \left(X\right)$ (perhaps with some specific separation properties) such that the interval $[\sigma ,\tau ]=\{\mu \in \Sigma \left(X\right)|\sigma ≔\mu ≔\tau \}$ is isomorphic to $L$.

This problem was solved for $T_1$ topologies $\sigma$ and $\tau$ by Valent and Larson and Rosický: Valent and Larson [7] proved that any finite distributive lattice can be realized as an interval between $T_1$ topologies, and Rosický [5] proved that any finite interval between $T_1$ topologies must be distributive. Hence a finite lattice can be realized as such an interval if and only if it is distributive.

Of course, Rosický's result implies that every finite interval between Hausdorff topologies must be distributive. In this paper we will show that the converse is true. This result has already been shown by Good, McIntyre and Watson in [3]—indeed, the topologies $\sigma$ and $\tau$ can be assumed to be $T_3$—under the assumption of the existence of infinitely many measurable cardinals. In this paper we will make no set-theoretic assumptions beyond ZFC.

2 Notation, terminology and basic results

For $P$ a partially ordered set, let $O\left(P\right)$ denote the set of down-closed subsets of $P$, partially ordered by inclusion. For $L$ a finite lattice let $J\left(L\right)$ denote the set of join-irreducible elements (i.e. elements $a$ such that $a$ is not the least element of the lattice and, if $a=b\vee c$ then $a=b$ or $a=c$). Recall that a finite lattice $L$ is distributive if and only if $L\cong O\left(P\right)$ for some $P$, which happens if and only if $L\cong O\left(J\left(L\right)\right)$.

If $F$ is a family of subsets of a set $X$ with the finite intersection property, let $⟨⟨F⟩⟩$ denote the filter on $X$ which has $F$ as a filter subbasis. We will say that $A$ is compatible with $F$ if $F\cup \left\{A\right\}$ has the finite intersection property.

Let $u$ denote the least cardinal of a filter subbasis for a free ultrafilter on $\omega$. It is easy to show that $\omega <u≔c$. For more details, and relationships between $u$ and other small cardinals, see [6]

If $p$ and $q$ are ultrafilters on sets $A$ and $B$ respectively, let $p·q$ denote the ultrafilter $\{S\subseteq A\times B|\{a\in A|S_a\in q\}\in p\},$ where $S_a=\{b\in B|⟨a,b⟩\in S\}$. We define ultrafilters $p^n$ on $A^n$ for $n\ge 1$ by $p^1=p$, $p^{n+1}=p^n·p$. Note that this product is associative: if $p$, $q$ and $r$ are ultrafilters, then $p·(q·r)=(p·q)·r$.

If $X$ is a topological space, then an o-filter on $X$ is a filter in the partial order of non-empty open sets of $X$, and an o-ultrafilter is a maximal o-filter.

For $\mu$ a topology on $X$ and $p\in X$, let $N_{\mu }\left(p\right)$ denote the neighbourhood filter at $p$ in the topology $\mu$, and let $N_{\mu }^o\left(p\right)$ denote the o-filter of open neighbourhoods of $p$ in the topology $\mu$ (in other words, $N_{\mu }^o\left(p\right)=N_{\mu }\left(p\right)\cap \mu$). For

$F$ a filter on $X$ and $A\subseteq X$, let $F\upharpoonright A$ be the trace of $F$ on $A$, in other words the family $\{F\cap A|F\in F\}$. Notice that $N_{\mu }\left(p\right)\upharpoonright A$ is a proper filter (i.e. does not contain $\varnothing$) if and only if $p\in A\mu$, and similarly for $N_{\mu }^o\left(p\right)\upharpoonright A$.

If $\sigma$ is a topology on $X$ and $A\subseteq X$, let $⟨\sigma ,A⟩$ denote the topology which has $\sigma \cup \left\{A\right\}$ as a subbasis.

Good, McIntyre and Watson proved the following result in [3]:

Suppose that $P$ is a finite partially ordered set, and $Q\subseteq P$. Let $X=\left\{p\right\}\cup {\cup }_{a\in P}{S}_a$ and $\sigma ,\tau$ be as above. Put $Y=\left\{p\right\}\cup {\cup }_{a\in Q}{S}_a$. Then $S_a\sigma \upharpoonright Y=S_a\sigma \cap Y$ and $N_{\sigma \upharpoonright Y}^o\left(p\right)\upharpoonright S_a=N_{\sigma }^o\left(p\right)$ for $a\in Q$. Thus $(Y,\sigma \upharpoonright Y)$ has the form required by Lemma 1 to ensure that $[\sigma \upharpoonright Y,\tau \upharpoonright Y]\cong O\left(Q\right)$.

Thus, to show that $O\left(P\right)$ can be realized, it is enough to show that $O\left(P\prime \right)$ can be realized for some $P\prime$ containing $P$. We will show in the next section that, for every $n\in \omega$, $O\left(2^n\right)$ can be realized as an interval between $T_2$ topologies.

Now, for each $\alpha \in u$, $B_{\alpha }\notin ⟨⟨\left\{B_{\beta }\right|\beta <\alpha \}⟩⟩$, so $\omega \backslash B_{\alpha }$ is compatible with $\left\{B_{\beta }\right|\beta <\alpha \}$. However, $\left\{B_{\beta }\right|\beta <\alpha \}\cup \{\omega \backslash B_{\alpha }\}$ is too small to be an ultrafilter subbasis, so there exist disjoint sets $A_0$ and $B_0$ which are compatible with $\left\{B_{\beta }\right|\beta <\alpha \}\cup \{\omega \backslash B_{\alpha }\}$. Put $S_{\alpha ,0}=(\omega \backslash B_{\alpha })\cap A_0$. Since $\left\{B_{\beta }\right|\beta <\alpha \}\cup \{\omega \backslash B_{\alpha },B_0\}$ is too small to be an ultrafilter subbasis, there exist disjoint sets $A_1$ and $B_1$ which are compatible with $\left\{B_{\beta }\right|\beta <\alpha \}\cup \{\omega \backslash B_{\alpha },B_0\}$. Put $S_{\alpha ,1}=(\omega \backslash B_{\alpha })\cap B_0\cap A_1$. Continuing in this way we construct a sequence $⟨S_{\alpha ,n}|n\in \omega ⟩$ of disjoint subsets of $\omega \backslash B_{\alpha }$, each of which is compatible with $\left\{B_{\beta }\right|\beta <\alpha \}$, as required.  _

3 The main result

Throughout this section, $A$ will denote some fixed finite set, and $P=\mathbb{P}A$, ordered by reverse inclusion. We will construct sets and ultrafilters as in Lemma 1 to show that $O\left(P\right)$ can be realized as an interval between $T_2$ topologies.

For each $a\in P$, let $X_a=u^{\left|a\right|}\times \omega \times \left\{a\right\}$. Let $p$, $⟨B_{\alpha }|\alpha <u⟩$ and $\left\{S_{\alpha ,n}\right|\alpha \in u,n\in \omega \}$ be as in Lemma 2. Let $f:\omega \times P\to \omega$ be a bijection. For each $\alpha \in u$, $n\in \omega$ and $a\in P$ let $S_{\alpha ,n,a}\prime =S_{\alpha ,f(n,a)}$ and let $q_{\alpha ,n,a}$ be a free ultrafilter extending $⟨⟨\left\{B_{\beta }\right|\beta <\alpha \}\cup \left\{S_{\alpha ,n,a}\prime \right\}⟩⟩$. Let $r$ be a uniform ultrafilter on $u$ (in other words, an ultrafilter extending the co-$<u$ filter on $u$). Let $x_0$ be some point which is not an element of ${\cup }_{a\in P}{X}_a$. For $n\in \omega$ and $s\in u^{n+1}$, let $l\left(s\right)=s\left(n\right)$ (the “last” element of the sequence $s$), and let $\underset{}{\overset{̄}{s}}=s\backslash \left\{⟨n,l\left(s\right)⟩\right\}$. For $a\in P\backslash \left\{\varnothing \right\}$ let $c\left(a\right)=\{a\backslash \left\{x\right\}|x\in a\}$ be the set of covers of $a$ in $P$.

We will specify a topology $\sigma$ on $X=\left\{x_0\right\}\cup {\cup }_{a\in P}{X}_a$ by specifying a weak neighbourhood system for $\sigma$, in other words a collection $B\left(x\right)$ of subsets of $X$ containing $x$ for each $x\in X$, such that a subset $U$ of $X$ is open if and only if for every $x\in U$ there is some $B\in B\left(x\right)$ with $B\subseteq U$.

For $x=⟨\varnothing ,n,\varnothing ⟩\in X_{\varnothing }$, let $B\left(x\right)=\left\{\left\{x\right\}\right\}$.

For $x=⟨s,n,a⟩\in X_a$ (where $a\in P\backslash \left\{\varnothing \right\}$), let $B\left(x\right)=\left\{N(x,S)\right|S\in q_{l\left(s\right),n,a}\}$, where $N(x,S)=\left\{x\right\}\cup \left\{⟨\underset{}{\overset{̄}{s}},m,b⟩\right|m\in S,b\in c\left(a\right)\}.$

Finally, let $B\left(x_0\right)=\left\{N(x_0,S)\right|S\in r^{\left|A\right|}·p\}$, where $N(x_0,S)=\left\{x_0\right\}\cup \left\{⟨s,n,A⟩\right|⟨s,n⟩\in S\}.$

It is easy to see that all the families $B\left(x\right)$ are closed under finite intersections, so $\left\{B\left(x\right)\right|x\in X\}$ forms a weak neighbourhood system for a topology $\sigma$ on $X$. It is also clear that for each $a\in P$, $X_a\sigma =\left\{x_0\right\}\cup {\cup }_{b≔a}{X}_b$.

For each $s\in u^n$ (where $n≔\left|A\right|$), let $Y_s=\left\{s\right\}\times \omega \times P_n$, where $P_n=\{a\in P|\left|a\right|=n\}$. We will prove by induction on $n$ that $Y_s$ can be covered by a disjoint family $\left\{V_x\right|x\in Y_s\}$ of open sets such that $x\in V_x\subseteq U_x$. The base step is trivial, since $Y_{\varnothing }=X_{\varnothing }$ contains only isolated points. So suppose the result holds for all $s\prime \in u^n$, and that $s\in u^{n+1}$. For each $x=⟨s,n,a⟩\in Y_s$ let $V_x=\left\{x\right\}\cup \cup \left\{V_{⟨\underset{}{\overset{̄}{s}},m,b⟩}\right|m\in S_{l\left(s\right),n,a}\prime ,b\in c\left(a\right)\}.$ Then $x\in V_x\subseteq U_x$, and $V_x$ is open. It remains only to show that these sets are disjoint. So let $x=⟨s,l,a⟩$ and $y=⟨s,m,b⟩$ be distinct elements of $Y_s$, and suppose that $z=⟨t,k,c⟩\in V_x\cap V_y$. Then we must have $z\ne x$ and $z\ne y$, so for some $i\in S_{l\left(s\right),l,a}\prime$ and some $d\in c\left(a\right)$, $z\in V_{⟨\underset{}{\overset{̄}{s}},i,d⟩}$ and similarly for some $j\in S_{l\left(s\right),m,b}$ and some $e\in c\left(b\right)$, $z\in V_{⟨\underset{}{\overset{̄}{s}},j,e⟩}$. By inductive hypothesis, then, we must have $⟨\underset{}{\overset{̄}{s}},i,d⟩=⟨\underset{}{\overset{̄}{s}},j,e⟩$, so $i\in S_{l\left(s\right),l,a}\prime \cap S_{l\left(s\right),m,b}\prime$. But these sets are disjoint, unless $⟨l,a⟩=⟨m,b⟩$, in which case $x=y$, a contradiction.

Now let $x=⟨s,m,a⟩$ and $y=⟨t,n,b⟩$ be distinct elements of $X\backslash \left\{x_0\right\}$. Without loss of generality we assume that $\left|a\right|≔\left|b\right|$. We will consider separately two cases, depending on whether or not $s=t\upharpoonright \left|a\right|$.

Case 1

$s=t\upharpoonright \left|a\right|$. Let $W_1={\cup }_{z\in Y_s\backslash \left\{x\right\}}{V}_x$, $W_2=\left\{⟨t\upharpoonright \left|c\right|,l,c⟩\right|l\in \omega ,\left|a\right|<\left|c\right|≔\left|b\right|\}$, and $W=W_1\cup W_2$. Notice that if $⟨u,l,c⟩\in W_2$ with $\left|c\right|>\left|a\right|+1$ then $\left\{\underset{}{\overset{̄}{u}}\right\}\times \omega \times c\left(c\right)\subseteq W_2$, and if $⟨u,l,c⟩\in W_2$ with $\left|c\right|=\left|a\right|+1$ then $\left\{\underset{}{\overset{̄}{u}}\right\}\times (\omega \backslash \left\{m\right\})\times c\left(c\right)\subseteq W_1$. Thus $W$ is open, so $x$ and $y$ are separated by the disjoint open sets $V_x$ and $W$.

Case 2

$s\ne t\upharpoonright \left|a\right|$. Let $l$ be the least element of $\left|a\right|$ such that $s\left(l\right)\ne t\left(l\right)$. Let $\alpha =s\left(l\right)$, $\beta =t\left(l\right)$. If $\alpha <\beta$, let $Z_0=\omega \backslash B_{\alpha }$ and let $W_0=B_{\alpha }$. If $\beta <\alpha$ let $Z_0=B_{\beta }$ and let $W_0=\omega \backslash B_{\beta }$. Either way, $Z_0$ and $W_0$ are disjoint subsets of $\omega$, and for every $⟨s\upharpoonright l+1,k,c⟩\in U_x\cap Y_{s\upharpoonright l+1}$, $Z_0\in q_{\alpha ,k,c}$, while for every $⟨t\upharpoonright l+1,k,c⟩\in U_y\cap Y_{t\upharpoonright l+1}$, $W_0\in q_{\beta ,k,c}$. Put $Z_1=\cup \left\{V_{⟨s\upharpoonright l,k,c⟩}\right|k\in Z_0,c\in P_l\}$, and put $W_1=\cup \left\{V_{⟨t\upharpoonright l,k,c⟩}\right|k\in W_0,c\in P_l\}$. Put $Z_2=\left\{⟨s\upharpoonright \left|c\right|,k,c⟩\right|l<\left|c\right|≔\left|a\right|\}$ and $W_2=\left\{⟨t\upharpoonright \left|c\right|,k,c⟩\right|l<\left|c\right|≔\left|b\right|\}$. Then, as in the previous case, $Z=Z_1\cup Z_2$ and $W=W_1\cup W_2$ are open sets. Moreover, since $Z_0$ and $W_0$ are disjoint, $Z_1$ and $W_1$ are disjoint, while since $s\left(l\right)\ne t\left(l\right)$, $Z_2$ and $W_2$ are disjoint. Thus $Z$ and $W$ are disjoint open sets separating $x$ and $y$.

To complete the proof that $\sigma$ is a $T_2$ topology, we must show that we can separate $x_0$ from any other element of $X$. So let $y=⟨s,n,a⟩\in X_a$. If $a=\varnothing$ then (since $\omega \backslash \left\{n\right\}$ in $q_{\alpha ,m,b}$ for all $\alpha$, $m$ and $b$, $X\backslash \left\{y\right\}$ is open, so $x_0$ and $y$ are separated by the disjoint open sets $X\backslash \left\{y\right\}$ and $\left\{y\right\}$. So suppose that $\left|a\right|>0$. Let $\alpha =s\left(0\right)$. Let $Z=\left\{x_0\right\}\cup \left\{⟨t,m,b⟩\right|b\in P\backslash \left\{\varnothing \right\},t\in u^{\left|b\right|},t\left(0\right)>\alpha \}\cup \left\{⟨\varnothing ,m,\varnothing ⟩\right|m\in B_{\alpha }\}$ and let $W=(U_y\backslash Y_{\varnothing })\cup \left\{⟨\varnothing ,m,\varnothing ⟩\right|m\in \omega \backslash B_{\alpha }\}.$ Then $Z$ and $W$ are disjoint open sets separating $x_0$ and $y$.  _

1. $T_l=T$;
2. if $l≔i<\left|A\right|$ then for every $⟨s,n⟩\in T_{i+1}$ and every $b\in P_{i+1}$, $\{m\in \omega |⟨\underset{}{\overset{̄}{s}},m⟩\in T_i\}\in q_{l\left(s\right),n,b}$; and
3. $T_i\in r^i·p$ for each $i$.

If we then take $U={\cup }_{\left|b\right|<a}{X}_b\cup \left({\cup }_{i=l}^{\left|A\right|},T_i,\times ,P_i,\cup ,\left\{x_0\right\}\right),$ then $U$ will be the open set we require.

So, suppose we have chosen $T_l,\dots ,T_k$ satisfying (1)-(3), and we wish to choose $T_{k+1}$. For $s\in u^k$ let ${\left(T_k\right)}_s=\{n\in \omega |⟨s,n⟩\in T_k\}$. Let $S=\{s\in u^k|{\left(T_k\right)}_s\}\in p$. Then, since $T_k\in r^k·p$, $S\in r^k$. For each $s\in S$ choose ${\alpha }_s\in u$ with $B_{{\alpha }_s}\subseteq {\left(T_k\right)}_s$. Let $S\prime =\{s\cup \left\{⟨k,\beta ⟩\right\}|s\in S,{\alpha }_s<\beta <u\}.$ Then, since $\left\{\beta \right|{\alpha }_s<\beta <u\}$ is in $r$ for each $s\in S$, $s\prime \in r^k·r=r^{k+1}$. Put $t_{k+1}=S\prime \times \omega$. Then $\{s\in u^{k+1}|{\left(T_{k+1}\right)}_s\in p\}=S\prime \in r^{k+1}$, so $T_{k+1}\in r^{k+1}·p$. Finally, suppose that $⟨s,n⟩\in T_{k+1}$ and $b\in P_{k+1}$. Then $s\in S\prime$, so $\underset{}{\overset{̄}{s}}\in S$ and ${\alpha }_{\underset{}{\overset{̄}{s}}}<l\left(s\right)$. Thus $B_{{\alpha }_{\underset{}{\overset{̄}{s}}}}\in q_{l\left(s\right),n,b}$ and $B_{{\alpha }_{\underset{}{\overset{̄}{s}}}}\subseteq {\left(T_k\right)}_{\underset{}{\overset{̄}{s}}}$, so $\{m\in \omega |⟨\underset{}{\overset{̄}{s}},m⟩\in T_k\}\in q_{l\left(s\right),n,b}$, as required.  _ Thus, putting $\tau =⟨\sigma ,\left\{x_0\right\}⟩$, we have that $[\sigma ,\tau ]$ is isomorphic to $O\left(P\right)$, as claimed. Now, since any finite partial order can be embedded in an order of the form $\mathbb{P}A$ (ordered by reverse inclusion) for some finite set $A$, we have proved the following:

4 Conclusions and questions

The construction given in this paper shows that any finite distributive lattice can be realized as an interval between $T_2$ topologies on a set of size $u$. It clearly does not yield a regular topology: the closure of any neighbourhood of $x_0$ will contain co-$<u$ many elements of each $X_a$, not just $r^{\left|a\right|}·p$ many. Two obvious directions for possible improvements on this result are to ask for more separation on the topologies or to ask for smaller sets.

In [3] it is shown that the distributive lattice

Acknowldegements

The work presented in this paper is one outcome of a mini workshop organised by the Mathematics Department of the University of Auckland in April 1996. The first author would like to acknowledge the financial assistance of the Mathematics Department of Auckland University, without which his attendance would not have been possible. The second author was present on a Royal Society Postdoctoral Fellowship. These authors wish to sincerely thank the Mathematics Department for their warm hospitality.

References