Let $\kappa$ be an infinite cardinal, and let $\text{Top}(\kappa)$ be the lattice of all topologies on $\kappa$, ordered by $\subseteq$. Let $\text{Top}^{T_1}(\kappa)$ be the lattice of all $T_1$-topologies on $\kappa$.

Is there an injective lattice homomorphism $\varphi: \text{Top}(\kappa)\to \text{Top}^{T_1}(\kappa)$?

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    $\begingroup$ Is there a reason you write "ordinal" instead of "cardinal", or indeed simply "set"? $\endgroup$ – Todd Trimble Mar 31 '18 at 18:41
  • $\begingroup$ There is no reason for preferring ordinal over cardinal - will change it. $\endgroup$ – Dominic van der Zypen Apr 1 '18 at 7:50

Is there an injective lattice homomorphism $\varphi: \text{Top}(\kappa)\to \text{Top}^{T_1}(\kappa)$?

The answer is Yes, there is such an embedding.

I will argue that if $\kappa$ is an infinite cardinal, then there is a complete lattice embedding $\varphi: \text{Top}(\kappa)\to \text{Top}^{T_1}(\kappa\times\kappa)$. This is enough to answer the question, for the following reason. Any bijection $\beta:\kappa\times\kappa\to\kappa$ induces a lattice isomorphism $\overline{\beta}: \text{Top}(\kappa\times\kappa)\to \text{Top}(\kappa)$ which maps the cofinite topology on $\kappa\times\kappa$ to the cofinite topology on $\kappa$. A topology is $T_1$ iff it contains the cofinite topology, so $\overline{\beta}$ restricts to a lattice isomorphism from $\text{Top}^{T_1}(\kappa\times\kappa)$ to $\text{Top}^{T_1}(\kappa)$. Thus, any (complete) lattice embedding $\text{Top}(\kappa)\to \text{Top}^{T_1}(\kappa\times\kappa)$ can be altered to a (complete) lattice embedding $\text{Top}(\kappa)\to \text{Top}^{T_1}(\kappa)$ by composing with such a $\overline{\beta}$.

If $U\subseteq \kappa$, then by a cofinite extension of $U$ I mean a subset $X\subseteq U\times \kappa$ where, for each $u\in U$, the set $\{\lambda<\kappa\;|\;(u,\lambda)\in X\}$ is cofinite in $\kappa$. To make sure this is clear, let me explain this a second way using the projection maps $\pi_1, \pi_2\colon \kappa\times\kappa\to\kappa$. $X$ is a cofinite extension of $U$ if (i) $\pi_1(X)=U$, and (ii) for every $u\in U$ we have $\pi_2((\{u\}\times\kappa)\cap X)$ is cofinite in $\kappa$.

If $\tau$ is a topology, let $\widehat{\tau}$ be the collection of all cofinite extensions of sets in $\tau$. I claim that

I. For any topology $\tau$ on $\kappa$, $\widehat{\tau}$ is a $T_1$ topology on $\kappa\times\kappa$.

II. The map $\tau\mapsto \widehat{\tau}$ is a complete lattice embedding of $\text{Top}(\kappa)$ into $\text{Top}^{T_1}(\kappa\times\kappa)$.

These are not hard to prove and they establish the result.

In the following justifications, if $X$ is a cofinite extension of $U$, then I may refer to the fibers of $X$, by which I mean fibers of $X$ under the first projection $\pi_1$. If $x\in\pi_1(X)$, then the fiber of $X$ over $x$ is $(\{x\}\times\kappa)\cap X$, which is a subset of $\kappa\times\kappa$. (So, a cofinite extension of $U\subseteq \kappa$ is a subset if $U\times \kappa$ with cofinite fibers.)

Sketch of proof of I. (Least and largest subsets) The least and largest subsets $\emptyset$ and $\kappa\times\kappa$ of the set $\kappa\times\kappa$ are cofinite extensions of the least and largest subsets $\emptyset$ and $\kappa$ of $\kappa$.

(Finite intersection) If $X, Y\in \widehat{\tau}$, then they are cofinite extensions of some $\pi_1(X)=U, \pi_1(Y)=V\in\tau$. Then $X\cap Y$ is a cofinite extension of $U\cap V\in\tau$, so $X\cap Y\in\widehat{\tau}$.

(Arbitrary union) If $X_i\in\widehat{\tau}$, then they are cofinite extensions of some $U_i\in\tau$. Then $\cup X_i$ is a cofinite extension of $\cup U_i\in\tau$, so $\cup X_i\in\widehat{\tau}$.

($T_1$) Every cofinite subset of $\kappa\times\kappa$ is a cofinite extension of $\kappa$, so any topology of the form $\widehat{\tau}$ on $\kappa\times\kappa$ contains all cofinite sets. This means that any such topology is $T_1$. \\\

Sketch of proof of II. Given topologies $\tau_i$ on $\kappa$ we must argue that
(Inj) the map $\tau\mapsto \widehat{\tau}$ is injective,
(M) $\widehat{\bigcap \tau_i}=\bigcap\widehat{\tau_i}$, and
(J) $\widehat{\bigvee \tau_i}=\bigvee\widehat{\tau_i}$.

The map $\tau\mapsto \widehat{\tau}$ is is easily seen to be order-preserving (and 1-1), so I focus on the claims
(M)' $\widehat{\bigcap \tau_i}\supseteq\bigcap\widehat{\tau_i}$, and
(J)' $\widehat{\bigvee \tau_i}\subseteq\bigvee\widehat{\tau_i}$.

Let's start with (M)'. Choose a set $X\in \bigcap\widehat{\tau_i}$ and let $U=\pi_i(X)$. Then $U\in \bigcap \tau_i$ for all $i$, and $X$ is a cofinite extension of $U$, so $X\in \widehat{\bigcap\tau_i}$.

Now (J)'. Suppose that $X\in \widehat{\bigvee\tau_i}$. Then $X$ is a cofinite extension of some set in $\bigvee\tau_i$, and a typical such set has the form $\bigcup_i (U_{i1}\cap \cdots\cap U_{ik_i})$ where $U_{ij}\in\tau_j$. In will now suffice for us to show that $X$ can also be represented in the form $\bigcup_i (\overline{U}_{i1}\cap \cdots\cap \overline{U}_{ik_i})$ where $\overline{U}_{ij}$ is a cofinite extension of some set in some $\tau_j$. Of course, we will choose $\overline{U}_{ij}$ to be a cofinite extension of the set $U_{ij}\in\tau_j$, but we must explain how to choose the fibers of $\overline{U}_{ij}$. If some $x\in U_{ij}$ also belongs to $\pi_1(X)$, then choose the fiber over $x$ in $\overline{U}_{ij}$ so that it agrees with the fiber over $x$ in $X$, which must be cofinite in $\kappa$. For any other $x\in U_{ij}$ it doesn't matter how you choose the fiber over $x$ in $\overline{U}_{ij}$ except that it must be cofinite. (To be specific, choose this fiber to be all of $\kappa$.)

We have now chosen $\overline{U}_{ij}\in\widehat{\tau_j}$ so that $\bigcup (\overline{U}_{i1}\cap \cdots\cap \overline{U}_{ik_i})$ has the same first projection and the same fibers as $X$, hence it equals $X$. This represents $X$ as an element of $\bigvee \widehat{\tau_i}$. \\\

  • $\begingroup$ Oh wow - I really didn't spot at all the mistake you pointed out! It is very subtle. Thanks for making this effort to write such a beautiful answer!! $\endgroup$ – Dominic van der Zypen Mar 31 '18 at 9:04

Claim: Any such $\varphi$ would have to map into a set on which all homomorphisms of $\text{Top}^{T_1}(\kappa)$ are constant.

Turing award winner Juris Hartmanis This follows from Theorems 1 and 2 of

Hartmanis, Juris, On the lattice of topologies, Can. J. Math. 10, 547-553 (1958). ZBL0087.37403. (See https://cms.math.ca/openaccess/cjm/v10/cjm1958v10.0547-0553.pdf)

Theorem 1 says that $\text{Top}(\kappa)$ has only trivial homomorphisms (they are either constant or embeddings).

Theorem 2 says that $\text{Top}^{T_1}(\kappa)$ does have some nontrivial ones; so let $\psi$ be one of them.

If the Claim failed then $\psi\circ\varphi$ would be a nontrivial homomorphism of $\text{Top}(\kappa)$, contradicting Theorem 1.

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    $\begingroup$ That's brilliant to go via (non-)trivial homomorphisms, thanks for this beautiful answer!! $\endgroup$ – Dominic van der Zypen Mar 14 '18 at 8:07
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    $\begingroup$ A very nice argument. $\endgroup$ – Joel David Hamkins Mar 14 '18 at 11:35
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    $\begingroup$ Sorry I have to unaccept this answer -- Keith Kearnes below spotted a subtle error that escaped me (and Bjorn and @JoelDavidHamkins) $\endgroup$ – Dominic van der Zypen Mar 31 '18 at 9:06
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    $\begingroup$ @BjørnKjos-Hanssen: The weaker claim is also refuted by my answer. $\endgroup$ – Keith Kearnes Mar 31 '18 at 18:11
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    $\begingroup$ @BjørnKjos-Hanssen: There is no contradiction. But let me add that what Hartmanis means is that the lattice of complete CONGRUENCES on $L_{T_1}(S)$ is isomorphic to ... ETC. $\endgroup$ – Keith Kearnes Mar 31 '18 at 20:46

This is a long comment about Bjørn's answer rather than an answer itself.

Hartmanis proves

Thm 1 If $\kappa>2$, then $\text{Top}(\kappa)$ is a simple lattice.

Thm 2 If $\kappa$ is infinite, then

(1) For any finite subset $F\subseteq \kappa$, the restriction map $\rho_F:\tau\mapsto \tau|_{\kappa-F}$ is a nonconstant, noninjective, complete lattice homomorphism from $\text{Top}^{T_1}(\kappa)$ to $\text{Top}^{T_1}(\kappa-F)\;(\cong\text{Top}^{T_1}(\kappa))$.

(2) For any finite subset $F\subseteq \kappa$, $\ker(\rho_F)$ is a proper, complete congruence on $\text{Top}^{T_1}(\kappa)$. Every proper, complete congruence has this form. If $F\neq G$ are distinct finite subsets of $\kappa$, then $\ker(\rho_F)\neq\ker(\rho_G)$. \\\

Theorem 1 implies that if $\varphi:\text{Top}(\kappa)\to\text{Top}^{T_1}(\kappa)$ is an embedding, then for any congruence $\theta$ on $\text{Top}^{T_1}(\kappa)$ one must have either
Case 1. the image of $\varphi$ is contained in a single $\theta$-class, or
Case 2. the image of $\varphi$ is contained in a $\theta$-transversal.
Otherwise $\varphi^{-1}(\theta)$ would be a nontrivial proper congruence of a simple lattice. Thus each congruence $\theta$ on $\text{Top}^{T_1}(\kappa)$ restricts the possibilities for $\varphi$.

Theorem 2 implies that there are lots of $\theta$'s to choose from, namely all of the form $\ker(\rho_F)$.

Thms 1&2 imply that there are lots of restrictions $\varphi$ must satisfy, but there are not enough restrictions to rule out the existence of $\varphi$. In my answer on this page I construct $\varphi$ whose image is contained in a $\theta$-transversal for every $\theta$ of the form $\ker(\rho_F)$.


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