# Is the associated order of a minimal $T_0$ space always total?

Let's call a space $(X,\tau)$ minimal $T_0$ if it is $T_0$ and for every topology $\sigma\subseteq \tau$ with $\sigma \neq \tau$ we have that $(X,\sigma)$ is not $T_0$ any more.

We say $x\leq y$ in a $T_0$-space $(X,\tau)$ if and only if every open neighborhood of $x$ is an open neighborhood of $y$. It's easy to verify that for $T_0$-spaces, this relation is a partial order. We call it the partial order associated to $(X,\tau)$.

Question: if $(X,\tau)$ is minimal $T_0$, is the associated order total?

• An interesting follow up question would be "if $(X,\mathcal{T})$ is a $T_{0}$-space, then does there exist a minimal $T_{0}$-space $(X,\mathcal{S})$ with $\mathcal{S}\subseteq\mathcal{T}$?" – Joseph Van Name Apr 5 '15 at 2:48

This conjecture is correct. I claim that if the topological space is minimal $T_{0}$, then the specialization ordering is total. We shall prove this result by contrapositive, so we shall show that if the specialization ordering is not linear, the the topology is not a minimal $T_{0}$. The idea in this proof is similar to the proof that Joel David Hamkins gave for finite spaces. If the specialization ordering is not a linear ordering, then we want to extend the specialization ordering to a larger partial ordering and give that larger partial ordering a smaller $T_{0}$ topology. However, instead of extending the partial ordering all the way to a linear ordering, we shall extend the partial ordering in a minimal way so that we can control the new partial ordering and new topology since Zorn's lemma does not give us much control over the new topology.

Suppose that $(X,\mathcal{T})$ is a $T_{0}$-space and let $\leq$ denote the specialization ordering on $X$. Suppose that $\leq$ is not a total ordering.

Since $X$ is not total ordering, there are two element $p,q\in X$ with $p\not\leq q$ and $q\not\leq p$. We shall now extend the partial ordering $\leq$ to a larger partial ordering $\preceq$ on $X$ so that $p\preceq q$ in the least way.

Define $\preceq$ to be the relation such that $x\preceq y$ if and only if $x\leq y$ or $x\leq p,q\leq y$. I claim that $\preceq$ is a partial ordering, and the proof that $\preceq$ is a partial ordering is a routine verification.

Reflexivity: The ordering $\leq$ is clearly reflexive.

Transitivity: Now assume that $x\preceq y$ and $y\preceq z$. Then we shall prove that $\preceq$ is transitive in the four following cases:

Case 1. $x\leq y;y\leq z$: Clearly $x\leq z$, so $x\preceq z$.

Case 2. $x\leq y;y\leq p,q\leq z$: In this case, $x\leq p,q\leq z$, so $x\preceq z$

Case 3. $x\leq p,q\leq y;y\leq z$: We have $x\leq p,q\leq z$, so $x\preceq z$

Case 4. $x\leq p,q\leq y;y\leq p,q\leq z$: We have $x\leq p,q\leq z$, so $x\preceq z$.

Symmetry: Suppose that $x\preceq y,y\preceq x$. Then we shall prove $x=y$ in the following four cases:

Case 1. $x\leq y;y\leq x$: Clearly $x=y$.

Case 2. $x\leq y;y\leq p,q\leq x$: This case is impossible for it implies that $q\leq x\leq y\leq p$ hence $q\leq p$ a contradiction.

Case 3. $x\leq p,q\leq y;y\leq x$: This case also impossible since it implies that $q\leq y\leq x\leq p$ which is also a contradiction.

Case 4. $x\leq p,q\leq y;y\leq p,q\leq x$: This case is impossible as well since it implies that $p\leq x\leq q$, a contradiction.

We therefore conclude that $\preceq$ is a partial ordering. Now give $X$ the topology $\mathcal{S}$ generated by the subbasis $(\downarrow^{\preceq}x)^{c}$ where $\downarrow^{\preceq}x=\{r\in X|r\preceq x\}$. Then $\preceq$ is the specialization ordering on $\mathcal{S}$. I now claim that $\mathcal{S}\subseteq\mathcal{T}$ and we shall prove this claim by showing that each downset $\downarrow^{\preceq}z$ is closed in $\mathcal{T}$ in two cases.

Case 1: $q\not\leq z$. Suppose that $q\not\leq z$. Then $x\preceq z$ if and only if $x\leq z$ or $x\leq p,q\leq z$. However, since $q\not\leq z$, we have $x\preceq z$ if and only if $x\leq z$. Therefore, $\downarrow^{\preceq}z=\downarrow^{\leq}z$, so $\downarrow^{\preceq}z$ is closed in $X$.

Case 2: $q\leq z$. Suppose that $q\leq z$. Then $x\preceq z$ if and only if $x\leq z$ or $x\leq p,q\leq z$. However, since $q\leq z$, we conclude that $x\preceq z$ if and only if $x\leq z$ or $x\leq p$. Therefore $\downarrow^{\preceq}z=\downarrow^{\leq}z\cup\downarrow^{\leq}p$, so $\downarrow^{\preceq}z$ is closed in $\mathcal{T}$.

We have now shown that each set $\downarrow^{\preceq}z$ is closed in $(X,\mathcal{T})$, so each set $(\downarrow^{\preceq}z)^{c}$ is open in $(X,\mathcal{T})$. Since $\mathcal{S}$ is generated by the subbasis consisting of the sets of the form $(\downarrow^{\preceq}z)^{c}$, we conclude that $\mathcal{S}\subseteq\mathcal{T}$. Furthermore, we have $\mathcal{S}\neq\mathcal{T}$ since these spaces have different specialization orderings.

$\textbf{Remark}$ The minimal $T_{0}$-spaces are in a one-to-one correspondence with the class of all totally ordered sets since every totally ordered set has a unique minimal $T_{0}$-topology. In particular, if $(X,\leq)$ is a totally ordered set, then the unique minimal $T_{0}$-topology on $X$ is generated by the basis consisting of sets of the form $\{x\in X|x>x_{0}\}$. In particular, if $A\subseteq X$, then $A$ is open if and only if $A$ is an upwards closed set and either $A$ has no least element or $A$ has a least element $a$ and $a$ has an immediate predecessor.

Let me show at least that for a finite space $X$, the answer is yes. Suppose that $(X,\tau)$ is a minimal $T_0$ topology on a finite set $X$, and that $\leq$ is the associated specialization order that you define, which since the space is $T_0$ we know is a partial order, rather than merely a pre-order, as it is for topological spaces in general. Note that every open set is an up-set, but also, the converse is true in a finite space: in a finite space, a set is open if and only if it is an up-set.

Every partial order extends to a total order, so let $\leq^*$ be a total order on $X$ extending $\leq$, and let $\tau^*$ be the associated topology, so that the $\tau^*$ open sets are the up-sets with respect to $\leq^*$. Since every set that is upward closed with respect to $\leq^*$ is also upward closed with respect to $\leq$, it follows that $\tau^*\subset\tau$. And $\tau^*$ is $T_0$, because $\leq^*$ is a total order. By minimality, therefore, we conclude $\tau=\tau^*$, and it follows that $\leq=\leq^*$ and so the specialization order is linear.

• Nice! Maybe the argument can be made to work for infinite spaces as well? – Dominic van der Zypen Mar 31 '15 at 12:44
• Yes, that is what I had hoped, but I haven't yet managed it. – Joel David Hamkins Mar 31 '15 at 13:34
• Joseph, please go ahead and post it! – Joel David Hamkins Mar 31 '15 at 14:07