Upper topology vs. Alexandrov topology

Question

Let $(P, \le)$ be a poset. We define the following two well-known topologies over $P$.

The upper topology has the principle upper sets, that is upper sets of the form $\{\uparrow x : x \in P\}$, as the subbase.

The Alexondrov topology has as base the upper sets, that is sets of the form $\{\uparrow U : \uparrow U \subseteq P\}$.

Now, it is clear that Alexandrov topology is at least as big as the upper topology (as every principle upper set is indeed an upper set, while the converse need not hold).

But I have the following confusion.

Since the upper topology will have all the principle upper sets as open sets, wouldn't then the arbitrary union of these open sets will end up generating the whole Alexandrov topology, as every upper set is a union of some set of principle upper sets?

Or, upper topology is simply presented with upper sets and their intersections, and nothing more? I am obviously missing something here.

Answer 1

The problem is that your definition of the upper topology is wrong: it’s the topology that has as a subbase all sets of the form $P\,\setminus\!\downarrow\!\!x$ for $x\in P$.

If $\langle X,\tau\rangle$ is a topological space, we can define a preorder order $\le_\tau$ on $X$, called the specialization order, by $x\le_\tau y$ if and only if $x\in\operatorname{cl}\{y\}$; $\le_\tau$ is a partial order if $X$ is $T_0$. The upper topology on $P$ is the coarsest topology on $P$ whose specialization order is $\le$, the original order on $P$, and the Alexandrov topology is the finest such topology.

For an example in which the two are distinct, let $P=\Bbb Z\times\Bbb Z$ with the natural product partial order, and let $U=\{\langle m,n\rangle\in P:m\ge -n\}$; clearly $U$ is an upper set and hence open in the Alexandrov topology. Let $p=\langle 0,0\rangle\in U$. If $U$ were open in the upper topology, there would be a finite $F\subseteq P$ such that

$$p\in\bigcap_{x\in F}\left(P\,\setminus\!\downarrow\!\!x\right)\subseteq U\;.$$

Say $F=\{\langle m_k,n_k\rangle:k=1,\ldots,\ell\}$, and let

$$m=\max\{m_k:k=1,\ldots,\ell\}$$

and

$$n=\max\{n_k:k=1,\ldots,\ell\}\;.$$

Then

$$P\,\setminus\!\downarrow\!\!\langle m,n\rangle\subseteq P\setminus\bigcup_{x\in F}\downarrow\!\!x=\bigcap_{x\in F}\left(P\,\setminus\!\downarrow\!\!x\right)\;,$$

but clearly $P\,\setminus\!\downarrow\!\!\langle m,n\rangle\nsubseteq U$, since $P\,\setminus\!\downarrow\!\!\langle m,n\rangle$ contains $\Bbb Z\times\{r\}$ for $r>n$.

Answer 2

Under your definitions, both topologies are the same. Note that the upper sets are non only a base, they form the whole topology. And the principal upper sets are only only a subbase, they from a base.

I have also found another definition of the upper topology: the principal lower sets generate the closed sets. I.e. the open sets has a subbase of form $\{P \setminus \mathop\downarrow x: x ∈ P\}$. This topology may be strictly coarser, but they are the same if the order is linear.