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Let be $\langle P,\preceq\rangle$ is a poset so that let's we put $$ {\uparrow\!\!x}=\{y\in P:x\preceq y\} \quad\text{and}\quad {\downarrow\!\!x}=\{y\in P:y\preceq x\} $$ for any $x$ in $P$: (it seems to me that) in this answer the professor Scott claims the collections $$ \mathcal U:=\{P\,\setminus\!\downarrow\!\!x:x\in P\} \quad\text{and}\quad \mathcal D:={P\,\setminus\uparrow\!\!x}=\{y\in P:x\preceq y\} $$ are a subbase for a topology on $P$ which are called respectively upper topology and lower topology.

Now if $\mathcal U$ was a subbase for then collection $$ \mathscr N(\mathcal U):=\biggl\{\bigcap\mathcal S:(\mathcal S\subseteq\mathcal U)\wedge(|\mathcal S|<\aleph_0)\biggl\} $$ was a base so that any $x$ in $P$ is element of some $U$ in $\mathcal U$; however if $\preceq$ is not linear then it seems to me that $\mathscr N(\mathcal U)$ is not a cover: indeed, a collection is a subbase for any topology iff it is a cover so that $\mathcal U$ is a subbase only if it is a cover but I was not able to prove it. Could someone help me, please?

Antonio Maria Di Mauro
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  • Maybe he meant that this collection generates the topology? This will not be a cover if $P$ is an infinite unbounded set. – Keen-ameteur Sep 03 '24 at 13:00
  • The definition of $\mathcal{D}$ is wrong, it should be analogous to the definition of $\mathcal{U}$. – Mentastin Sep 03 '24 at 13:07
  • @Mentastin You are right: I just edited the question. – Antonio Maria Di Mauro Sep 03 '24 at 13:23
  • @Keen-ameteur Unfortunately I cannot say what Scott actually want mean: indeed, if $P$ is ${1,2,3}$ then $3$ is not an element of $\mathscr N(\mathcal U)$, right? – Antonio Maria Di Mauro Sep 03 '24 at 13:25
  • Maybe you want ${ 3}$ to be in $\mathcal{N}(\mathcal{U})$? $\mathcal{N}(\mathcal{U})$ should be a subset of the power set of $P$. Also, assuming you took the usual upper topology on ${1,2,3}$, you have ${1,2,3}\in \mathcal{N}(\mathcal{U})$. So $\mathcal{N}(\mathcal{U})$ should in fact be a cover then. – Keen-ameteur Sep 03 '24 at 13:53
  • You're right: indeed ${2,3}={1,2,3}\setminus(\leftarrow,1]$, but $1$ is not in $\mathscr N(\mathcal U)$. – Antonio Maria Di Mauro Sep 03 '24 at 13:56

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The upper topology is the smallest topology that contains $\mathcal{U}$. There is no need for $\mathcal{U}$ to cover $P$.

Note that $\mathcal{U}$ covers $P$ if and only if $P$ does not have a least element.

Edit: Here is some more information about the meaning of subbase:

  • Every subset $X\subseteq \mathcal{P}(P)$ of the powerset is contained in a least topology.
  • To construct this topology, the first step is to add finite intersections of sets in $X$. Then we add arbitrary unions of such sets.
  • This yields a set $Y\supseteq X$ that is indeed closed finite intersections and arbitrary unions. For it to be a topology, we need $\emptyset, P \in Y$.
  • If we want, we can just add $\emptyset, P$ which necessarily gives us a topology. Clearly every set we added in this construction is necessary, so we have also constructed the least topology that contains $X$.
  • But actually, this last step is often superfluous, since $\emptyset, P$ are often already in $Y$. We have $\emptyset \in Y$ since $\emptyset = \bigcup \emptyset$ and by convention "arbitrary union" also means "union over zero sets". I do not particularly like this convention (even though it is standard) because the corresponding convention for $\bigcap$ does not work.
  • The step of adding $P$ is superfluous if and only if $\bigcup X = P$. Hence your covering requirement. This is purely a matter of convenience. Some people assume it when talking about "subbase" and some not. The difference is immaterial since we can always just add $P$ manually.
  • Note that adding $\emptyset, P$ does not allow us to create any new sets and so the result is still closed under arbitrary unions and finite intersections.
Mentastin
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  • Here is explained why a collection must be formally a cover to be a subbase: indeed, the symbol $\bigcap\emptyset$ has no meaning in ZFC. – Antonio Maria Di Mauro Sep 03 '24 at 13:22
  • There is no covering requirement. Every set of subsets of $P$ is contained in a least topology on $P$. The only reason people add the covering requirement is so that this topology can be constructed by taking finite intersections and arbitrary unions. If it makes you happier, feel free to use the subbase $\mathcal{U} \cup {P}$. – Mentastin Sep 03 '24 at 16:45
  • Well, I use Engelking's subbase definition: if $\mathcal T$ is a topology for a set $X$ then any subcollection $\mathcal S$ of $\mathcal T$ is a subbase for $\mathcal T$ iff the family of finite intersections is a base for $\mathcal T$ -see General Topology 1.1 by Ryszard Engelking. So with respect the previous definition $\mathcal S$ must be a cover -provided you are using ZF: – Antonio Maria Di Mauro Sep 03 '24 at 19:11
  • however we can relax the previous definition saying that $\mathcal S$ is a subbase for $\mathcal T$ if the collection given by $X$ and the finite intersections of $\mathcal S$ is a base for $\mathcal T$ (right?) and actually is what you above implicitely done, right? – Antonio Maria Di Mauro Sep 03 '24 at 19:11
  • P.S. I don't understand completely what $Y$ is in your answer since first you wrote $X\susbeteq Y$ and then $X\in Y$. – Antonio Maria Di Mauro Sep 03 '24 at 19:13
  • Indeed. There are two common definitions of subbase (as was highlighted in the link you gave). A subbase in the more general sense becomes one in the more restrictive sense if you add the full set $P$. – Mentastin Sep 03 '24 at 20:35
  • The set $Y$ is the set $X$ closed under first finite intersections and then arbitrary unions. Both are subsets of $\mathcal{P}(P)$. We do not have $X\in Y$. (When I say that the topology contains $X$ I mean $X \subseteq \mathcal{T}$, not $X\in \mathcal{T}$). – Mentastin Sep 03 '24 at 20:37
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    Finally, I am slightly worried the fixation on ZF(C) imight be hiding a misunderstanding. Using a set theory with proper classes (like MK or BNG) might give meaning to $\bigcap \emptyset$, but not in a way that would be relevant here. The convention $\bigcap \emptyset = P$ actually arises when $\bigcap, \bigcup$ are are considered the meet and join in the complete lattice $\mathcal{P}(P)$. This does not have much to do with the underlying set theory. – Mentastin Sep 03 '24 at 20:44