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TLDR:

(1) what condition defines the class of sets $S$ such that for all $s\in S$ there exists topologies $T_1$ and $T_2$ on $s$ such that $T_1\cup T_2$ is not a topology on $s$?

(2) what condition defines the class of sets $U$ such that for all $u\in U$ the union of any [possibly finite] number of topologies on $u$ is a topology?

(3) for any set $X$, what condition defines the class/set of topologies $T$ on $X$ such that the union of any [possibly finite] number of members of $T$ is a topology on $X$?



Let $\mathbf{set}$ be the class of sets. For all $X\in\mathbf{set}$, let $\mathbf{top}(X)$ be the class/set of all topologies on $X$.

For all $X\in\mathbf{set}$, and topologies $T_1,T_2\in\mathbf{top}(X)$, we have that $T_1\cap T_2\in\mathbf{top}(X)$. However, it is not necessarily the case that $T_1\cup T_2\in\mathbf{top}(X)$.

(Paraphrased from my notes)

Not necessarily, i.e. it may be the case that the union of two topologies is a topology, or it may not. This begs the question when is the union of two topologies a topology, and when isn't it?


Here's my reasoning:

Suppose that every class of topologies contains at least two topologies such that their union is not a topology.

Proposition 1 (false):

$$\forall X\in \mathbf{set}.\exists T_1,T_2\in\mathbf{top}(X):T_1\cup T_2\notin\mathbf{top}(X)$$

Counterexample:

Let $X=\emptyset$.

$\mathbf{top}(X)=\{\{\emptyset\}\}$

$\forall T_1,T_2\in\mathbf{top}(X).T_1=T_2=\{\emptyset\}$

$\therefore \forall T_1,T_2\in\mathbf{top}(X).T_1\cup T_2\in \mathbf{top}(X)\quad\square$

Okay, so maybe it's just the empty set for which this is the case. Not a problem.

Proposition 2 (false):

$$\forall X\in\mathbf{set}\setminus\emptyset.\exists T_1, T_2\in\mathbf{top}(X):T_1\cup T_2\notin\mathbf{top}(X)$$

Counterexample:

Let $X=\{0,1\}$.

$\mathbf{top}(X)=\{\{\emptyset,X\},\{\emptyset,\{0\},X\},\{\emptyset,\{1\},X\},\{\emptyset,\{0\},\{1\},X\}\}$

$\forall T_1,T_2\in \mathbf{top}(X).T_1\cup T_2\in\mathbf{top}(X)\quad\square$

(This also covers the case of singleton sets, which I forgot to mention)


So... if not every set has topologies such that their union isn't a topology, then there must be a collection of sets $S$ such that for all $s\in S$ the union of any two topologies in $\mathbf{top}(s)$ is a topology. I would suppose that there is likewise a class of sets $P$ such that for all $p\in P$ the union of at least two topologies $\mathbf{top}(p)$ is not a topology. The question is: how do I define these classes?

(It is unclear at present whether the collections of such sets constitutes a set or a proper class. Please excuse the inconsistent use of 'set', 'class', and similar set-theoretic terms. I will correct this as soon as I am able.)

Specifically, I would like to show that each of the following is true, and define the class of all sets for which each holds:

(1) There exists a class of sets $S$ such that for all sets $X\in S$, there exist topologies $T_1$ and $T_2$, on $X$, such that $T_1\cup T_2$ is not a topology on $X$.

$$\exists S\subset\mathbf{set}:\forall X\in S.\exists T_1,T_2\in\mathbf{top}(X):T_1\cup T_2\notin\mathbf{top}(X)$$

(2) There exists a class of sets $U$, such that for all sets $X\in U$, the union of any number of topologies on $X$ is a topology on $X$. $$\exists U\subset\mathbf{set}:\forall X\in U.\forall T\in\mathcal{P}(\mathbf{top}(X))\setminus\emptyset.\bigcup_{\tau\in T}\tau\in\mathbf{top}(X)$$

(3) For every set $X$, there is a collection of collections of topologies $V$, on $X$, such that for every element $T\in V$, the union of all members of $T$ is a topology on $X$. $$\forall X\in\mathbf{set}.\exists V\subseteq\mathcal{P}^2(\mathbf{top}(X)):\forall T\in V.\bigcup_{T\in V} T\in\mathbf{top}(X)$$

(1) and (2) can easily be proven via example ((1) using your choice of infinite set and (2) using the examples of the emptyset and/or $\{0,1\}$), but I am unsure about (3).

Edit: Following Henno Brandsma's answer the class of sets specified in (1) is the class of all sets whose cardinality is $>2$, and the class of sets $U$ specified in (2) is its complement, the class of sets whose cardinality is $\leq2$.


Bonus question:

Is there a class of sets $W$ such that for all $X\in W$ the union of any [possibly finite] number of [nontrivial] topologies on $W$ is not a topology on $W$? My gut says 'no', but I'm not sure how to proceed with this one.

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Observation: for any two point set $X$ the set of topologies on $X$ has $4$ members (discrete, indiscrete and two Sierpinski variants) and has the property that any pair of them has a union that is a topology again. So the set of topologies on $X$ is closed under (all/finite) unions.

For a three point set $X$ you can find incompatible topologies whose union is no longer a topology like $\{0\}$ (as the only non-trivial open set on $X=\{0,1,2\}$) for $\mathcal{T}_1$ and for $\mathcal{T}_2$ the topology with non-trivial open sets $\{1\}, \{2\}, \{1,2\}$: the union has both $\{0\}$ and $\{1\}$ in it but not $\{0,1\}$ so is not a topology. This example also shows the same for any larger set, of course, using only finite topologies like these...

So the only sets on which topologies are closed under finite unions are those of size at most $2$...

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