Countably close topological spaces

Question

Does every uncountable topological space has a different and mutually "countably close" space?

Define "countably close" space $S$ with respect to another space $S'$ both on the same set iff the difference between any open set $U\in S$ and some open set $V\in S'$ is countable $|U\setminus V|\leq\aleph_0$

I tried disproving the negation by taking some space and some open set from any other so that their difference is uncountable but still couldn't get it.

Answer 1

The cocountable topology provides a counter-example. In this topology a set is open iff it is cocountable, i.e., its complement is countable. Removing a countable set from a cocountable subset of an uncountable set gives you a cocountable set, so a topology that is countably close to the cocountable topology must be the same as the cocountable topology.

[EDIT] The above was based on a possible misunderstanding that $V \subseteq U$ was required in the original question. I will revisit the answer.

Answer 2

This answer is about symmetric differences rather than differences $U\setminus V$. Let $(X,\tau)$ be an infinite topological space.

Case $1$: $\tau$ is the discrete topology and fix $x\in X$. Let $\sigma_1$ consists of those $U\subset X$ such that $x\notin U$. Let $\sigma_2$ consist of those $U\subset X$ such that $X\setminus U$ is finite. Let $\tau'=\sigma_1\cup \sigma_2$. Then one easily checks that $\tau'$ is a topology.Any member of $\tau'$ is in $\tau$, because everything is in $\tau$. Any member of $\tau$ is at most one point away from being in $\tau'$. Essentially what we've done is taken a topology with no non-isolated points and made a topology with exactly one isolated point.

Case $2$: $\tau$ is not the discrete topology. Then there exists $x\in X$ such that $\{x\}$ is not open. Create a new topology $$\tau'=\tau\cup \{A\cup \{x\}:A\in \tau\}.$$ One easily checks that this is a topology. Every $\tau$-open set is $\tau'$ open. Every $\tau'$ open set is at most one point away from a $\tau$-open set. Essentially what we've done is isolate a non-isolated point.

Answer 3

It is possible to verify that for any topological space $G$ there are some other topologies which make them countably close to each other, if not then by such assumption it would happen that the axioms of any such topology aren't preserved.

Assume by negation that for $G$ there are no different topologies which are countably close, then take another topological space $G'$ and suppose it's countably close to $G$, it must appear that one of the axioms don't hold.

Each of the open sets $U_1, U_2\subset G'$ are countably close to open sets $V_1, V_2\subset G$ and their union must also be countably close, indeed $\big|(U_1\cup U_2)\vartriangle (V_1\cup V_2)\big|\leq \big|(U_1\vartriangle U_2)\cup(V_1\vartriangle V_2)\big|\leq\aleph_0$

Same thing can be said about the intersection again countably close, hence the axioms of a topology are preserved meaning that there must be countably close topological space $G'$ and the initial assumption is refuted.

tl;dr for any topological space $G_{\tau}=(G, \tau)$ "deform" the topology $\tau$ by removing or adding countably many elements from each set $V\in\tau$ turning it into a new topology $\tau'$ and it's axioms will still be satisfied.