https://en.wikipedia.org/wiki/Topological_indistinguishability
If two points are indistinguishable in a topological space, then they must share the exact same neighborhoods. I'm trying to visualize what it means to be arbitrarily close in a topological sense. It seems to me that in metric spaces, no matter how close an element $y$ is to another element $x$, as long as they're distinct, they're distinguishable. Therefore, arbitrarily close elements are distinguishable.
Therefore, is it true that in metric spaces, two points are indistinguishable iff the distance between them is $0$? The wiki page says that this is true for pseudometric spaces, but not necessarily for metric spaces.
Suppose $X$ is a metric space and $x, y \in X$. I'll also use "$\equiv$" to denote indistinguishable.
As @AnneBauval mentioned, for metric spaces, $x \neq y \implies x \not\equiv y$, which means the contrapositive must be true $x \equiv y \implies x = y$
Therefore, if $x = y \Leftrightarrow x \equiv y$ is not true for metric spaces, it must be the forward implication that throws it off. Therefore, $x = y$ can yield $x \not\equiv y$ or $x \not\equiv y$ can yield $x = y$ to throw off the bi-implication.
How can it possibly be true that two non-distinct elements are distinguishable?
If $U_x$ is an open neighborhood of some $x \in X$, due to singletons not being open in metric spaces, the following statement is true:
$$\forall U_x, \exists y \in U_x: x \neq y$$
In other words, every time an element appears in a topology, it appears with some surrounding points and this must hold true for arbitrarily small $U_x$. However, the following statement is false:
$$\exists y: x \neq y \:\:\text{and}\:\: \forall U_x, y \in U_x$$
, which can be verified due to metric spaces being $T_2$. Because of the false statement, I believe this implies that $x \neq y \implies x \not\equiv y$ and thus $x \equiv y \implies x = y$
Therefore, $x = y \Leftrightarrow x \equiv y$ is true for metric spaces. Thus, for metric spaces, a point can only be indistinguishable to itself. Am I wrong? This sounds like a discrete topology to me, but I'm pretty sure metric topologies can have more indistinguishable elements than the discrete topology. Maybe it's true for both metric and discrete topologies, it's just that discrete topologies have no arbitrarily close points. This makes me wonder if
$$\exists y: x \neq y \:\:\text{and}\:\: \forall U_x, y \in U_x$$
would be a topological generalization of arbitrarily close (though impossible in metric spaces) and
$$\exists x,y: \forall U_x, y \in U_x \:\:\text{and}\:\: \forall U_y, x \in U_y$$
, which is the same as $x \equiv y$, would be a topological generalization of zero distance.
No mystery: