Jtdylktuy

For a topological space $T=(X, tau)$, then $A subset X$ is $ boxed{quadvphantom{A}quad}$ if there is ${mathscr{T}_alpha } subset tau $ such that

• for every $a in A$ there is $mathscr{T}_a in {mathscr{T}_alpha }$ such that $a in mathscr{T}_a$, and




• for every $a, b in A$ with $a neq b$, then $mathscr{T}_a bigcap mathscr{T}_b = emptyset$.

For example, under the usual topology with $X=[0,1]$, then

$ { frac{1}{n+1} mid n in mathbb{N} }$ is $boxed{quadvphantom{A}quad}$, but

$mathbb{Q} cap (0,1)$ is not $boxed{quadvphantom{A}quad}$.

It implies $A$ is (relatively) discrete, i.e. discrete in the subspace topology. It's clear that $mathcal{T}_a cap A = {a}$. (BTW, "stylistically" it's better to name the sets $O_a$, say, as $mathcal{T}_a$ will suggest a family of sets rather than just one set.)

If we have totally (i.e. also outside of $A$) disjoint $O_a$ as you demand, you could say that the $O_a$ are "simultaneously separated" (there is no standard term): it's the conclusion when a space is so-called collectionwise Hausdorff: a space is called that when every relatively discrete set $A$ has such a separating family of pairwise disjoint open sets.

E.g. all metric spaces have this property: they are paracompact (which implies collectionwise normal which implies collectionwise Hausdorff).
So in a metric space all discrete sets are "simultaneously separated", and there is no real difference. In general a Hausdorff space need not be collectionwise Hausdorff, and there could be a discrete subspace without such a simultaneous separation.