I've been interested lately in topologies whose definitions are given by generating sets, and I was trying to ascertain a few properties of theirs. However since I am unsure how a general open set looks like, I was wondering whether I can test some of the properties just by using base elements, for example separation axioms. I came to the conclusion that one can check whether a space is $T_0$\ $ T_1$\ $T_2$ just by checking the definition with basis elements. But I tend to find out later that conclusions which I come up with, are sometimes not true so I was hoping to see whether the argument seems valid to other people.

For example, written below is an argument for the case of the space being $T_1$:

Let $(X,\tau)$ be a toplogical space with a basis $\mathcal{B}$. Then $X$ is $T_1$ if and only if for any distinct $x,y\in X$ there exist $U,V\in \mathcal{B}$ such that: (i) $x\in U$ and $y\notin U$, (ii) $y\in V$ and $x\notin V$.

This is different from the standard definition just by the fact that we can restrict ourselves to talking about sets in $\mathcal{B}$. Also one implication is trivial since $\mathcal{B}\subset\tau$, but the other implication is also short by the properties of a topological basis.

I was wondering if this argument is true? Also, this sort of argument seem to translate naturally to determining $T_0$ and $T_2$ if it is true, but would appreciate anyone pointing out if this is not case.


Indeed, the similar test can be formulated (and holds) for $T_0$ and $T_2$. Basically because of the fundamental property of a base

$$\forall O \text{ open } \forall x \in O: \exists B \in \mathcal{B}: x \in B \subseteq O\tag{1}$$

and smaller sets (from the base) often work just as well as the larger one.

So if $X$ is $T_0$ and $x \neq y$ we have some open set $O$ that contains $x$ but not $y$ or vice versa. Suppose WLOG $x \in O$, $y \notin O$. Then the same holds for the $B$ from the base that fulfills $x \in B \subseteq O$.

So $$X \text{ is } T_0 \iff \forall x \neq y: \exists B \in \mathcal{B}: |B \cap \{x,y\}| = 1$$

And similarly, for $T_2$, disjoint open sets around $x$ and $y$ also give us smaller (so still disjoint) base sets around the same points. Hence

$$X \text{ is } T_2 \iff \forall x \neq y: \exists B_x,B_y \in \mathcal{B}: B_x \cap B_y = \emptyset$$

Even $T_3$ has a base-formulation:

$X$ is $T_3$ iff for all $x \in X$ and for every basic open $B$ with $x \in B$ we can find a basic open $B'$ such that $x \in \overline{B'} \subseteq B$.

The proof is obvious.

I don't think we can go higher than that with pure base reformulations (so $T_4,T_5,T_6$) but I could be proven wrong: I have no concrete counterexample yet. Those properties are more complex and do need consideration of all open sets and not just basic ones. Their behaviour wrt products is also more complex, which hints to that. Vague, I know, but that's my hunch.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.