Topology - Separation Axioms

Question

A quick question that will help better clarify the separation axioms:

On a set with three elements, there are nine inequivalent topologies (listed below). Which of these topologies are $T_{0}$, $T_{1}$, $T_{2}$, $T_{3}$ (regular, or not), and $T_{4}$ (normal, or not)?

Let $X=\{a, b, c\}$ be a set with 3 elements.

1. {$\emptyset$, {a, b, c}}

2. {$\emptyset$, {c}, {a, b, c}}

3. {$\emptyset$, {a, b}, {a, b, c}}

4. {$\emptyset$,{c}, {a, b}, {a, b, c}}

5. {$\emptyset$, {c}, {b, c}, {a, b, c}}

6. {$\emptyset$, {c}, {a, c}, {b, c}, {a, b, c}}

7. {$\emptyset$, {a}, {b}, {a, b}, {a, b, c}}

8. {$\emptyset$, {b}, {c}, {a, b}, {b, c}, {a, b, c}}

9. {$\emptyset$, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}

So, I already know that $T_{2}$ implies $T_{1}$, which implies $T_{0}$. Also, that $T_{3}$ and $T_{4}$ do not ensure the previous three separation axioms (unless these two are regular and normal, respectively). I think that identifying these simple topologies would better clarify the meaning of each, though. Thoughts?

Here is a quick link to the separation axioms for ease.

Answer 1

For example, a space is $T_0$ if for every pair of distinct points in the space, at least one has an open neighborhood not containing the other.



There are three unordered pairs, $\{a,b\}, \{a,c\}, \{b,c\}$

$1)$ not $T_0$
$2)$ not $T_0$
$3)$ not $T_0$
$4)$ not $T_0$
$5)$ $T_0$ $\huge !$ Why??

$c$ has a neighborhood that doesn't contain $a$ or $b$ $\checkmark$,
$b$ has a neighborhood that doesn't contain $a$ $\checkmark$ $T_0$-ness is satisfied!

Just apply definitions, and see what you can come up with.

Answer 2

Hint that covers some cases:

If you can find two elements that are not topologically distinguishable in the sense that no open set exists containing exactly one of them then (even) $T_0$ is not true.

Example: in 4) $\left\{ a,b\right\} \subseteq O\vee\left\{ a,b\right\} \cap O=\emptyset$ is true for any open set $O$.

Answer 3

Following your link, you use the standard definitions, where each $T$-axiom implies the lower one.

For a finite space $T_1$ means discrete (as then all finite, hence all, subsets are closed), and only 9. is discrete, so $T_1$-$T_4$.

So 1-8 are not $T_1$ or higher, being non-discrete.

So decide $T_0$-ness for 1-8. Regular and normal are indeed more case to case: e.g. the indiscrete space 1. is trivially regular and normal, as there are no disjoint closed sets to separate at all (so logic says we can separate all such pairs by open sets ...), nor is there a pair $x, F$ with $F$ closed non-empty and $x \notin F$.

E.g. for 2., as $\{c\}$ is open, $\{a,b\}$ is closed, and it's the only non-trivial closed set. So normality is already trivial. But we cannot separate $c$ from $\{a,b\}$ by disjoint open sets, because the only open set that contains $\{a, b\}$ is the whole space. So space 2. is not regular, but is normal. It's not $T_0$ because every open set that contains $a$ also contains $b$ and vice versa.

Now you look at the others.