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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.

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    $\begingroup$ My only thought about this is that if you find that working out these examples is helpful, you should do it. $\endgroup$
    – Carsten S
    Mar 4 '14 at 11:13
  • $\begingroup$ I don't know how to show it completely. I meant to comment right after posting, but got caught up. @Rustyn So, we know that 1 through 4 are not $T_{0}$. 5 through 9 are $T_{0}$. 1 isn't any of them, correct? $\endgroup$
    – kathystehl
    Mar 4 '14 at 15:51
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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.

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  • $\begingroup$ $T_{3} \rightarrow T_{2}$ and $T_{4} \rightarrow T_{2}$ only hold if $T_{3}$ and $T_{4}$ are regular and normal (i.e. that $T_{2}$ is also true), correct? $\endgroup$
    – kathystehl
    Mar 4 '14 at 15:58
  • $\begingroup$ Also, my book explicitly says that $T_{2}$ does not imply $T_{1}$ or $T_{0}$. I don't mean to be contradictory, but could you explain your reasoning please? Thanks! $\endgroup$
    – kathystehl
    Mar 4 '14 at 15:59
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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 $O$.

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  • $\begingroup$ So, in 2, 3 and 4, points a and b are topologically indistinguishable. Does this imply that, for each of these topologies, none of the separation axioms hold? Or, am I inappropriately extrapolating? $\endgroup$
    – kathystehl
    Mar 4 '14 at 16:00
  • $\begingroup$ Yes, that implies indeed than none of the separation axioms holds. That is also the case in 1 (you did not mention). $\endgroup$
    – drhab
    Mar 4 '14 at 17:40
  • $\begingroup$ Ah, you're right. And, none of these should ever hold for the trivial case. $\endgroup$
    – kathystehl
    Mar 4 '14 at 17:42
  • $\begingroup$ Could $T_{3}$ or $T_{4}$ (excluding $T_{0}$ through $T_{2}$) hold for 2 through 4? $\endgroup$
    – kathystehl
    Mar 4 '14 at 17:43
  • $\begingroup$ You have $T_{i+1}\Rightarrow T_i$ in general for $i=0,1,2,3$. $\endgroup$
    – drhab
    Mar 4 '14 at 17:56
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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.

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