Let $X=\{1,2,3\}$. Find all topology of $X$ which are either $T_0$, or $T_1$, or $T_3$

Here is what I got

$T_1$ is discrete topology

$T_0$ are all 1-one point topologies including

$\{1\},\{2\},\{3\}, \{\{1\},\{1,2\}\},\{\{1\},\{1,3\}\},\{\{1\},\{2,3\}\},\{\{2\},\{1,2\}\},\{\{2\},\{1,3\}\},\{\{2\},\{2,3\}\},\{\{3\},\{1,2\}\},\{\{3\},\{1,3\}\},\{{3\},\{2,3\}\}, \{\{1\},\{1,2\},\{1,3\}\},\{\{2\},\{1,2\},\{2,3\}\},\{\{3\},\{1,3\},\{2,3}\}\}$

$T_3$ I'm not so sure.

I also have a question. I know that if $X$ is $T_1$ then $X$ is $T_0$, but is it true that if $X$ is $T_2$ then $X$ is $T_1$?

  • $\begingroup$ $T_2$ implies $T_1$, yes. If you get no answers, consider making you explanations clearer. $\endgroup$
    – GEdgar
    Feb 24, 2014 at 14:36
  • $\begingroup$ It's "discrete", not "discreet" - two different words. Your answers for $T_0$ are not well-formed, i.e. the brackets are off. $\endgroup$
    – Arno
    Feb 24, 2014 at 19:18
  • $\begingroup$ is that there is no $T_3$ in this problem? because there is no open set in topology of $X$ contain a closed set of $D$ and disjoint to another open set contain an element of $X$? $\endgroup$ Feb 24, 2014 at 19:43

2 Answers 2


Generally, $T_i$ implies $T_j$ for $j < i$. (Although some care is needed, because some people would not necessarily understand a $T_i$ space for $i > 0$ to be $T_0$, too)

So, there are two settings here: $T_0$ spaces with 3 elements, and $T_1$ spaces with 3 elements. Any finite $T_1$ space is already discrete, as open sets are closed under finite intersections. Thus, there is only one $T_1$ topology on $\{1,2,3\}$.

For $T_0$, we can try to construct a minimal $T_0$ topology up to permutation of the elements. It is ok if one element has only $X$ as neighborhood, wlog let that one be 3. Both 1 and 2 need smaller neighborhoods, in particular this implies that $\{1, 2\}$ is open. Then again, for one point this can be the smallest neigherborhood, wlog let that be 2. Then, $\{1\}$ is open, too.

So, the $T_0$ topologies are those that (after some permutation) include the open sets $\{1\}$ and $\{1,2\}$.

  • $\begingroup$ It is not true that $T_i$ implies $T_j$ for $j<i$. For example, $T_3$ does not imply $T_2$. $\endgroup$
    – MJD
    Feb 24, 2014 at 14:53
  • $\begingroup$ Well, as I mentioned, this depends on your favourite terminology, and in particular on whether $T_3$ implies $T_0$ (because in any case, $T_3 \wedge T_0$ implies $T_2$. If you do follow e.g. Wikipedia's terminology, then $T_0$ is included in any higher separation axioms, and they do become linearly ordered. $\endgroup$
    – Arno
    Feb 24, 2014 at 15:17
  • $\begingroup$ I fixed $T_1$ and $T_0$, can you check if I got all topologies correct? $\endgroup$ Feb 24, 2014 at 19:12

\begin{align*} \\T1=\{\{1\},¤, X\} \\T2=\{\{2\},¤,X\} \\T3=\{\{3\},¤, X\} \\T4=\{\{1,2\},¤, X\} \\T5=\{\{2,3\},¤, X\} \\T6=\{\{1,3\},¤, X\} \\T7=\{\{1\}, \{1,2\},¤, X\} \\T8=\{\{2\}, \{1,2\},¤, X\} \\T8=\{\{1\},\{1,3\},¤, X\} \\T10=\{\{3\},\{1,3\},¤, X\} \\T11=\{\{2\}, \{2,3\},¤, X\} \\T12=\{\{3\}, \{2,3\},¤, X\} \\T13=\{\{1\}, \{2,3\},¤, X\} \\T14=\{\{2\}, \{1,3\},¤, X\} \\T15=\{\{3\},\{1,2\},¤,X\} \\T16=\{\{1\}, \{2\}, \{1,2\},¤, X\} \\T17=\{\{2\}, \{3\}, \{2,3\},¤, X\} \\T18=\{\{1\},\{3\}, \{1,3\},¤, X\} \\T19=\{\{1,2\}, \{1,3\}, \{1\},¤, X\} \\T20=\{\{1,2\}, \{2,3\},\{2\},¤, X\} \\T21=\{\{1,3\}, \{2,3\}, \{3\},¤, X\} \\T22=\{\{1\}, \{2\}, \{1,2\}, \{2,3\},¤, X\} \\T23=\{\{1\}, \{3\}, \{1,3\}, \{3,2\},¤, X\} \\T24=\{\{2\}, \{3\}, \{2,3\}, \{3,1\},¤, X\} \\T25=\{\{1\}, \{2\}, \{1,2\}, \{1,2,3\},¤, X\} \\T26=\{\{2\}, \{3\}, \{2,3\}, \{1,2,3\},¤, X\} \\T27=\{\{1\}, \{3\}, \{1,3\}, \{1,2,3\},¤, X\} \\T28=\{¤, X\} \\T29=\{\{1,2,3\},¤, X\} \end{align*}

So there are 29 topologies of X={1,2,3}

  • $\begingroup$ Could you edit this to answer the question? $\endgroup$
    – Tommi
    Sep 2, 2015 at 8:38
  • $\begingroup$ T1=:{{1},¤X} T2={{1,2},¤, X} and T3={{1 }, {2}, {1,2},¤, X} here T1 is contain in T2 $\endgroup$
    – nav
    Sep 2, 2015 at 8:46
  • $\begingroup$ I assume that $T_i$ in the question refer to the separability axioms, see e.g. sidebar here: en.wikipedia.org/wiki/Kolmogorov_space $\endgroup$
    – Tommi
    Sep 2, 2015 at 9:07

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