The following is taken from wikipedia: http://en.wikipedia.org/wiki/Finite_topological_space

2 points

Let $X = \{a,b\}$ be a set with 2 elements.
There are four distinct topologies on $X$:

  • $T_1$: $\{\emptyset, \{a,b\}\}$ (the trivial topology)
  • $T_2$: $\{\emptyset, \{a\}, \{a,b\}\}$
  • $T_3$: $\{\emptyset, \{b\}, \{a,b\}\}$
  • $T_4$: $\{\emptyset, \{a\}, \{b\}, \{a,b\}\}$ (the discrete topology)

The second and third topologies above are easily seen to be homeomorphic. The function from $X$ to itself which swaps $a$ and $b$ is a homeomorphism. A topological space homeomorphic to one of these is called a Sierpiński space. So, in fact, there are only three inequivalent topologies on a two point set: the trivial one, the discrete one, and the Sierpiński topology. The specialization preorder on the Sierpiński space $\{a,b\}$ with $\{b\}$ open is given by: $a \le a$, $b \le b$, and $a \le b$.

I am looking for homeomorphic topologies apart from this "easily seen" one $(X,T_2) \to (X,T_3)$.

These are bijections that are continuous with continuous inverses.

Is $(X,T_1) \to (X,T_2)$
$(X,T_2) \to (X,T_1)$

Are there any others?


No there aren't any others. (If I understand your question correctly.) How many bijections can you have on $X$? Two, namely $f_1(x) = id_X$ and $f_2: a \mapsto b, b \mapsto a$.

Why is neither of these a homeomorphism between $T_1$ and $T_2$? Because $f^{-1}(\{a\}) = \{a\}$ is not open in $T_1$ hence $f_1$ (and similarly $f_2$) is not continuous in this case.

What about $T_2 \to T_1$? Well, a homeomorphism has to be open but $f_1(\{a\}) = \{a\}$ is not open. So again, neither of the $f_i$ is a homeomorphism.

  • $\begingroup$ In case I misunderstand your question please ping me and I'll edit my answer. $\endgroup$ – Rudy the Reindeer Apr 23 '12 at 18:38
  • 2
    $\begingroup$ A different way to put it is that a homeomorphism yields a bijection of the topologies. Since $T1$, $T2$ and $T4$ have different cardinalities, no homeomorphism can exist. (thanks for the bug fix in the other answer :)) $\endgroup$ – t.b. Apr 23 '12 at 18:40
  • $\begingroup$ Ok, i understand the cardinalities ! so the only homeomorphisms are T3 -> T2 T2 -> T3 $\endgroup$ – Kiv Efehe Apr 23 '12 at 19:49
  • $\begingroup$ @KivEfehe Yes, the only two homeomorphic spaces in your question are $T_3$ and $T_2$. $\endgroup$ – Rudy the Reindeer Apr 23 '12 at 19:52
  • $\begingroup$ thank you Matt N and t.b. . (also, I am quite new to this website, what is ping? I will delete this after appropiately) $\endgroup$ – Kiv Efehe Apr 23 '12 at 19:54

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