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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 $(X,T_1) \to (X,T_2)$
and
$(X,T_2) \to (X,T_1)$
homeomorphic?

Are there any others?

1 Answers 1

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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.

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    In case I misunderstand your question please ping me and I'll edit my answer.2012-04-23
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    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 :))2012-04-23
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    Ok, i understand the cardinalities ! so the only homeomorphisms are T3 -> T2 T2 -> T32012-04-23
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    @KivEfehe Yes, the only two homeomorphic spaces in your question are $T_3$ and $T_2$.2012-04-23
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    thank you Matt N and t.b. . (also, I am quite new to this website, what is ping? I will delete this after appropiately)2012-04-23
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    @KivEfehe A ping is when you write `@user` at the beginning of a comment. If you do that `user` will get a message notification in their inbox. If you don't use `@` then they won't know that you want to send them a message and may possibly never notice it.2012-04-23
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    @MattN. thank you. how is our 'easily seen' homeomorphism T3 and T2 open ?2012-04-23
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    @KivEfehe Well, $f(\varnothing) = \varnothing, f(\{a\}) = \{a\}, f(\{b\}) = \{b\}$ and $f(X) = X$. So all open sets are mapped to open sets.2012-04-23
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    Great ! Thank you @MattN. I think I understand this completely now. I was mainly struggling on the continuity part... let me try it all out for a set with three elements !2012-04-23
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    @BenjaminLim Thanks Ben : )2012-04-24