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    Hi Everyone

    The Fort topology is defined by:

    T = T_{x_0} = \{U \subset X : U^c is finite, or x_0 \notin U \}

    If X is non empty and x_0, x_1 \in X I need to show that T_{x_0} and  T_{x_1} are equivalent.

    The homeomorphism  f: T_{x_0} \rightarrow T_{x_1} I constructed is as follows:

     f(x) = \begin{cases}

 & x_1\text{ if } x= x_0  \\ 

 & x_0\text{ if } x= x_1 \\ 

 & x\text{ if } x= x 

\end{cases}

    This is useful (I hope!) because now  f = f^{-1} but how does one show that f is continuous?

    Usually we use this theorem: "f is continuous iff  f^*V \in T_{x_0} \text{ whenever } V\in T_{x_1}

    So if we assume  V\in T_{x_1} then either:

    Case 1:  x_1 \in V or
    Case 2:  V^c is finite with x_1 \notin T_{x_1} .

    Case 1 I think I've got sorted, but what about case 2? If the function simplified to f(x) = x then it would be easy:  (f^*V)^c = (f_*V)^c = V^c which is finite. But how do you show (f^*V)^c is finite in general?

    Thanks
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    Note that your f is a bijection. So it shouldn't be hard to prove stuff about the size of f*(V)...
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    (Original post by DFranklin)
    Note that your f is a bijection. So it shouldn't be hard to prove stuff about the size of f*(V)...
    Is this kind of reasoning ok:

    The cardinality of V = the cardinality of f^*V so this implies that the cardinality of (f^*V)^c = the cardinality of V^c? And hence finite
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    No, you can't do that. Because V and f*(V) might both be infinite, and this doesn't let you deduce anything abou the relative sizes of V^c and (f*V)^c.

    On the other hand, you can confidently say that V^c and f*(V^c) have the same size...
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    (Original post by DFranklin)
    No, you can't do that. Because V and f*(V) might both be infinite, and this doesn't let you deduce anything abou the relative sizes of V^c and (f*V)^c.

    On the other hand, you can confidently say that V^c and f*(V^c) have the same size...
    Oooh thank you so much

    And f*(V^c) = (f*(V))^c? ...Surely... ? And hence the pre image is finite?
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    Yes.
 
 
 
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