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Showing f(0)=f(1)

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Let [itex]X=\{0,1\}[/itex] with the Sierpinski topology [itex]\tau = \{ \emptyset , \{0\} ,\{0,1\} \}[/itex].

    Suppose [itex]f:X\to \mathbb{R}[/itex] is continuous.

    Show [itex]f(0)=f(1)[/itex].

    [Potentially useful observation: [itex]\{f(0)\}[/itex] is closed in [itex]\mathbb{R}[/itex].]

    3. The attempt at a solution

    [itex]f:X\to\mathbb{R}[/itex] is continuous [itex]\iff[/itex] for every open (closed) set [itex]A\subseteq \mathbb{R},\;f^*(A)[/itex] is open (closed) in [itex]X[/itex].

    How to show f(0)=f(1)?
     
  2. jcsd
  3. Dec 6, 2011 #2
    Suppose the contrary, and find an open set whose preimage is {1}.
     
  4. Dec 6, 2011 #3
    Does [itex]f^*(\{0\}) = \{1\}[/itex]?
     
  5. Dec 6, 2011 #4

    Dick

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    Who could say until you define what f is?? Follow Citan Uzuki's advice and define f(0)=a and f(1)=b where a and b are unequal real numbers. Is f continuous?
     
  6. Dec 6, 2011 #5
    So [itex]\{a\} \subseteq \mathbb{R}[/itex] and [itex]\{b\} \subseteq \mathbb{R}[/itex] are closed, and [itex]f^*(\{a\})=\{0\} \subseteq X[/itex] is closed in [itex]X[/itex] but [itex]f^*(\{a\})=\{1\}[/itex] is not closed in [itex]X[/itex] so [itex]f[/itex] is not continuous - contradiction; therefore [itex]f(0)=f(1)[/itex]?
     
  7. Dec 6, 2011 #6
    ^ That should say [itex]f^*(\{b\})=\{1\}[/itex] but I can't find an edit button!
     
  8. Dec 6, 2011 #7

    Dick

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    Right. Now to finish up show that if a=b then f IS continuous.
     
  9. Dec 6, 2011 #8
    That approach will work, but you're doing it backwards. {0} is not closed in X, because its complement is {1}, which is not in the topology on X. So the conclusion that f is not continuous comes from the fact that f*({a}) = 0, not the fact that f*({b}) = 1.
     
  10. Dec 6, 2011 #9

    Dick

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    Yeah, actually I didn't notice, but you right. My 'Edit' button disappeared for a while yesterday. Seems to be back now. At least for me. Citan Uzuki is also right. I didn't notice you'd swapped the open and closed either. Guess I should read more carefully.
     
  11. Dec 6, 2011 #10
    If we have a contradiction after supposing [itex]a\neq b[/itex] doesn't that mean we must have [itex]a=b[/itex]?
     
  12. Dec 6, 2011 #11

    Dick

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    No! All you've shown is that if a not equal b is then f is not continuous. How can that be a proof that if a=b then f is continuous? There might not be ANY continuous functions.
     
  13. Dec 6, 2011 #12
    But we're assuming all along that [itex]f:X\to \mathbb{R}[/itex] is continuous (as the question tells us this!)
     
  14. Dec 6, 2011 #13

    Dick

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    Ok, this is really not my day. You are right. I thought you had to prove f continuous iff a=b. I saw the double arrow in your original post and fixated on it.
     
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