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Establishing that a function is discontinous

  1. Jan 6, 2014 #1
    According to the definition of continuity in topology a function f:X -> Y is continous if for every open set V in Y, the set ##f(V)^{-1}## is open in X. I have found this definition powerfull in dealing with proofs of a more general nature, but when presented with the trivially discontinous function from the reals to the reals

    $$f(x) = 1 \ \text{for} \ x< 0, \ 0 \ \text{for} \ \ x \geq 0 $$

    I have problems to argue why this function is obviously not continuous from the above definition.
    Which open set do I choose in it's codomain which is not open in it's domain?
     
  2. jcsd
  3. Jan 6, 2014 #2

    Office_Shredder

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    The set (-1/2,1/2) (or any open set containing zero and not 1) has as its inverse image [itex] [0,\infty) [/itex] which is not open.

    Basically as long as you include a small open set around a discontinuity you are good to go if the set is small enough.
     
  4. Jan 7, 2014 #3
    Thanks! It's clear now. My confusion lied in trying to construct open sets in the image of f instead of the codomain.
     
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