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Continuous Functions

  1. Apr 14, 2012 #1
    Hi, can someone please check if my proof is correct

    1. a) Assume f : R -> R is continuous when the usual topology on R is
    used in the domain and the discrete topology on R is used in the range. Show
    that f must be a constant function.

    My attempt :

    Let f: R --> R be continuous. Suppose that f is not constant, then f assumes
    at least 2 values, p and q.
    In R we can find two disjoint open intervals I and J such that
    p is in I and q is in J.
    By continuity f^{-1} and f^{-1}[J] are open.
    They are disjoint as I and J are, and non-empty as p and q have preimages.
    This contradicts the fact above. So f must be constant.

    #1 b) Prove that for any two topological spaces (X; T1) and (Y; T2), if
    y0 is in Y then the constant function f : X -> Y defi ned by f(x) = y0 is continu-
    ous.

    My Attempt:

    f:x->y is continuous if for every v in T2 , f inverse (v) is in T

    let y0 in Y then f(x) = y0 for every x in X is continuous and X is open set

    If y0 is not in v, then f inverse (v) = empty set so its always open....therefore every constant function is always continuous.

    thanks
     
  2. jcsd
  3. Apr 14, 2012 #2

    morphism

    User Avatar
    Science Advisor
    Homework Helper


    Contradicts what fact?
    This is fine, but your write-up is sloppy.
     
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