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Local Continuity and Restriction

  1. Aug 27, 2011 #1

    Let f :X-->Y ; X,Y topological spaces is any map and {Ui: i in I} is a cover for X
    so that :

    f|_Ui is continuous, i.e., the restriction of f to each Ui is continuous, then:

    1) If I is finite , and the {Ui} are all open (all closed) , we can show f is continuous:

    taking W open in Y, f^-1(W)= \/ f^-1(W /\ Ui) = is the union of open sets in X;

    each W/\Ui is open, and W/\Ui is contained in Ui.

    ( where \/ is the union over I ; /\ is intersection over I ); same for V closed in Y.

    2) If I is infinite, the argument can break down , i.e., if {Ui } is a closed cover

    for X (each Ui is closed) ; f|_Ui is continuous and I is infinite, then this result

    fails. Does anyone know of an example of this last?
  2. jcsd
  3. Aug 27, 2011 #2
    As an example where it fails, take any discontinuous function [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex] and consider the closed cover [itex](\{x\})_{x\in \mathbb{R}}[/itex], doesn't that work?
  4. Aug 27, 2011 #3
    Micromass: but how do we then define f|_x to be continuous?
  5. Aug 27, 2011 #4
    Just define [itex]f\vert_{\{x\}}:\{x\}\rightarrow \mathbb{R}:x\rightarrow f(x)[/itex]. This is clearly continuous since [itex]\{x\}[/itex] is indiscrete...
  6. Aug 27, 2011 #5
    Right, good point; thanks.
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