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Help understanding equivalent definitions for continuity

  1. Apr 18, 2013 #1
    I was hoping someone could help me understand the equivalence between the definitions for functions to be continuous between topological spaces, ie:

    For X and Y topological spaces, and f:X-->Y a function, my notes don't prove why these definitions are equivalent (possibly because I'm missing something pretty obvious!):

    1. f continuous IFF for U open in Y f^-1(U) is open in X
    2. f continuous IFF for C closed in Y, f^-1(C) is closed in X

    I can see why this is true when f is surjective, because then f^-1(Y) = X, so for F closed in Y, U=Y\F is open in Y and f^-1(Y) = f^-1(U u F) = f^-1(U) u f^-1(F) (because U and F are disjoint) so X = f^-1(U) u f^-1(F) implies f^-1(U) = X\f^-1(F) open in X (and then it is also easy to see the converse here). But if f is not surjective, then all that follows is that f^-1(U) is in X\f^-1(F) (open in X), so why does it follow in this case that f^-1(U) is open in X, given definition 2?
  2. jcsd
  3. Apr 18, 2013 #2


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    You can show [itex]f^{-1}(Y\setminus C) = X\setminus f^{-1}(C)[/itex], because [itex]x\in f^{-1}(Y\setminus C) \Leftrightarrow f(x) \in Y\setminus C \Leftrightarrow f(x) \notin C \Leftrightarrow x\notin f^{-1}(C) \Leftrightarrow x\in X\setminus f^{-1}(C)[/itex]. Now let's say f is continuous by the open set definition. Then [itex]C \text{ closed } \Rightarrow Y\setminus C \text{ open } \Rightarrow f^{-1}(Y\setminus C) \text{ open } \Rightarrow X\setminus f^{-1}(C) \text{ open } \Rightarrow f^{-1}(C) \text{ closed}[/itex] using continuity at the second step and the statement from above at the third. The other way around works the same.
  4. Apr 22, 2013 #3
    Hey belated thanks for your help. Quite obvious in the end!
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