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Math Help - Function proof

  1. #1
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    Function proof

    Let g be defined on a set containing the range of a function f. If f is continuous at z0 and g is continuous at f(z0), then g(f(z)) is continuous at z0.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jzellt View Post
    Let g be defined on a set containing the range of a function f. If f is continuous at z0 and g is continuous at f(z0), then g(f(z)) is continuous at z0.
    what have you tried? what definition of continuity are you working with?
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  3. #3
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    The definition of continuity we're unsing involves limits and is NOT the epsilon delta method.

    I havn' really gotten anywhere because I havn't a clue how to do this..
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    Does it involve sequences? I mean (given f:X->R, a in the set of limit points of X), did you define limits like :

    \lim_{x\to a}{f(x)}=L

    if, and only if, for all sequence (x_n)_{n\in\mathbb{N}} such that x_n\in X\backslash{\{a\}}\forall n\in\mathbb{N} and \lim_{n\to \infty}x_n=a we have \lim_{n\to\infty}f(x_n)=L?
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  5. #5
    Member mabruka's Avatar
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    Maybe his definition of continuity is

    "f is continuous in a iff \lim_{x\to a}{f(x)}=f(a) "
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by jzellt View Post
    Let g be defined on a set containing the range of a function f. If f is continuous at z0 and g is continuous at f(z0), then g(f(z)) is continuous at z0.
    This is a classic and easy to find on the internet theorem...but let met give you an outline of the proof.

    Let f:A\mapsto B and let g:B\mapsto C where f,g are both continuous. (we actually only have to have \mapsto C" alt="g'\mapsto C" /> where g'|B=g but that is neither here nor there). Then gf:A\mapsto C is also continuous.

    Proof: Let B_{\varepsilon}(g(f(c))) be arbitrary. Since f(c)\in B and g is continuous there exists some B_{\delta}(f(c)) such that g\left(B_{\delta}(f(c)\right)\subseteq B_{\varepsilon}(g(f(c))). And since f is continuous there exists some B_{\sigma}(c) such that f\left(B_{\sigma}(c)\right)\subseteq B_{\delta}(f(c)) and so g\left(f\left(B_{\sigma}(c)\right)\right)\subseteq g\left(B_{\delta}(f(c))\right)\subseteq B_{\varepsilon}(g(f(c)))


    Or much nicer, if you are using the more topological definition (which is equivalent to the regular in metric spaces) that k:X\mapsto Y is continuous iff k^{-1}(O) is open in X whenever O is open in Y merely note that (gf)^{-1}(O)=f^{-1}(g^{-1}(O)) and g^{-1}(O) is ope in B and so f^{-1}(g^{-1}(O)) is open in A. Done.
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