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Thread: need help

  1. #1
    Junior Member
    Feb 2009

    need help

    consider a function $\displaystyle f: Real \rightarrow Real$. Prove that f is continuous on R iff for every subset B of Real
    $\displaystyle f^{-1}(int B) \subseteq f^{-1} (B)$
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  2. #2
    Senior Member
    Nov 2008

    This is always true.

    I think you meant: $\displaystyle f^{-1}(intB)\subseteq int(f^{-1}(B))$

    I $\displaystyle f$ is continuous, then for any open set $\displaystyle V,\ f^{-1}(V)=U$ is an open set. So let $\displaystyle x$ be an element of $\displaystyle f^{-1}(intB),$ i.e. there exists an open set $\displaystyle V$ such that $\displaystyle f(x)\in V\subseteq B.$ Then $\displaystyle f^{-1}(V)$ is an open subset of $\displaystyle f^{-1}(B)$ and contains $\displaystyle x,$ therefore $\displaystyle x\in int(f^{-1}(B))$ and we can conclude: $\displaystyle f$ continuous $\displaystyle \Rightarrow f^{-1}(intB)\subseteq int(f^{-1}(B))$

    Conversely, assume f$\displaystyle ^{-1}(intB)\subseteq int(f^{-1}(B))$ . Let $\displaystyle U$ be an open set. Then intU=U, and we can write:

    $\displaystyle f^{-1}(U)=f^{-1}(intU)\subseteq int(f^{-1}(U))$ i.e. $\displaystyle f^{-1}(U) \subseteq int(f^{-1}(U))$

    But, for any subset $\displaystyle A,\ A\subseteq intA$ means that $\displaystyle A$ is open (if you never saw that, try to prove it)

    Thus for any open subset $\displaystyle U$, $\displaystyle f^{-1}(U)$ is open so f is continuous, we just proved: $\displaystyle f^{-1}(intB)\subseteq int(f^{-1}(B)) \Rightarrow$ $\displaystyle f$ continuous.
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