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Math Help - f is not continuous on R and discrete metric

  1. #1
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    f is not continuous on R and discrete metric

    f:R->R
    f(x) is 1 on Q and 0 elsewhere

    Show f is not continuous at any point of R.
    What about if the usual metric on R is replaced by the discrete metric?

    I know that if f(x) takes 1 on Q and 0 elsewhere(Q complement), then by the density property there's infinite number of points between any values of Q, so the graph will keep jumping from 1 to 0 and back, so that's obviously not continuous. Have no idea how to show that in an actual proof though.

    Also, I don't really understand replacing a metric by some other metric. What happens id R is replaced by the discrete metric?
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  2. #2
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    what happens if you use de descrete metric is that any set is open(descrtete toplogy) so any function f:R->R will be continous.(look for definition of continuity in topological spaces)
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cp05 View Post
    f:R->R
    f(x) is 1 on Q and 0 elsewhere

    Show f is not continuous at any point of R.
    What about if the usual metric on R is replaced by the discrete metric?

    I know that if f(x) takes 1 on Q and 0 elsewhere(Q complement), then by the density property there's infinite number of points between any values of Q, so the graph will keep jumping from 1 to 0 and back, so that's obviously not continuous. Have no idea how to show that in an actual proof though.

    Also, I don't really understand replacing a metric by some other metric. What happens id R is replaced by the discrete metric?
    I assume that \mathbb{R} is intially under the usual topology. Then, suppose that f is continuous at x_0\in\mathbb{R}. Then, since \mathbb{Q},\mathbb{R}-\mathbb{Q} are both dense in \mathbb{R} there exists sequences \{q_n\},\{i_n\} in each such that q_n,i_n\to x_0. Thus, 0=\lim\text{ }0=\lim\text{ }f(i_n)=f(x_0)=\lim\text{ }f(q_n)=\lim\text{ }1=1. That looks strange, huh?


    Also, if X is a space with the discrete topology then any mapping f:X\to Y is continuous regardless of f or Y. This is clear since if U\subseteq Y is open then f^{-1}(U)\subseteq X but since every subset of X is open we see in particular that f^{-1}(U) is open.
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    Also, if X is a space with the discrete topology then any mapping f:X\to Y is continuous regardless of f or Y. This is clear since if U\subseteq Y is open then f^{-1}(U)\subseteq X but since every subset of X is open we see in particular that f^{-1}(U) is open.[/quote]

    How do I know that every subset of X is open? Thanks so much.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cp05 View Post
    Also, if X is a space with the discrete topology then any mapping f:X\to Y is continuous regardless of f or Y. This is clear since if U\subseteq Y is open then f^{-1}(U)\subseteq X but since every subset of X is open we see in particular that f^{-1}(U) is open.
    How do I know that every subset of X is open? Thanks so much.[/QUOTE]

    That's what the discrete topology means. The discrete metric gives rise to the discrete topology. To see this merely let X have the discrete metric and let E\subseteq X be arbitrary. Then, given any x\in X we have that B_{\frac{1}{2}}(x)\cap E contains at most one point of E and so x\notin D(E) ( D(E) is the derived set...the set of limit points). Since x\in X was arbitrary it follows that D(E)=\varnothing. Thus, trivially D(E)\subseteq E and thus E is closed....BUT! This implies that if G\subseteq X then G is open since X-G is closed (since any subset of X is closed)
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