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Thread: 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 $\displaystyle \mathbb{R}$ is intially under the usual topology. Then, suppose that $\displaystyle f$ is continuous at $\displaystyle x_0\in\mathbb{R}$. Then, since $\displaystyle \mathbb{Q},\mathbb{R}-\mathbb{Q}$ are both dense in $\displaystyle \mathbb{R}$ there exists sequences $\displaystyle \{q_n\},\{i_n\}$ in each such that $\displaystyle q_n,i_n\to x_0$. Thus, $\displaystyle 0=\lim\text{ }0=\lim\text{ }f(i_n)=f(x_0)=\lim\text{ }f(q_n)=\lim\text{ }1=1$. That looks strange, huh?


    Also, if $\displaystyle X$ is a space with the discrete topology then any mapping $\displaystyle f:X\to Y$ is continuous regardless of $\displaystyle f$ or $\displaystyle Y$. This is clear since if $\displaystyle U\subseteq Y$ is open then $\displaystyle f^{-1}(U)\subseteq X$ but since every subset of $\displaystyle X$ is open we see in particular that $\displaystyle f^{-1}(U)$ is open.
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    Also, if $\displaystyle X$ is a space with the discrete topology then any mapping $\displaystyle f:X\to Y$ is continuous regardless of $\displaystyle f$ or $\displaystyle Y$. This is clear since if $\displaystyle U\subseteq Y$ is open then $\displaystyle f^{-1}(U)\subseteq X$ but since every subset of $\displaystyle X$ is open we see in particular that $\displaystyle f^{-1}(U)$ 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 $\displaystyle X$ is a space with the discrete topology then any mapping $\displaystyle f:X\to Y$ is continuous regardless of $\displaystyle f$ or $\displaystyle Y$. This is clear since if $\displaystyle U\subseteq Y$ is open then $\displaystyle f^{-1}(U)\subseteq X$ but since every subset of $\displaystyle X$ is open we see in particular that $\displaystyle f^{-1}(U)$ 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 $\displaystyle X$ have the discrete metric and let $\displaystyle E\subseteq X$ be arbitrary. Then, given any $\displaystyle x\in X$ we have that $\displaystyle B_{\frac{1}{2}}(x)\cap E$ contains at most one point of $\displaystyle E$ and so $\displaystyle x\notin D(E)$ ($\displaystyle D(E)$ is the derived set...the set of limit points). Since $\displaystyle x\in X$ was arbitrary it follows that $\displaystyle D(E)=\varnothing$. Thus, trivially $\displaystyle D(E)\subseteq E$ and thus $\displaystyle E$ is closed....BUT! This implies that if $\displaystyle G\subseteq X$ then $\displaystyle G$ is open since $\displaystyle X-G$ is closed (since any subset of $\displaystyle X$ is closed)
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