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Math Help - Function from Metric Space to R is Continuous

  1. #1
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    Function from Metric Space to R is Continuous

    Let (M, d) be a metric space. Let f:M\longrightarrow M be a continuous function. Let g:M\longrightarrow\mathbb{R} be defined by x\longmapsto d(x,f(x)). Show that g is continuous.

    So what we know is that f is continuous, so for any \eta>0, there exists \mu>0 such that:

    d(x,y)<\mu \Rightarrow d(f(x),f(y))<\eta

    But what we want to do for any \epsilon>0 is to produce \delta>0 such that:

    d(x,y)<\delta \Rightarrow |g(x)-g(y)|=|d(x,f(x))-d(y,f(y))|<\epsilon

    I have a feeling we'll want to produce \delta in terms of \epsilon, \eta, \mu. I was thinking breaking something up using a triangle inequality, but I've tried a few different ways that have all led to dead ends.
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  2. #2
    Super Member girdav's Avatar
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    You can use the inequality |d(x,z)-d(y,z)|\leq d(x,y) and write d(x,f(x))-d(y,f(y))=d(x,f(x))-d(x,f(y))+d(x,f(y))-d(y,f(y)).
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  3. #3
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    I already tried that and came up with a dead end. The result of that yields:

    |g(x)-g(y)|\leq d(f(x),f(y))+d(x,y)

    Which if we let those be less than \eta, \mu respectively from the previous part, we get:

    |g(x)-g(y)|<\eta+\mu

    Which I just don't see how you could possibly get \delta out of something of that sort.
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  4. #4
    Super Member girdav's Avatar
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    Given \eta, we can find \mu such that d(f(x),f(y))\leq \frac{\eta}2 if d(x,y)\leq \mu. We can choose this \mu smaller than \frac{\eta}2.
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