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Homework Help: Show that f(E) is dense in f(X) if E is dense in X

  1. Jun 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Let [itex]f[/itex] be a continuous mapping of a metric space [itex]X[/itex] into a metric space [itex]Y[/itex]. Let [itex]E[/itex] be a desnse subset of [itex]X[/itex]. Prove that [itex]f(E)[/itex] is dense in [itex]f(X)[/itex].

    3. The attempt at a solution
    Choose any [itex]x\in X[/itex].

    Since [itex]x\in X[/itex] it follows that [itex]f(x)\in f(X)[/itex]. We need to show that [itex]f(x)[/itex] is a limit point of [itex]f(E)[/itex] or is a point of [itex]f(E)[/itex]. If [itex]f(x)\in f(E)[/itex] then we are done so suppose not. Let [itex]\epsilon>0[/itex] be given. [itex]f[/itex] is continuous so [itex]f(p)\in N_{\epsilon}(f(x))[/itex] whenever [itex]p\in N_{\delta}(x)[/itex] for [itex]\delta>0[/itex] and we know that there exists a [itex]p\in N_{\delta}(x)[/itex] such that [itex]p\not=x[/itex] and [itex]p\in E[/itex] since [itex]x[/itex] is a limit point of [itex]E[/itex]. [itex]p\in E[/itex] so [itex]f(p)\in f(E)[/itex] and from continuity we saw that [itex]f(p)[/itex] is contained in the neighborhood of [itex]f(x)[/itex], [itex]N_{\epsilon}(f(x))[/itex], whenever [itex]p\in N_{\delta}(x)[/itex]. The neighborhood of [itex]f(x)[/itex] and the point [itex]p[/itex] were chosen arbitrarily. We have shown that for any neighborhood of [itex]f(x)[/itex] there exists a point [itex]f(p)[/itex] contained in that neighborhood such that [itex]f(p)\not=f(x)[/itex] and [itex]f(p)\in f(E)[/itex]. Hence [itex]f(x)[/itex] is a limit point of [itex]f(E)[/itex]. It follows that [itex]f(E)[/itex] is dense in [itex]f(X)[/itex].

    Could someone take a look over this proof and correct me on any errors?
    Rereading this I think I already see a problem because I haven't shown that [itex]f(x)\not=f(p)[/itex] but this can be easily justified because we assumed that [itex]f(x)\not\in f(E)[/itex]. Since [itex]f(p)\in f(E)[/itex] it can't be that [itex]f(x)=f(p)[/itex] or else we contradict ourselves.
    Last edited: Jun 6, 2013
  2. jcsd
  3. Jun 6, 2013 #2


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    Looks ok to me.
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