Show that f(E) is dense in f(X) if E is dense in X

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SUMMARY

The discussion centers on proving that if E is a dense subset of a metric space X, then the image f(E) under a continuous mapping f from X to another metric space Y is also dense in f(X). The proof demonstrates that for any point f(x) in f(X), there exists a point f(p) in f(E) arbitrarily close to f(x), thus establishing that f(x) is a limit point of f(E). The proof is validated by the continuity of f and the properties of dense sets, confirming that f(E) is indeed dense in f(X).

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Homework Statement


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].

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.
 
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Looks ok to me.
 

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