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

  • Thread starter DeadOriginal
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In summary, the proof shows that for any x in X, if f(x) is not in f(E), then there exists a point f(p) in f(E) that is contained in any neighborhood of f(x). Therefore, f(E) is dense in f(X).
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DeadOriginal
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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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  • #2
Looks ok to me.
 

1. What does it mean for a set to be dense?

A set is considered dense if it contains points that are arbitrarily close to every point in a given space. In other words, there are no gaps in the set and it fills up the entire space.

2. How is density related to continuity?

A function is continuous if it preserves the closeness of points. In other words, if two points are close together in the domain, their images under the function will also be close together. This concept is closely related to density, as a dense set in the domain ensures that the function is continuous throughout the entire space.

3. How does the density of a set affect the density of its image under a function?

If a set is dense in a space, then its image under a continuous function will also be dense in the image space. In other words, if the set fills up the entire space, then its image will also fill up the entire image space.

4. Can you provide an example of how density in the domain translates to density in the image space under a function?

Consider the function f(x) = 1/x, where the domain is the interval (0,1) and the image space is the real numbers. The set of rational numbers in the domain is dense, as there are infinitely many rational numbers between any two irrational numbers. The image of these rational numbers under f(x) is also dense in the real numbers, as the function maps them to the entire real number line.

5. How can we prove that f(E) is dense in f(X) if E is dense in X?

To prove that f(E) is dense in f(X), we need to show that for any point y in f(X), there exists a sequence of points in f(E) that converge to y. This can be done by first choosing a point x in E that is arbitrarily close to the preimage of y under f. Then, by the continuity of f, we can find a sequence of points in E that converge to x, and therefore, also converge to y under f. This shows that f(E) is dense in f(X).

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