Simple Analysis Question: Showing a Set is Closed


by tylerc1991
Tags: analysis, showing, simple
tylerc1991
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#1
Mar8-12, 06:46 PM
P: 166
1. The problem statement, all variables and given/known data

Suppose [itex]S[/itex] is a nonempty closed subset of [itex]\mathbb{R}^n[/itex], and let [itex]x \in \mathbb{R}^n[/itex] be fixed. Show that [itex]A = \{d(x, y) : y \in S\}[/itex] is closed.

2. Relevant equations

A set is closed if its complement is open, or if it contains all of its limit points.

3. The attempt at a solution

I first defined a function [itex]f : S \to \mathbb{R}[/itex] by [itex]f(y) = d(x, y)[/itex]. Notice that [itex]f[/itex] is continuous. Then [itex]A[/itex] is not open because [itex]S[/itex] is closed (if [itex]A[/itex] is open then [itex]f^{-1}(A)[/itex] is open). However, this doesn't show that [itex]A[/itex] is closed.

I feel like I have the intuition, but actually showing this is frustrating. Help would be greatly appreciated!
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micromass
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#2
Mar8-12, 06:54 PM
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So, what definition/characterization of closed would you like to use?? I like the sequence definition, do you know that one?
tylerc1991
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#3
Mar8-12, 06:59 PM
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Quote Quote by micromass View Post
So, what definition/characterization of closed would you like to use?? I like the sequence definition, do you know that one?
I think so. That definition states that the limit of a convergent sequence is contained in [itex]A[/itex] if and only if [itex]A[/itex] is closed. So to show that [itex]A[/itex] is closed, I would start with an arbitrary convergent sequence of points of [itex]A[/itex], say [itex](p_n) \to p \in \mathbb{R}[/itex], where [itex](p_n) \subset A[/itex]. I then need to show that [itex]p \in A[/itex]. I am probably going to use the closedness of [itex]S[/itex] in this right?

micromass
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Mar8-12, 07:07 PM
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Simple Analysis Question: Showing a Set is Closed


Quote Quote by tylerc1991 View Post
I think so. That definition states that the limit of a convergent sequence is contained in [itex]A[/itex] if and only if [itex]A[/itex] is closed. So to show that [itex]A[/itex] is closed, I would start with an arbitrary convergent sequence of points of [itex]A[/itex], say [itex](p_n) \to p \in \mathbb{R}[/itex], where [itex](p_n) \subset A[/itex]. I then need to show that [itex]p \in A[/itex]. I am probably going to use the closedness of [itex]S[/itex] in this right?
Yes. So take a convergent sequence [itex](x_n)_n[/itex] in A. We know that we can write [itex]x_n=d(x,y_n)[/itex] for some [itex]y_n\in S[/itex]. Can you show that the [itex](y_n)_n[/itex] is Cauchy?
tylerc1991
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#5
Mar8-12, 07:20 PM
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Quote Quote by micromass View Post
Yes. So take a convergent sequence [itex](x_n)_n[/itex] in A. We know that we can write [itex]x_n=d(x,y_n)[/itex] for some [itex]y_n\in S[/itex]. Can you show that the [itex](y_n)_n[/itex] is Cauchy?
How about this: Since [itex](d(x, y_n))[/itex] is convergent, for all [itex]\varepsilon > 0[/itex], there exists an [itex]N > 0[/itex] such that
[itex]d(x, y_n) < \frac{\varepsilon}{2}[/itex]
when [itex]n > N[/itex]. Therefore, when [itex]m, n > N[/itex], we have
[itex]d(y_n, y_m) \leq d(y_n, x) + d(x, y_m) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon[/itex],
which shows that [itex](y_n)[/itex] is Cauchy. Since [itex]S[/itex] is complete, we have that [itex](y_n)[/itex] converges to a point of [itex]S[/itex]. Then since [itex]f[/itex] is continuous, [itex](x_n)[/itex] converges to a point of [itex]A[/itex]?
micromass
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Mar8-12, 07:24 PM
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Right. That's ok, I think.


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