
#1
Mar812, 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! 



#2
Mar812, 06:54 PM

Mentor
P: 16,703

So, what definition/characterization of closed would you like to use?? I like the sequence definition, do you know that one?




#3
Mar812, 06:59 PM

P: 166





#4
Mar812, 07:07 PM

Mentor
P: 16,703

Simple Analysis Question: Showing a Set is Closed 



#5
Mar812, 07:20 PM

P: 166

[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]? 


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