Simple Analysis Question: Showing a Set is Closed

In summary, the conversation discusses how to show that a nonempty closed subset of \mathbb{R}^n, denoted as S, is closed. The participants consider using the definition of closed as the complement being open or containing all of its limit points. They also mention the sequence definition, where the limit of a convergent sequence is contained in the set if and only if it is closed. The conversation then delves into how to use the closedness of S to show that a set A = \{d(x, y) : y \in S\} is closed. Ultimately, the participants come to the conclusion that since S is complete and the function f is continuous, A is closed.
  • #1
tylerc1991
166
0

Homework Statement



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.

Homework Equations



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

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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  • #2
So, what definition/characterization of closed would you like to use?? I like the sequence definition, do you know that one?
 
  • #3
micromass said:
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?
 
  • #4
tylerc1991 said:
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?
 
  • #5
micromass said:
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]?
 
  • #6
Right. That's ok, I think.
 

FAQ: Simple Analysis Question: Showing a Set is Closed

What is a closed set?

A closed set is a set of numbers or points that includes all of its boundary points. This means that all of the points on the edge of the set are also included in the set itself.

How do you show that a set is closed?

To show that a set is closed, you must prove that it contains all of its boundary points. This can be done by showing that for any point on the edge of the set, there exists a sequence of points within the set that converges to that point.

What is the boundary of a set?

The boundary of a set is the set of points that are on the edge of the set, but not necessarily included in the set. These points are neither completely inside nor completely outside the set.

Can a set be both open and closed?

No, a set cannot be both open and closed. A set is considered open if it does not contain its boundary points, while a set is considered closed if it contains all of its boundary points. Therefore, a set cannot have both open and closed properties at the same time.

Why is it important to show that a set is closed?

It is important to show that a set is closed because it is a fundamental concept in topology and analysis. Closed sets have many properties and are used to define other concepts, such as continuity and compactness. Showing that a set is closed also allows for more precise mathematical reasoning and proofs.

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