Simple Analysis Question: Showing a Set is Closed

[tylerc1991]

Homework Statement

Suppose S is a nonempty closed subset of \mathbb{R}^n, and let x \in \mathbb{R}^n be fixed. Show that A = \{d(x, y) : y \in S\} 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 f : S \to \mathbb{R} by f(y) = d(x, y). Notice that f is continuous. Then A is not open because S is closed (if A is open then f^{-1}(A) is open). However, this doesn't show that A is closed.

I feel like I have the intuition, but actually showing this is frustrating. Help would be greatly appreciated!

 

[micromass]

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

 

[tylerc1991]

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 A if and only if A is closed. So to show that A is closed, I would start with an arbitrary convergent sequence of points of A, say (p_n) \to p \in \mathbb{R}, where (p_n) \subset A. I then need to show that p \in A. I am probably going to use the closedness of S in this right?

 

[micromass]

I think so. That definition states that the limit of a convergent sequence is contained in A if and only if A is closed. So to show that A is closed, I would start with an arbitrary convergent sequence of points of A, say (p_n) \to p \in \mathbb{R}, where (p_n) \subset A. I then need to show that p \in A. I am probably going to use the closedness of S in this right?

Yes. So take a convergent sequence (x_n)_n in A. We know that we can write x_n=d(x,y_n) for some y_n\in S. Can you show that the (y_n)_n is Cauchy?

 

[tylerc1991]

Yes. So take a convergent sequence (x_n)_n in A. We know that we can write x_n=d(x,y_n) for some y_n\in S. Can you show that the (y_n)_n is Cauchy?

How about this: Since (d(x, y_n)) is convergent, for all \varepsilon > 0, there exists an N > 0 such that
d(x, y_n) < \frac{\varepsilon}{2}
when n > N. Therefore, when m, n > N, we have
d(y_n, y_m) \leq d(y_n, x) + d(x, y_m) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon,
which shows that (y_n) is Cauchy. Since S is complete, we have that (y_n) converges to a point of S. Then since f is continuous, (x_n) converges to a point of A?

 

[micromass]

Right. That's ok, I think.