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.