0
$\begingroup$

I came upon this when trying to understand the proof to "Theorem on Existence of Best Approximations in a Metric Space" as given by Cheney (1981).

Let $\mathcal K$ denote a compact set in a metric space. To each point $p$ of the space there corresponds a point in $\mathcal K$ of minimum distance from $p$.

The proof begins with assuming a sequence $x_1, x_2, \cdots, x_n$ in $\mathcal K$ and continues as below.

"By the compactness of $\mathcal K$, we may assume that the sequence converges to a point $x^*$ of $\mathcal K$; for if necessary we may extract from the given sequence a subsequence with this property."

Now this last sentence comes quite problematic to me; as I imagine we may extract two (or arbitrarily many) different convergent sub-sequences from the original $\{ x_n\}$ each converging to a different $x^*.$ (Am I right about this?)

Thus I believe the answer to the question in the title is No. The proof then proceeds: "We will show that $x^*$ is a point of $\mathcal K$ of minimum distance from $p$." But if I am right in what I described earlier, then we may have two (or arbitrarily many) $x^*$s. Am I missing something here?

$\endgroup$
3
$\begingroup$

You are right that any two different convergent subsequences you extract may have different limits. This doesn't make a difference for the proof because the proof only needs a subsequence converging to some limit $x^*$ and then it shows $x^*$ has the properties you need. It doesn't matter at all that you could have chosen a different subsequence with a different limit $y^*$.

Notice that you shouldn't be worried by the possibility of arbitrarily many points $x^*$ at minimum distance from $p$ since such points aren't unique. For example, consider the unit circle $K = \{x \in \mathbb{R}^2: |x| = 1\}$ and $p = 0$. Then every point of $K$ minimises the distance to $p$.

$\endgroup$
  • 1
    $\begingroup$ @neutrino (you've deleted the comment I replied to, but I'll leave this here in case it's useful) Ah, I think I understand your confusion now. You won't start by picking an arbitrary sequence in $K$ but rather one that minimises the distance to $p$. That is, by basic properties of $\inf$, you can find a sequence $(x_n)$ in $K$ such that $d(x_n,p) \to \inf \{d(x,p) : x \in K\} = \delta$. Then you can show that if $x^*$ is the limit of any convergent subsequence of $(x_n)$, you have $d(x^*,p) = \delta$ $\endgroup$ – Rhys Steele Mar 26 at 13:44
  • $\begingroup$ Oops, sorry. So this was the original comment: Thanks. Yes, I understand that this is an "existence proof" and does not concern uniqueness. But then again, if the number of the convergent sub-sequences that we may construct and their corresponding $x^*$s is arbitrary, what guarantees that their distance from $p$ are the same? $\endgroup$ – Neutrino Mar 26 at 13:50
1
$\begingroup$

The (sequential) compactness tells us that a sequence has at least one convergent subsequence with limit in $\mathcal{K}$. The proof just picks one of those and goes on with the corresponding subsequence (as if that were the original sequence; an often recurring idea in proofs), and I suppose for the proof it does not matter which one it chose, but we know there is at least one. The "property" of the chosen sequence that the proof talks about, must then be one that is preserved by going to a subsequence, of course.

The sequence $x_n = q_n$ where $q_n, n \in \mathbb{N}$ is an enumeration of the rationals in $[0,1]$, say, has the property that all $x \in [0,1]$ can be limits of subsequences of $(x_n)$, so the answer to the question in the title is indeed no, there can be many subsequential limits.

And in the application (nearest point), many such nearest points are not unique. E.g. $\mathcal{K}=[0,1]\cup[2,3]$ and $x=\frac32$ is a simple example, both $1$ and $2$ are nearest.

$\endgroup$
  • 1
    $\begingroup$ @Neutrino that depends in how the $x_n$ are chosen. You haven’t given the full construction. $\endgroup$ – Henno Brandsma Mar 26 at 13:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.