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"If a sequence $(x_n: n\in\mathbb N)$ of a metric space does not contain any convergent subsequence, then the set $\{x_n: n\in\mathbb N\}$ is closed in the metric space"

I am trying to figure out which definitions make this proof the easiest. Since we are talking about sequences, closed should be best defined as "every sequence in the closed set converges to a point in the set." Now since the sequence $(x_n)$ does not contain any convergent subsequence, it must be unbounded by the Bolzano theorem.

Now can we say that the subset $\{x_n\mid n\in\Bbb N\}$ is closed because you cannot find a sequence in it which converges to a point outside it? (since no sequence even converges to anything at all)

Anne Bauval
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Bill
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2 Answers2

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You cannot use Heine-Borel for any arbitrary metric space $X$.

We wish to show that the set $A = \{x_n \mid n \in \Bbb N\}$ is closed in $X$. Note that if $y$ is a limit point of $A$, then there exists a sequence of distinct points in $A$ converging to $y$. However, given a sequence of distinct points in $A$, you can find a corresponding subsequence of $(x_n)_n$.1 However, we know that $(x_n)_n$ has no convergent subsequences. Thus, $A$ has no limit points and is closed.

Note that I've also shown that no point of $A$ is a limit point.

  1. To elaborate more:
    Suppose $(z_n)_n$ is a sequence of distinct points in $A$. Then, each $z_n$ is equal to some $x_{m(n)}$. Points being distinct implies that $n \mapsto m(n)$ is one-one.
    Now, there are only finitely many naturals which are smaller than $m(1)$. Thus, we can find $n_2 > 1$ large enough so that $m(n_2) > m(1).$ A similar argument shows that we can find $n_3 > n_2$ large enough so that $m(n_3) > m(n_2)$ and so on.
    Putting $n_1 = 1$, we see that $\left(z_{n_i}\right)_i$ is indeed a subsequence of $(x_n)_n$. Now, being a subsequence, $\left(z_{n_i}\right)_i$ has the same limit as $(z_n)_n$ and the argument goes through.
Aryaman Maithani
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  • What is the correspondence between "sequences in $A$" and "subsequences of $(x_n)$"? – azif00 Feb 14 '21 at 04:16
  • There is no natural nice correspondence. But given a sequence (of distinct points!) $(z_n)$ in $A$, one can get a subsequence of $(z_n)$ which is also a subsequence of $(x_n)$. – Aryaman Maithani Feb 14 '21 at 04:18
  • @azif00 I had not mentioned "distinct points" in my answer earlier; have fixed that now. – Aryaman Maithani Feb 14 '21 at 04:19
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    @azif00: Let $\sigma=\langle x_n:n\in\Bbb N\rangle$ and $A={x_n:n\in\Bbb N}$. If you look at closure in terms of sequences, you’ll have to show that every sequence in $A$ has a subsequence that is also a subsequence of $\sigma$. This is not entirely trivial: if $\langle x_{n_k}:k\in\Bbb N\rangle$ is a sequence in $A$, there is no reason to think that the subscripts $n_k$ are increasing with $k$, and if they’re not, then $\langle x_{n_k}:k\in\Bbb N\rangle$ is not a subsequence of $\sigma$, and your hypothesis tells you nothing about it. However, it is possible to show ... – Brian M. Scott Feb 14 '21 at 04:22
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    ... that $\langle x_{n_k}:k\in\Bbb N\rangle$ has a subsequence $\langle x_{n_{k_i}}:i\in\Bbb N\rangle$ such that the indices $n_{k_i}$ are increasing with $i$, so that this is also a subsequence of $\sigma$. With experience you can take this for granted, but at some point in your education you should actually spell it out in detail. – Brian M. Scott Feb 14 '21 at 04:23
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    I have made yet another edit to clarify it. I think that that's what @BrianM.Scott was alluding to. – Aryaman Maithani Feb 14 '21 at 04:26
  • By the way, is it easier to just prove the complement is open? – Bill Feb 14 '21 at 04:30
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Let's have a sharper statement:

If a sequence $(x_n: n\in\mathbb N)$ of a metric space does not contain any subsequence convergent to any point different from any of $\ x_n,\ $ then the set $\{x_n: n\in\mathbb N\}$ is closed in that metric space.

Proof   For every point $\ y\ $ different from all $\ x_n\ $ there exists a radius $\ r_y>0\ $ such that open ball

$$ B_y\ :=\ \{z: d(y\ z)<r_y\} $$

does not contain any of $\ x_n.\ $ Then the union of such balls,

$$ \bigcup_y\, B_y $$

is an open set, and $\ \{x_n: n\in\mathbb N\} $ is closed because it is the complement of the said union.

Wlod AA
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  • Where was the assumption that there is no convergent subsequence used? – Bill Feb 14 '21 at 08:16
  • A convergent subsequence to y would enter an arbitrary ball around y (of a positive radius; by the very definition). – Wlod AA Feb 15 '21 at 03:33