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Let $(u_n)\in \mathbb R ^{\mathbb N}$ a sequence of real numbers. Suppose it is bounded.

There are two cases:

  1. either $u(\mathbb N)$ is infinite.
  2. either $u(\mathbb N)$ is finite.

In the first case, we can use the Weierstrass theorem (in $\mathbb R^k$, every infinite bounded subspace has one accumulation point) to see that $(u_n)$ has a limit point, and hence, has a subsequence converging to that point.

I cannot come to the same conclusion in the second case. Nevertheless, I have learnt in school the "Bolzano-Weierstrass" theorem that says "every bounded real sequence has a converging subsequence". So I should come to the same conclusions in the second case.

I have in mind the example of $u_n =(-1)^n$, its terms are finite ($2$ terms) and there are two converging subsequences. But can we say $1$ and $-1$ are accumulation points of $u_n$ ?

niobium
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    In the second case a constant subsequence will exist and if $c$ is this constant then $c$ a limit point of sequence $(u_n)_n$. However $c$ is not a limit point of set $u(\mathbb N)$. If it comes to limit points then must keep sequences and sets apart. Indeed $1$ and $-1$ in your example are limit points of sequence of $(u_n)$ but not of set ${-1,1}$. Also see here. – drhab Feb 07 '23 at 11:02

1 Answers1

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If $u(\mathbb{N})$ is finite, then there is $v\in u(\mathbb{N})$ such that $u^{-1}(\{v\})$ is infinite (because otherwise $\mathbb{N}$ would be a finite union of finite sets). In particular the restriction of $u$ to $u^{-1}(\{v\})$ is a constant subsequence of $u$, hence convergent.

And indeed, this version of Bolzano-Weierstrass that you've mentioned (btw, it should be "in $\mathbb{R}$", not "in a metric space", it doesn't hold for abitrary metric space) is a different variant of what we typically consider as Bolzano-Weierstrass theorem. Both are very similar though.

But can we say $1$ and $-1$ are accumulation points of $u_n$ ?

No, they are not. At least not under the standard definition of accumulation points. And so as you can see, the limit of some subsequence does not have to be an accumulation point. I get that this can be confusing if you use the term "limit point" instead of "accumulation point", that's why I prefer the latter.

freakish
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  • The last part of your answer confuses me. This SE post and this one indicate that $(u_n)$ admits a limit point $p$ iff there exists a subsequence converging to $p$ – niobium Feb 07 '23 at 12:01
  • @niobium the term "limit point" has multiple meanings, depending on the context. A limit point of a sequence is typically defined differently from the limit point of a subset (which typically is the same thing as accumulation point). Even if that subset happens to be (the image of) a sequence, these don't coincide. Its a subtlety. – freakish Feb 07 '23 at 12:40
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    A limit point of a sequence $(x_n)$, in a metric space $(X,d)$ is typically defined as a an element $y\in X$ such that for any $\epsilon>0$ and any $M>0$ there is $m>M$ such that $d(y,x_m)<\epsilon$. Its similar to simple convergence, except with convergence you have "there exists $M>0$ such that for any $m>M$" instead. – freakish Feb 07 '23 at 12:57
  • Ok so if I sum up: if $u(\mathbb N)$ is infinite, $(u_n)$ will have an accumulation point in the "set" sense; and if $u(\mathbb N)$ is finite, $(u_n)$ will have nonetheless an accumulation point, but here in the "sequence" sense. – niobium Feb 07 '23 at 14:58
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    @niobium yes, correct. – freakish Feb 07 '23 at 15:02