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Given a bounded sequence ${X_n}$ with infinite range $[a,b]$, is it possible to prove that the set of all sub-sequential limits is also $[a,b]$.

My reasoning is as follows:

Every point in range $[a,b]$ is a limit point. Take any point from $[a,b]$, e.g. $c$. $c$ is a limit point. Choose $n_1$ such that $d(c,X_{n_1})<1$. Choose $n_i$ where $i>1$, such that $d(c,X_{n_i})<\frac{1}{i}$. We show that there is a sub-sequence converging to $c$.

Update: Following Coffeemath's comment, I realise that the domain is a countable infinity while the range is an uncountable infinity. However, reading Rudin's Principles of Mathematical Analysis proof for theorem 3.6(a), an infinite range $E$ is possible but I wonder how can we ensure that $E$ is a countable infinity especially since we can show that there are limit points in $E$ hence are there not uncountable number of punctured neighbourhoods around the limit point that is non-empty?

Rudin's theorem 3.6(a) - If ${p_n}$ is a sequence in a compact metric space $X$, then some sub-sequence of ${p_n}$ converges to a point of $X$.

His proof -

Let $E$ be the range of ${p_n}$. If $E$ is finite then there is a $p∈E$ and a sequence ${n_i}$ with $n_1<n_2<n_3<...$, such that

$p_{n_1} = p_{n_2} = p_{n_3} = ... = p$.

The subsequence $\{p_{n_i}\}$ so obtained converges evidently to $p$.

If $E$ is infinite, Theorem 2.37 (see below) shows that $E$ has a limit point $p∈X$. Choose $n_1$ so that $d(p,p_{n_1})<1$. Having chosen $n_1,...,n_{i-1}$, we see from theorem 2.20 (below) that there is an integer $n_i>n_{i-1}$ such that $d(p,p_{n_i})<\frac{1}{i}$. Then $\{p_{n_i}\}$ converges to $p$.

Theorem 2.37 - If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.

Theorem 2.20 - If $p$ is a limit point of a set $E$, then every neighborhood of $p$ contains infinitely many points of $E$.

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  • $\begingroup$ Assuming $a<b,$ since $[a,b]$ is uncountable there is no bounded sequence with that range. $\endgroup$ – coffeemath Oct 4 '20 at 16:01
  • $\begingroup$ Apologies, it did not occur to me that this is not possible, do you happen to have a link that could elaborate on that a bit more or if you could outline the reason why such a bounded sequence does not exist, thank you! $\endgroup$ – sunnydk Oct 4 '20 at 16:06
  • $\begingroup$ A sequence $\{X_n\}$ has a finite or countably infinite range because that range can be covered by the real numbers $X_1,X_2,\cdots .$ So it cannot be all of $[a,b]$ assuming $a<b,$ since the latter is not countable. (one doesn't need boundedness for this argument) $\endgroup$ – coffeemath Oct 4 '20 at 16:41
  • $\begingroup$ @coffeemath thank you for the reply, I got it, but I also wonder does Rudin's proof for theorem 3.6(a) in his Principles of Mathematical Analysis assume a range that is uncountable since a limit point in an infinite range would assume an uncountable number of punctured neighbourhoods that are non-empty? $\endgroup$ – sunnydk Oct 4 '20 at 17:04
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    $\begingroup$ No problem, I will include his proof in the question $\endgroup$ – sunnydk Oct 4 '20 at 17:09
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Example: Let $X$ be the set of reals consisting of $0$ along with each number $1/n$ ($n=1,2,\cdots .$) Then let $p_n=1/n$ and note the only limit point of $p_n$ is $0$ which by construction lies in the set $X$ we defined. [$X$ is a metric space under restriction of the usual metric on the reals.]

For Rudin's second part of proof, $p$ is $0$ (only thing it could be). The rest of Rudin's argument is clear, in this example.

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  • $\begingroup$ Thank you for the example, I see how the range can be countably infinite now. However, I imagine other $p$ might be possible such as a trivial example of $p_n = (1+(1/n))$ converging to $p=1$? $\endgroup$ – sunnydk Oct 4 '20 at 20:15
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    $\begingroup$ Yes, one could make an example by letting $X$ be the set of terms of any convergent sequence $\{x_n\}$ of distinct reals along with its limit, and $p_n=x_n.$ $\endgroup$ – coffeemath Oct 4 '20 at 21:13

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