# Find the set of sub-sequential limits of a bounded sequence with infinite range

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, 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 $$E$$.

• Assuming $a<b,$ since $[a,b]$ is uncountable there is no bounded sequence with that range. – coffeemath Oct 4 '20 at 16:01
• 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! – sunnydk Oct 4 '20 at 16:06
• 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) – coffeemath Oct 4 '20 at 16:41
• @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? – sunnydk Oct 4 '20 at 17:04
• No problem, I will include his proof in the question – sunnydk Oct 4 '20 at 17:09

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.
• 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 $p=1$? – sunnydk Oct 4 '20 at 20:15
• 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 $p_n=x_n.$ – coffeemath Oct 4 '20 at 21:13