an example that the limit of a converging sequence is not unique

Question

The following is an example that the limit of a converging sequence is not unique.

There are uncountably many numbers in Cantor Set. Thus, there are uncountably many numbers in the segments in Cantor Set. The length of each segment is the same. Because the number of the endpoints of the segments is countably infinite, the number of the segments is countably infinite. Thus, there has to be more than finitely many numbers in each segment even though the length of the segment is calculated as 0. Otherwise, it leads to a contradiction that finitely many numbers of countably infinite is uncountably infinite. Thus, there has to be more than finitely many numbers between the endpoints in each segment even though the length of the segment is calculated as 0.

Then, let’s consider a sequence of endpoint numbers in each step leading to one segment at its limit in Cantor Set. One may say that Cantor set has no “segments” because its length is 0. A “segment” is just a figurative term because there are endpoints and many internal points within each segment. This sequence converges into that segment at its limit in Cantor Set. Even though the length of the segment is calculated as 0, there are more than finitely many numbers between the endpoint numbers in each segment at its limit. The endpoint numbers are the numbers in the sequence. Therefore, this sequence cannot lead to a single number even at its limit. Thus, a sequence of endpoint numbers in each step leading to one segment at its limit in Cantor Set converges, but its limit is not unique.■

I would like to get critical comments on this example.

Answer 1

The standard proof that the limit of a convergent sequence is unique goes as follows.

Definition: A sequence $\{a_n\}$ converges to a limit $L$ if, for any open set $U$ containing $L$, all but finitely-many points in the sequence are contained in $U$.

Claim: If $\{a_n\}$ converges to $L_1$ and also to $L_2$, then $L_1 = L_2$.

Proof: Let $\{a_n\}$ converge to both $L_1$ and $L_2$ and define $\delta = |L_1-L_2|$. If $\delta>0$, then $I_1=(L_1-\delta/2, L_1+\delta/2)$ and $I_2=(L_2-\delta/2,L_2+\delta/2)$ are open sets containing $L_1$ and $L_2$, respectively. In this case, since $\{a_n\}$ converges to $L_1$, infinitely many of the $a_n$ must lie in $I_1$. However, since $I_1\cap I_2 = \emptyset$, that means that infinitely many $a_n$ must lie outside of $I_2$. This is a contradiction, since we said that $\{a_n\}$ converges to $L_2$. Therefore, we must conclude that $\delta=0$, meaning that $L_1 = L_2$.

This proof is very straightforward and (hopefully) easy to follow. Your proposed counterexample is not - it relies on some misconceptions (e.g. "However, my argument holds because Cantor set is not a set along the way but the limit set, and there are more than finitely many numbers between the endpoints in each segment in this limit set even though the length is calculated as 0." The limit set does not contain endpoints, because it does not consist of intervals).

However, rather than engaging with this and other issues with your proposed example, I would suggest you find a way to show that the simple proof written above fails, since it must fail if you are able to produce a counterexample.