Converse of Bolzano-Weierstrass theorem

Question

Bolzano-Weierstrass theorem states that every bounded sequence has a limit point. But, the converse is not true.

That is, there are some unbounded sequences which have a limit point. In my course book, I found an example for this claim, but it doesn't make sense.

Here's the example give in the book: The set: {1, 2, 1, 4, 1, 6, ...} is unbounded, but has a limit point of 1. I can't understand how this set has a limit point as 1. According to the book definition of limit point, 'x' is the limit point of a sequence, if every neighborhood of 'x' has infinitely many elements of the sequence. If I apply it here, then I get only infinity as the limit point. Am I missing something?

Answer 1

It seems there is some confusion between sets and sequences. In this example, what you have written as "{1, 2, 1, 4, 1, 6, ...}" is not meant to be a set but rather a sequence $(a_n)$ with $a_0=1,a_1=2,a_2=1,a_3=4,\dots$.

In particular, then, when we say a point $x$ is a limit point of $(a_n)$, this means that for every neighborhood $U$ of $x$, there exist infinitely many $n\in\mathbb{N}$ such that $a_n\in U$. It does not mean there are infinitely many different numbers in the sequence which are in $U$, since these values of $a_n$ for different $n$ might actually be the same. So in this case, since every neighborhood of $1$ contains $1$, it contains $a_0,a_2,a_4,\dots$, and so $1$ is a limit point of the sequence.

(In contrast, $1$ is not a limit point of the set $\{1, 2, 1, 4, 1, 6, \dots\}=\{1, 2, 4, 6, \dots\}$ because $(0,2)$ is a neighborhood of $1$ that contains only one element of this set, namely $1$.)

Answer 2

We have

$$a_n = \begin{cases} 1, & n \text{ is odd} \\ n, & n \text{ is even}\end{cases}$$

$1$ appears infinitely often. Hence there is a subsequence that always take value $1$. That subsequence converges to $1$. Hence $1$ is a limit to the subsequence.

Answer 3

In my point of view, there are two types of confusions. First, a set is being mixed with a sequence. Secondly, limit point of a sequence is being confused with the limit of a sequence. If we take care of these two ( limit of a sequence and limit point of a sequence), then the things are very much clear. Obviously, in this problem, if we regard this as a sequence 1,2,1,4,1,6,... this sequence is clearly unbounded. But, it has a subsequence {a_n}, where a_n=1 for every natural number n. This subsequence has limit 1, because it converges to 1. If we regard {a_n} as a set , then {a_n }={ 1}, for all n. This set has no limit point. Because, we take a neighborhood (1/2, 2) of 1, this neighborhood doesn't contain any other point (different from 1) of the set {1}, so 1 is not a limit point of the set {1}. By the same argument, it shows that 1 is not limit point of the set { 1,2,4,6,...}... Also, 1 is not the limit of the sequence 1,2,1,4,1,6,1,...

The things will be cleared if we focus on the definitions of limit point of a set and limit point of a sequence.

Usually, we talk about the limit of a sequence not about limit point of a sequence. This is the basic fact that creates confusion.

NOTE: Importantly, whenever we talk about limit point of a sequence, it is meant limit point of range set of that sequence.