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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?

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  • $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ – José Carlos Santos Oct 6 '18 at 6:32
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    $\begingroup$ As others have pointed out already, the book's example works fine. Having said that, here is an example you might like more: $1, \frac{1}{2}, 3, \frac{1}{4}, 5, \frac{1}{6}, \ldots$ . Clearly it is an unbounded sequence which has a (possibly more satisfying) limit point at $0$. $\endgroup$ – John Coleman Oct 6 '18 at 13:37
  • $\begingroup$ I know this isn't answering your question (there are other posts below for that), but as some additional information, note that if a sequence is convergent (not just "has a limit point", ie "has a subsequence that converges"), then it is in fact bounded -- the proof is a little exercise :) $\endgroup$ – Sam T Oct 6 '18 at 19:46
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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 $1$.)

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  • $\begingroup$ Thanks for clearing my confusion! It was indeed between sets and sequences. $\endgroup$ – Sankalp Oct 6 '18 at 6:10
  • $\begingroup$ Eric, very clear and nice as usual $\endgroup$ – Peter Szilas Oct 6 '18 at 6:13
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    $\begingroup$ @Sankalp: That's why I think we should never use curly-braces "{ }" for sequences... $\endgroup$ – user21820 Oct 6 '18 at 18:04
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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.

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  • $\begingroup$ But, according to the limit point definition, 'x' is the limit point of a sequence, if every neighborhood of 'x' has infinitely many elements of the sequence. $\endgroup$ – Sankalp Oct 6 '18 at 5:59
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    $\begingroup$ I think Bolzano states that every bounded sequence has a convergence subsequence. $\endgroup$ – Siong Thye Goh Oct 6 '18 at 6:03
  • $\begingroup$ The third paragraph of wikipedia might be of interest to you $\endgroup$ – Siong Thye Goh Oct 6 '18 at 6:06

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