# *|*Interesting Limit Set*|*

1. Oct 23, 2005

### bomba923

Not homework! Just curious ! :

$$\text{Is} \; \mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^n {\frac{i} {n}} = \mathbb{Q} \cap \left[ {0,1} \right] \; ?$$

If so, then I will:
$${\text{Prove that }}\mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^n {\frac{i}{n}} = \mathbb{Q} \cap \left[ {0,1} \right]$$

If not, then I will:
$${\text{Prove that }}\mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^n {\frac{i}{n}} \ne \mathbb{Q} \cap \left[ {0,1} \right]$$

Last edited: Oct 23, 2005
2. Oct 23, 2005

### AKG

By:

$$\text{Is} \; \mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^n {\frac{i}{n}} = \mathbb{Q} \cap \left[ {0,1} \right] \; ?$$

I suppose you mean:

$$\text{Is} \; \mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^n \left \{\frac{i}{n}\right \} = \mathbb{Q} \cap \left[ {0,1} \right] \; ?$$

Anyways, I believe the answer is "No". I'm not entirely certain on the definition of the limit of a sequence of sets, but I would suspect that if the limit were to be what you hypothesize it is, then for all q in $\mathbb{Q} \cap [0,1]$, there exists N > 0 such that q is in

$$\bigcup\limits_{i = 0}^n \left \{\frac{i}{n}\right \}$$

for all n > N.

Last edited: Oct 23, 2005
3. Oct 23, 2005

### bomba923

!Hey, are you referring to the set-theoretic limit ?

Last edited: Oct 24, 2005
4. Oct 24, 2005

### AKG

Yes. So unless I made an error, the answer is "No," the limit is not $\mathbb{Q} \cap [0,1]$.

5. Dec 17, 2005

### bomba923

Hmm...it seems we have quite a spirited debate regarding that question on http://www.intpcentral.com/forums/showthread.php?t=7954 [Broken].

However, we don't use the set-theoretic limit...and thus comes the unruly debate

(just thought I'd refer this thread here :shy:)

Admittedly, most members of that site seem to be a little too..well, imprecise

Last edited by a moderator: May 2, 2017
6. Dec 18, 2005

### shmoe

I only looked briefly at that other website, but replacing the "n" in the upper limit of the union to "n!" changes the limit (this is assuming you mean the usual set-theoretic limit like the one you linked to). With n! your limit will be all the rationals in [0,1], with n you will just get {0,1}.

7. Dec 18, 2005

### bomba923

Last edited: Dec 18, 2005
8. Dec 18, 2005

### shmoe

Help what out? If you understand what's going on, it looks like you need to explain to them what a set theoretic limit is. It looks like this is where the trouble is, and without understanding what you mean by the limit there is no hope at all of understanding your question.

If you are having difficulty understanding something yourself, post here and I'll do my best.

9. Dec 19, 2005

### bomba923

Thanks
From my understanding of the set-theoretic limit, in order for
$$\mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} = \mathbb{Q} \cap \left[ {0,1} \right]$$

I must show that
$$\begin{gathered} \mathop {\lim }\limits_{n \to \infty } \inf \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} = \mathop {\lim }\limits_{n \to \infty } \sup \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} \Rightarrow \hfill \\ \mathop {\lim }\limits_{k \to \infty } \bigcup\limits_k {\bigcap\limits_{n \geqslant k} {\bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} } } = \mathop {\lim }\limits_{k \to \infty } \sup \bigcap\limits_k {\bigcup\limits_{n \geqslant k} {\bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} } } \hfill \\ \end{gathered}$$

However,
Obviously, we know that
$$\bigcup\limits_n^\infty {\bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i} {{n!}}} \right\}} } = \mathbb{Q} \cap \left[ {0,1} \right]$$

Needless to say, we can also state that:
$$\bigcup\limits_n^\infty {\bigcup\limits_{i = 0}^n {\left\{ {\frac{i} {n}} \right\}} } = \mathbb{Q} \cap \left[ {0,1} \right]$$

*For the limit supremum,
$$\begin{gathered} \because \forall k \in \mathbb{N},\;\bigcup\limits_{n \geqslant k} {\bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} } = \mathbb{Q} \cap \left[ {0,1} \right] \Rightarrow \hfill \\ \therefore \mathop {\lim }\limits_{k \to \infty } \sup \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} = \bigcap\limits_k {\bigcup\limits_{n \geqslant k} {\bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} } } = \mathbb{Q} \cap \left[ {0,1} \right] \hfill \\ \end{gathered}$$

The trouble is with the limit infinitum; although it is indeed intuitively plausible , how do I show mathematically that

$$\bigcup\limits_k {\bigcap\limits_{n \geqslant k} {\bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} } } = \mathbb{Q} \cap \left[ {0,1} \right]$$

?

10. Dec 19, 2005

### AKG

No, you need to show that:

$$\mathop {\liminf} _{n \to \infty} \bigcup _{i=0} ^{n!} \left \{\frac{i}{n!}\right \} = \mathop {\limsup} _{n \to \infty} \bigcup _{i=0} ^{n!} \left \{ \frac{i}{n!}\right \} = \mathbb{Q} \cap [0,\, 1]$$

and to show that, you need to show:

$$\bigcup _{k = 0} ^{\infty} \bigcap _{n = k} ^{\infty} \bigcup _{i=0} ^{n!} \left \{ \frac{i}{n!}\right \} = \bigcap _{k = 0} ^{\infty} \bigcup _{n = k} ^{\infty} \bigcup _{i=0} ^{n!} \left \{ \frac{i}{n!}\right \} = \mathbb{Q} \cap [0,\, 1]$$

Alternatively, you can use the first method on the link given, by looking at what they call "indicator variables." However, I think the method you started with is easy enough.

$$\bigcup _{k = 0} ^{\infty} \bigcap _{n = k} ^{\infty} \bigcup _{i=0} ^{n!} \left \{ \frac{i}{n!}\right \} = \bigcup _{k = 0} ^{\infty} \bigcup _{i=0} ^{k!} \left \{ \frac{i}{k!}\right \} = \mathbb{Q} \cap [0,\, 1]$$

The first equality holds because if m > n, then

$$\bigcup _{i=0} ^{n!} \left \{ \frac{i}{n!}\right \} \subset \bigcup _{i=0} ^{m!} \left \{ \frac{i}{m!}\right \}[/itex] since if m > n, then 1/n! is just a multiple of 1/m!. The second equality holds because every fraction will occur, because the fraction a/b in [0, 1] will occur when k = b, since a/b is just a multiple of 1/b!, specifically a/b = [a(b-1)!] * (1/b!) and clearly [a(b-1)!] < b! otherwise a/b > 1. Also: [tex]\bigcap _{k = 0} ^{\infty} \bigcup _{n = k} ^{\infty} \bigcup _{i=0} ^{n!} \left \{ \frac{i}{n!}\right \} = \bigcap _{k = 1} ^{\infty} (\mathbb{Q} \cap [0,\, 1]) = \mathbb{Q} \cap [0,\, 1]$$

The second equality holds for obvious reasons. The first is also obvious, especially given what I said about the liminf just before this, and you can figure that out on your own.

11. Dec 19, 2005

### shmoe

Let A(n)={0, 1/n!, 2/n!,...,n!/n!}. Then A(n) is contained in A(n+1). The limit is then simply the union of these sets. Easy enough to show this is the rationals in [0,1]

12. Dec 19, 2005

### shmoe

I thought about this a little more, and with an "n" the limit doesn't exist. The page you linked to gives an incorrect definition of the limit of a sequence of sets. The one with "indicator variables" is not equivalent to the limsup/liminf version.

The liminf/limsup version is correct, the indicator variable one actually gives the liminf, which is all elements not in only a finite number of sets in your sequence. The limsup gives all elements in infinitely many of the sets in your sequence.

This doesn't change my response with "n!" as it's a nested increasing sequence. With "n", the liminf and limsup are different (can you find them?).

Last edited: Dec 19, 2005
13. Dec 19, 2005

### AKG

shmoe

I thought the very same thing about the indicator variable formulation (although this was a while back when I saw this definition on wikipedia) and the two definitions actually are the same. You have to pay attention to the fine print:

If the limit as i goes to infinity of xi exists for all x

You will find that:

$$\{x : \lim _{i \to \infty}x_i = 1\} = \liminf _{i \to \infty} A_i$$

and

$$\{x : \lim _{i \to \infty}x_i = 0\} = (\limsup _{i \to \infty} A_i)^C$$

So the $\lim _{i \to \infty}x_i$ is defined for all x iff limsup = liminf.

14. Dec 19, 2005

### shmoe

Ahh, good point. I missed the bit about the limit existing for all x. The limsup of the sets is all elements whose "indicator sequence" has limsup=1 (the usual limsup of a sequence here), the liminf of the sets is all elements whose indicator sequence has liminf=1. If the limits exist for all x, then the limsup and liminf of the sets are obviously equal, and vice versa.

15. Dec 31, 2005

### bomba923

Hmm...
Is this correct?
$$\begin{gathered} \mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} = \mathbb{Q} \cap \left[ {0,1} \right] \Rightarrow \hfill \\ \mathop {\lim }\limits_{n \to \infty } \inf \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} = \mathop {\lim }\limits_{n \to \infty } \sup \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} = \mathbb{Q} \cap \left[ {0,1} \right] \Rightarrow \hfill \\ \bigcup\limits_{k \geqslant 0} {\bigcap\limits_{n \geqslant k} {\bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} } } = \bigcap\limits_{k \geqslant 0} {\bigcup\limits_{n \geqslant k} {\bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} } } = \mathbb{Q} \cap \left[ {0,1} \right] \Rightarrow \hfill \\ \bigcup\limits_{k \geqslant 0} {\bigcup\limits_{i = 0}^{k!} {\left\{ {\frac{i}{{k!}}} \right\}} } = \bigcap\limits_{k \geqslant 0} {\left( {\mathbb{Q} \cap \left[ {0,1} \right]} \right)} = \mathbb{Q} \cap \left[ {0,1} \right] \Rightarrow \hfill \\ \mathbb{Q} \cap \left[ {0,1} \right] = \mathbb{Q} \cap \left[ {0,1} \right] = \mathbb{Q} \cap \left[ {0,1} \right]. \hfill \\ \therefore \mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} = \mathbb{Q} \cap \left[ {0,1} \right] \hfill \\ \end{gathered}$$

16. Dec 31, 2005

### shmoe

As you've written it, not really. You're starting by assuming what you want to prove then have a bunch of one-way implications. These are all actually two way implications, "if and only ifs", but you don't mention this. Take a look again at how AKG organized it, considering the lim inf and lim sup seperately, and his justifications for each step.

Incidently, if you can show that the lim inf is the rationals in [0,1] since you also know lim inf is contained in lim sup (which is clearly contained in the rationals in [0,1] here), you must have lim inf=lim sup.

17. Jan 1, 2006

### bomba923

1) For any sequence of sets where lim inf and lim sup exist, will lim inf always be a subset of lim sup ?

2) Ok...notation-wise:

*Is it:
$$\mathop {\lim }\limits_{n \to \infty } \inf A_n$$

*Or:
$$\mathop {\lim \inf }\limits_{n \to \infty } A_n$$

Which notation is correct? Where do we place the "$n \to \infty$" ?

3) Can I prove my statement otherwise, using the worded definitions of liminf and limsup, as such:

$$\begin{gathered} \forall A \subset \mathbb{Q} \cap \left[ {0,1} \right],\; \hfill \\ \exists N \in \mathbb{N}\;{\text{such that }}\forall n > N,\;A \subset \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} , \hfill \\ {\text{and }}\forall N \in \mathbb{N},\;\exists n > N\;{\text{such that }}A \subset \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} \hfill \\ \end{gathered}$$

Last edited: Jan 2, 2006
18. Jan 2, 2006

### shmoe

lim sup and lim inf always exist, and yes lim inf will be a subset of lim sup. lim inf is all elements that are excluded from a finite number of sets only, lim sup is all elements that are in infinitely many of the sets, excluded from finitely many=> included in infinitely many.

Compare with the usual lim inf and lim sup of real sequances which always exist and are finite when the sequences are bounded. Any sequence of sets is "bounded" below by the empty set and above by the union of the sets (with the containment partial ordering).

The second way, under both the lim and the inf. think of "lim inf" as one symbol. It's sometimes written as $$\underline{\lim}$$ for normal sequences (of real numbers) and I'd expect for sets as well ($$\overline{\lim}$$ for lim sup).

It's maybe worth noting how latex handles it: $$\liminf _{n \to \infty}$$

I'm not sure what you're getting at here, since this is an increasing sequence the first statement follows from the second, and the second is a little silly-why bother with the N bound? Though these are certainly false if A is not finite.

Last edited: Jan 2, 2006
19. Mar 7, 2006

### bomba923

This may seem rather strange...
Without any reference to a set-theoretic limit, someone on a different forum (not physicsforums) suggested that

$$\mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} = \left[ {0,1} \right]$$

As it contains all Cauchy sequences whose terms are within the unit interval.

For example, let
$$A_n = \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}}$$

and let Cn represent the nth term in the Cauchy sequence (beginning with C1)
$$\left\{ {0.3,0.31,0.314, \ldots } \right\} = \left\{ {\frac{3}{{10}},\frac{{31}}{{100}},\frac{{314}}{{1000}}, \ldots } \right\}$$
which converges to $\pi / 10$.

As such,
$$\forall n > 24,\;C_n \in A_n$$
-----------------------------------------------
*Does, then,
$$\mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} = \left[ {0,1} \right]\;?$$

(due to Cauchy sequences that converge to reals in [0,1])

Last edited: Mar 7, 2006
20. Mar 7, 2006

### shmoe

So they are taking those symbols (you limit of sets) to mean something else (who knows what, it looks like set closure for some reason) and get a different answer that has nothing to do with your problem? So what?