*|*Interesting Limit Set*|*

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

 

[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.

 

[bomba923]

I'm not entirely certain on the definition of the limit of a sequence of sets ...

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

 

[AKG]

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

 

[bomba923]

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

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

 

[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}.

 

[bomba923]

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}.

Thank you! Please sign into INTPC (that forum) and help me out!

 

[shmoe]

Thank you! Please sign into INTPC (that forum) and help me out!

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.

 

[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,

The supremum of a sequence of sets is the smallest set containing all the sets, i.e., the countable union of the sets.

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

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

?

 

[AKG]

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

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

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

[tex]\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:

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

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.

 

[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]

 

[shmoe]

With n! your limit will be all the rationals in [0,1], with n you will just get {0,1}.

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?).

 

[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 x_i 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.

 

[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.

 

[bomba923]

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

 

[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.

 

[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:

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

 

[shmoe]

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

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).

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 "$n \to \infty$" ?

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: $$\liminf _{n \to \infty}$$

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

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

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.

 

[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 C_n represent the n^th 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])

 

[shmoe]

Without any reference to a set-theoretic limit, someone on a different forum (not physicsforums) suggested that

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?

 

[bomba923]

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?

Are there any other (official) definitions of a "limit of a sequence of sets" other than the set-theoretic limit (http://en.wikipedia.org/wiki/Set-theoretic_limit)?

 

[Hurkyl]

The only other thing I recall seeing is just a mild generalization: instead of having the sets actually be subsets of one another, you have a chain of maps:

$$S_0 \rightarrow S_1 \rightarrow S_2 \rightarrow S_3 \rightarrow \cdots$$

and then you can take the limit of this diagram:

$$\begin{tabular}{ccccccccc} S_0 & \rightarrow & S_1 & \rightarrow & S_2 & \rightarrow & S_3 & \rightarrow & \cdots \\ \downarrow & & \downarrow & & \downarrow & & \downarrow & &\cdots \\ S & = & S & = & S & = & S & = & \cdots \end{tabular}$$

Of course, this limit won't be unique, since I could replace S with anything of the same cardinality, and just compose all of those downward arrows with a corresponding bijection.

(If all of the rightward arrows are identity maps, then you get something isomorphic to the ordinary set limit)

 

[bomba923]

But where
$$A_n = \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}}$$

Aren't the sets subsets of one another, since
$$\forall n > 0,\;A_n \subset A_{n + 1}$$
?

 

[Hurkyl]

Yes they are. And their limit is Q.

 

[shmoe]

Are there any other (official) definitions of a "limit of a sequence of sets" other than the set-theoretic limit (http://en.wikipedia.org/wiki/Set-theoretic_limit)?

Does it matter? _You_ posed the question and I thought you agreed that was what you meant by this limit of sets?

I had a look at the post in the other forum, surely you can find many things wrong with it? e.g. what does it mean for differences to "be differential"? My guess-absolutely nothing sensible in that case.

 

[bomba923]

I had a look at the post in the other forum, surely you can find many things wrong with it? e.g. what does it mean for differences to "be differential"? My guess-absolutely nothing sensible in that case.

Indeed,
About when I was to post my response about the set-theoretic limit,
I thought of the set sequence containing Cauchy sequence...but that did not work, as I see, in reference to the set-theoretic limit.

*Btw, how did you know which forum & post I was referring to?
(I didn't provide any http://www.intpcentral.com/forums/showthread.php?t=7954&page=9 )

 

[shmoe]

Indeed,
About when I was to post my response about the set-theoretic limit,
I thought of the set sequence containing Cauchy sequence...but that did not work, as I see, in reference to the set-theoretic limit.

*Btw, how did you know which forum & post I was referring to?
(I didn't provide any http://www.intpcentral.com/forums/showthread.php?t=7954&page=9 )

I'm magic. (not really, you linked in page 1 of this thread)

Another problem with this attempted argument, the set limit says nothing about this set containing limit points of it's elements. If you took A(n)=Q, the rationals, you hopefully wouldn't think that the set theoretic limit of A(n) was the reals (assuming we were thinking of the rationals as being a subeset of the reals here).