**Interesting Limit Set**
Not homework! Just curious ! :
[tex]\text{Is} \; \mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^n {\frac{i} {n}} = \mathbb{Q} \cap \left[ {0,1} \right] \; ? [/tex] If so, then I will: [tex] {\text{Prove that }}\mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^n {\frac{i}{n}} = \mathbb{Q} \cap \left[ {0,1} \right] [/tex] :smile: If not, then I will: [tex] {\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] [/tex] :smile: 
By:
[tex]\text{Is} \; \mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^n {\frac{i}{n}} = \mathbb{Q} \cap \left[ {0,1} \right] \; ? [/tex] I suppose you mean: [tex]\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] \; ? [/tex] 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 [itex]\mathbb{Q} \cap [0,1][/itex], there exists N > 0 such that q is in [tex]\bigcup\limits_{i = 0}^n \left \{\frac{i}{n}\right \}[/tex] for all n > N. 
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Yes. So unless I made an error, the answer is "No," the limit is not [itex]\mathbb{Q} \cap [0,1][/itex].

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 settheoretic limit...and thus comes the unruly debate :redface: (just thought I'd refer this thread here :shy:) Admittedly, most members of that site seem to be a little too..well, imprecise 
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 settheoretic 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}.

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If you are having difficulty understanding something yourself, post here and I'll do my best. 
Thanks :smile:
From my understanding of the settheoretic limit, in order for [tex]\mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} = \mathbb{Q} \cap \left[ {0,1} \right][/tex] I must show that [tex]\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} [/tex] However, Quote:
[tex] \bigcup\limits_n^\infty {\bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i} {{n!}}} \right\}} } = \mathbb{Q} \cap \left[ {0,1} \right] [/tex] Needless to say, we can also state that: [tex] \bigcup\limits_n^\infty {\bigcup\limits_{i = 0}^n {\left\{ {\frac{i} {n}} \right\}} } = \mathbb{Q} \cap \left[ {0,1} \right] [/tex] *For the limit supremum, [tex] \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} [/tex] The trouble is with the limit infinitum; although it is indeed intuitively plausible :biggrin:, how do I show mathematically that [tex]\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] [/tex] ? 
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[tex]\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][/tex] and to show that, you need to show: [tex]\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][/tex] 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. [tex]\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][/tex] 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(b1)!] * (1/b!) and clearly [a(b1)!] < 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][/tex] 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. 
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]

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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?). 
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: [tex]\{x : \lim _{i \to \infty}x_i = 1\} = \liminf _{i \to \infty} A_i[/tex] and [tex]\{x : \lim _{i \to \infty}x_i = 0\} = (\limsup _{i \to \infty} A_i)^C[/tex] So the [itex]\lim _{i \to \infty}x_i[/itex] is defined for all x iff limsup = liminf. 
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.

Hmm...
Is this correct? [tex] \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} [/tex] 
As you've written it, not really. You're starting by assuming what you want to prove then have a bunch of oneway 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. 
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...notationwise: *Is it: [tex]\mathop {\lim }\limits_{n \to \infty } \inf A_n [/tex] *Or: [tex]\mathop {\lim \inf }\limits_{n \to \infty } A_n [/tex] Which notation is correct? Where do we place the "[itex]n \to \infty[/itex]" ? 3) Can I prove my statement otherwise, using the worded definitions of liminf and limsup, as such: [tex] \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} [/tex] 
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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). Quote:
It's maybe worth noting how latex handles it: [tex]\liminf _{n \to \infty}[/tex] Quote:

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