List all Rationals in [a,b]: a + lim x->0 x{n}_0^(floor((b-a)/x))

[bomba923]

Given that ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3} \right\}}$,
Can I list all the rational numbers in $\left[ {a,b} \right]$ as the sequence represented by
$$a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor }$$
----------------------------------------------------------------
My notation is rather strange, but here
$$\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor }$$
really just represents $\left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } \right\}$
(I used floor brackets b/c indexes of terms in a sequence must be whole numbers,
and to not let the last term exceed $b$ when multiplied by $x$ and added to $a$)
----------------------------------------------------------------
For example, let $a = 3$ and $b = 5$. Then, for $x = 0.5$, the sequence will be $\left\{ {3,3.5,4,4.5,5} \right\}$.
Similarly, for $x = 0.1$, the sequence will be $\left\{ {3,3.1,3.2,3.3, \ldots ,5} \right\}$.
And basically, the idea is to list all of the rationals within $\left[ {a,b} \right]$ as $x$ approaches zero (from the right!).
------------------------------------------------------------------
Well then, (for ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$ ,
with $\mathbb{Q}$ being the set of all rationals, that is),
|Will this work?
$${a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor }$$
(Btw, I expressed notation here that may be strange/non-standard.
How would I rewrite the expression to be more understandable--more standard??)

 

[rachmaninoff]

Given that ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3} \right\}}$,
Can I list all the rational numbers in $\left[ {a,b} \right]$ as the sequence represented by
$$a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor }$$

Please explain what you mean. I don't see how

$$a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor }$$

could be a well-defined sequence. If you say it is a sequence, then what are its elements in the second, third, nth positions?

 

[bomba923]

If you say it is a sequence, then what are its elements in the second, third, nth positions?

Ok, but I am NOT trying to make a sequence!--the goal is a "LisT", just a mere "list" of all the rationals in $\left[ {a,b} \right]$. I'm sorry I called it a sequence!---but its NOT a sequence that we're aiming for, just a mere list of all the rationals, surrounding by braces.
---------------
$${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$$
just means that both $a$ and $b$ must be rational, that $b > a$, and that $x$ is a positive rational less than $\left( {b - a} \right)$.
---------------------------
You agree that:
*The sum or difference of any two rationals is always rational: $\forall \left( {a,b} \right) \in \mathbb{Q},\;\left( {a \pm b} \right) \in \mathbb{Q}$
*The product of any two rationals is always rational: $\forall \left( {a,b} \right) \in \mathbb{Q},\;\left( {ab} \right) \in \mathbb{Q}$
*There is a countably infinite quantity of nonequivalent rationals between any two nonequivalent rationals.
*Because there is a countably infinite quantity of nonequivalent rationals between any two nonequivalent rationals, I can make a one-to-one correspondence between $\mathbb{N}$ and all the rationals in [a,b], where both a & b are rational (both $\mathbb{Q}$ and $\mathbb{N}$ have cardinality $\aleph _0$).
*For ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$, we can say $\forall n \in \mathbb{N},\;\left( {a + n \cdot x} \right) \in \mathbb{Q}$.
*(and this one is just for fun: there is a one-to-one correspondence between a rational number and its decimal representation)
---------------------------
Now the formula, using my strange notation:
The list must begin from $a$ to $b$. In accordance with ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$, and the formula
$$a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor }$$:

1) Let's say I want three rationals in the list. Thus, $n = 0,\;{\text{then 1, then 2}}$. Thus--I will have to select an $x$ so that only three values appear---which means list n=0 to n=2. By
$${\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } }$$
we see that the limits of "listing" are from zero to
$$\left\lfloor {\frac{{b - a}}{x}} \right\rfloor$$
Because we want n=0 to n=2, we have to make
$$\left\lfloor {\frac{{b - a}}{x}} \right\rfloor = 2$$
And so we select a possible $x$ value. By the formula, we will get $\left\{ {a + 0 \cdot x,\;a + 1 \cdot x,\;a + 2 \cdot x} \right\}$, which is really the same as saying $a + x \cdot \left\{ {0,1,2} \right\}$. Needless to say, because we want only rationals in our list, we will have to keep $x \in \mathbb{Q}$ (not really an issue, but just so you know!)

2) However, we will want much more than just three rationals!--which means the upper limit of our "listing," a.k.a
$$\left\lfloor {\frac{{b - a}}{x}} \right\rfloor$$
will have to be much more than two---which means that $x \ll \left( {b - a} \right)$, (closer to zero that is, farther from $b - a$). To review, our "list" will look like
$$a + x\left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{b - a}}<br /> {x}} \right\rfloor } \right\}$$
and the last term will be
$$a + x\left\lfloor {\frac{{b - a}}{x}} \right\rfloor = a + \left( {b - a} \right) = b$$
(which is want we want--we want our first term to be $a$ and our last term to be $b$)
-----------------------------------------------------------
Now, for example:
*Well, let $a = 3,\;b = 7$, and let's make $x = 0.1$. Then, our list shall be
$$\left\{ {3,\;3 + 0.1,\;3 + 2 \cdot 0.1,\;3 + 3 \cdot 0.1, \ldots ,3 + 0.1 \cdot \left\lfloor {\frac{{7 - 3}}{{0.1}}} \right\rfloor } \right\} = 3 + 0.1\left\{ {0,1,2,3, \ldots ,40} \right\} = \left\{ {3,3.1,3.2,3.3, \ldots ,7} \right\}$$
And will have 41 terms (or list elements, if you prefer), all of which are rational and within $\left[ {a,b} \right]$.
------------------------------
But! We will obviously want more than just 41 terms--so let's make $x = 0.01$. Well then, we'll have something like this:
$$\left\{ {3,3.01,3.02,3.03, \ldots ,7} \right\}$$
------------------------------
And so, you can make $x$ even smaller---and you'll have more terms, all of which are rationals, in [a,b].
----------------------------------------------------------------
And so then (consistently with ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$), we come to this expression:
$$a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor }$$
(by now rachmaninoff, I think you understand the idea)
-------|Now the question:
And my question is: will this expression indeed work? Will it really list out (not sequence!*remember) all the rationals in [a,b] ?
--------------------------------------------------
Or...
If you accept my awfully disrespectful notation, I'll rephrase the question:
In consistency with ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$,
Does
$$\mathbb{Q}_{\,a}^{\,b} = {a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } } \;\,?$$

 

[TenaliRaman]

Limit of anything (if it exists) is a finite value. I think you are using the idea of "x approaches 0" a bit too liberally. Still if we go out of our way and accept it, then note that as x approaches zero, it can take up irrational values as well and rational+irrational = irrational. So your notation will actually give all reals between a and b, which is concisely denotes as [a,b]

-- AI

 

[bomba923]

Limit of anything (if it exists) is a finite value. I think you are using the idea of "x approaches 0" a bit too liberally. Still if we go out of our way and accept it, then note that as x approaches zero, it can take up irrational values as well and rational+irrational = irrational. So your notation will actually give all reals between a and b, which is concisely denotes as [a,b]

-- AI

Ok, now what part of
$${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$$
do you NOT understand?? Have you read my previous post:

$${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$$
just means that both $a$ and $b$ must be rational, that $b > a$, and that $x$ is a positive rational less than $\left( {b - a} \right)$.

------------------------------------------------
And this?

as x approaches zero, it can take up irrational values as well

(Needless to say, zero is rational)
Obviously you do not understand set notation. OK--but think about this:
Provided that $$n \in \mathbb{N}$$, let $$A_n = n^{ - 1}$$. Now, I think you agree that: $$\mathop {\lim }\limits_{n \to \infty } A_n = 0$$
**!BUT, is $$A_n$$ ever irrational??
|NO!---and yet $$A_n$$ approaches zero! Surprised?
----------------------------------------------------
Also:

So your notation will actually give all reals between a and b, which is concisely denotes as [a,b]

Are you nuts? Do you understand that $$\mathbb{R} = \mathbb{Q} \cup \mathbb{F}$$ and that $$\mathbb{F} \cap \mathbb{Q} = \emptyset$$ ? Good--you can follow along:

You can agree that $$\left[ {\left( {{\text{any }}a \in \mathbb{Q}} \right) \pm \left( {{\text{any }}x \in \mathbb{Q}} \right)} \right] \in \mathbb{Q}$$ and that $$\left[ {\left( {{\text{any }}a \in \mathbb{Q}} \right) \pm \left( {{\text{any }}x \in \mathbb{F}} \right)} \right] \in \mathbb{F}$$.
Thus, knowing that $$\mathbb{F} \cap \mathbb{Q} = \emptyset$$, you cannot, for any one $x$ value, have a sequence of reals (with both rationals and irrationals).
In other words, (sorry for my crazy LaTex manipulation)
$$\left[ {\left( {{\text{any }}a \in \mathbb{Q}} \right) \pm x} \right] \in \left( {{\text{either }}\mathbb{Q}{\text{ or }}\mathbb{F} \ldots {\text{but not BOTH!}}} \right)$$

Thus, whether you get a sequence of rationals or irrationals will
depend on the identity of $x$. Want all reals? Unite all the rationals in [a,b] and irrationals in [a,b]
--the union shall give you all reals in [a,b] (both rationals & irrationals).

(I am Very Sorry if I seem rude or impolite--today was just a bad day (sorry!))
------------------------------------------------
Finally,

I think you are using the idea of "x approaches 0" a bit too liberally

Mmm?? Explain mathematically what's wrong with my use of limits.
(I'm not saying you're wrong, I just want you to expand on this mathematically!)
--------------------------------------------------------------------
But aside from that, for ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$,
what do you other guys think of this expression:
$${a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } }$$
Does it indeed list out all the rationals within $$\left[ {a,b} \right]$$ ?

 

[TenaliRaman]

Ok, now what part of
$${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$$
do you NOT understand??

Sorry, somewhere in my thoughts i forgot you quoted (a,b,x) belongs to Q^3.

Well, however, still the idea of lim does not seem right!
When we say lim_n->00 A_n -> 0, we are saying as n gets larger, A_n tends to zero. It does not describe a sequence, rather it just says that the extreme end values are more towards zero.

You could rather use Union $$\cup_{x \in (0,1)}$$.
This seems more appropriate to me atleast.

(I am Very Sorry if I seem rude or impolite--today was just a bad day (sorry!))

No bother, i can see that you are having a bad day, when you went out of your way to explain to me rationals + irrationals = [a,b]. Its fine, i would be angry with me too, if i were in your place.

-- AI

 

[bomba923]

extreme end values

By the Cauchy definition of a limit of a sequence:
"$$A_n$$ has limit $L$ if $$\exists \,\varepsilon > 0$$ such that $\forall n > {\text{N}}$, $$\left| {A_n - L} \right| < \varepsilon$$ ."

I suppose your "extreme end values" are just values of $$A_n$$, for any $n > N$ ??

$$\cup_{x \in (0,1)}$$

Never seen this notation; what does it mean? (the set of reals exclusively between zero and one??)
ANd

No bother, i can see that you are having a bad day. Its fine, i would be angry with me too, if i were in your place.

Thanks for understanding
-------------------------------------------
Finally,

Well, however, still the idea of lim does not seem right!
When we say lim_n->00 A_n -> 0, we are saying as n gets larger, A_n tends to zero. It does not describe a sequence, rather it just says that the extreme end values are more towards zero.

(Just because it doesn't seem right does not make it wrong or any less right, unless you show mathematically why it is incorrect.)

It does not describe a sequence, rather it just says that the extreme end values are more towards zero.

What is the "it" you are referring to?
-If you are referring to the part
$$x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor }$$
this simply means
$$x \cdot \left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } \right\}$$
The limit applies to $x$, which helps determine the upper bound of the sequence and multiplies with each term in the sequence. In the expression
$$a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor }$$
I am not taking the "limit of the sequence"--I'm just taking the limit of $x$, who's purpose I mentioned in the previous sentence (as you see).
---------------------------------------------------------
|With that said, what do you guys think of the expression:
$$a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor }$$
For ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$, does it indeed list all the rationals within $$\left[ {a,b} \right]$$ ?

 

[rachmaninoff]

You've defined a sequence of sequences: $\left( A_n \right)$; but each element sequence is of a different size (!). If you apply the limit to your sequence, each index of your element sequences converges to zero: {0,0,0...}. You do not get a limiting sequence with the properties you want because there does not exist a 'smallest' rational number greater than zero. If you want to order the rational numbers on [a,b], you have to start with a rational number on [a,b], and continue on listing only rational numbers, as in (if [a,b]=[0,1]):
{1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5...}

Hope this clears things up.

 

[bomba923]

You've defined a sequence of sequences: $\left( A_n \right)$; but each element sequence is of a different size (!).

I specifically said:

If you say it is a sequence, then what are its elements in the second, third, nth positions?

Ok, but I am NOT trying to make a sequence!--the goal is a "LisT", just a mere "list" of all the rationals in [a,b] ...I'm sorry I called it a sequence!---but its NOT a sequence that we're aiming for, just a mere list of all the rationals, surrounding by braces.

*??-If you read,
I never intended for a "sequence"--you must understand; the idea is only to list all the rationals with [a,b]. And no, I do not have a "sequence of sequences" because
(1) I never defined $x$ to be an element of any sequence at all, and
(2) the idea is to make a list, not a sequence

*Also, where did you get $A_n$ ?? I just used $A_n = n ^ {-1}$ as a completely different example to demonstrate to TenaliRaman that values can approach zero without oscillating between rationals and irrationals.

If you apply the limit to your sequence, each index of your element sequences converges to zero: {0,0,0...}

OK--but you understand
$${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$$
which also says $x \ne 0$!

Let me give you an example:
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{x} = \infty$$
!BUT! does
$$\frac{1}{0} = \infty \;?$$
NO!-->
$$\frac{1}{0} = {\text{undefined}}$$
You see, the domain of Reals for $$1 / x$$ is $\left\{ {x|\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)} \right\}$, because $x \ne 0$ (!)

For $\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}$, the expression
$${a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } }$$
takes the limit as $x \to 0$ WITHOUT "$x = 0$".

there does not exist a 'smallest' rational number greater than zero

Right-->which is why there are an infinite quantity of rational numbers in $$\left[ {a,b} \right]$$,
which why I have $x \to 0$, (but $x \ne 0$)
which is why I take $\mathop {\lim }\limits_{x \to 0} \left\lfloor {\frac{{b - a}}{x}} \right\rfloor$,
to get the appropriate infinite quantity of terms.

What exactly is the problem here ??
----------------------------------------------

If you want to order the rational numbers on [a,b], you have to start with a rational number on [a,b], and continue on listing only rational numbers

OK--well u see that

*${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$ already defines $a$ and $b$ to be rational numbers (i.e., $\left( {a,b} \right) \in \mathbb{Q}^2$).
*$\forall \left( {a,x} \right) \in \mathbb{Q}^2 \,{\text{and }}\forall n \in \mathbb{N},\;\left( {a + nx} \right) \in \mathbb{Q}$, so the list will only contain the rationals!

Again, what exactly is the problem here??
------------------------------------------------
That said, I'll end this post like I usually do (like I did for the past three !)
:$\downarrow$
*For ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$, does the expression
$${a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } }$$
Really list out all the rationals in $$\left[ {a,b} \right]$$ ?

 

[rachmaninoff]

Ignoring your attitude problems -

It looks like your question hinges on this: can all rational numbers in [a,b] be expressed as

$$a+\lim_{x\rightarrow \infty}n \frac{b-a}{x}, \, n\in \mathbb{N}$$

The answer is unambigously no.

$$\begin{align*} \lim_{x\rightarrow \infty}n \frac{b-a}{x}&=n \lim_{x\rightarrow \infty}\frac{b-a}{x} \\<br /> &= n \cdot 0 \\<br /> &= 0 \end{align}$$

For all natural numbers n. Not only do you not get an ordering of $\mathbb{Q}$, (restricted to [a,b]), you don't even get the set $\mathbb{Q}$ itself; in fact, you don't get any elements except a + 0 = a. You could rearrange your definition in some way to force

$$a+\lim_{x\rightarrow \infty}\frac{x m (b-a)}{n x}=a+\frac{m}{n} (b-a), \, \forall m,n \in \mathbb{N}$$,

which illustrates how limits can be used unamigously, and let's you get the set Q by plugging in Q itself. Not useful at all.

A couple more comments - there's no important difference between a "seqence" and a "list", sequences are just infinite lists (both are ordered!). For your purposes, your finite lists (L1...Ln) are no different from the corresponding sequences
(L1...Ln, 0, 0...). The important feature is elementwise convergence - in each index, that element converges to zero.

The usual $\mathbb{R}$ (and also Q) does not contain 'infinitissimeals' and 'infinites'. There exist alternative algebras, such as that of the hyperreals, which do contain the 'infiniteisimals' and could (I don't know anything about them) contain a common infiniteisimal denominator for all rational numbers (don't quote me on that). 'Standard' analysis does not allow for this.

And about limits, the definition of the limit applies to sequences/lists, like
$${a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } }$$,
in specific ways - in particular, you take the limit of each index element individually (if they converge). In your example, the elements all go to zero indexwise, so your 'limit' sequence does not have the properties you are trying to force from it.

-rachmaninoff

 

[rachmaninoff]

Let me make this even more clear.

$${a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } }$$

'Limit' implies a sequence. The implied sequence is

$$A_0 = \left( a, b \right)$$
$$A_1 = \left( a, a+\frac{1}{2}(b-a), b \right)$$
$$A_2 = \left( a, a+\frac{1}{3}(b-a), a+\frac{2}{3}(b-a), b \right)$$
$$A_x = \left( a, a+\frac{1}{x}(b-a), ..., a+\frac{x-1}{x}(b-a), b \right)$$

and by every tortured meaning of the word 'limit' you can imagine,

$$\begin{align*} lim_{x \rightarrow \infty} ( A_x ) &= \left( lim_{x \rightarrow \infty}a, lim_{x \rightarrow \infty} a+\frac{1}{x}(b-a), lim_{x \rightarrow \infty} a+\frac{2}{x}(b-a), ... \right) \\ <br /> &= (a,a,a,...,a,a,...) \end{align}$$.

 

[EnumaElish]

For example, let $a = 3$ and $b = 5$. Then, for $x = 0.5$, the sequence will be $\left\{ {3,3.5,4,4.5,5} \right\}$.
Similarly, for $x = 0.1$, the sequence will be $\left\{ {3,3.1,3.2,3.3, \ldots ,5} \right\}$.
And basically, the idea is to list all of the rationals within $\left[ {a,b} \right]$ as $x$ approaches zero (from the right!).

It seems to me the answer is "yes." You will obtain a set with countably infinite number of elements as long as x does not reach 0 (stays away from zero); but you define 0 < x so I think you're okay. Sketch: Suppose your list "skipped over" rational r, a < r < b. You can always express r in terms of a, b, n, and an x that is arbitrarily close to zero. Contradiction.

 

[rachmaninoff]

It seems to me the answer is "yes." You will obtain a set with countably infinite number of elements as long as x does not reach 0 (stays away from zero); but you define 0 < x so I think you're okay.

Dead wrong. All x > 0 yield only a finite number of rational numebers in [a,b] - particularly those rational numbers in [a,b] whose integral denominators are divisble by 1/x. Hence x = 0.1 yields only rational multiples of 1/10, such as {3.0, 3.1, 3.2... 5.0}. Clearly only a finite number of rationals can be represented in this way.

Sketch: Suppose your list "skipped over" rational r, a < r < b. You can always express r in terms of a, b, n, and an x that is arbitrarily close to zero. Contradiction

You got this totally backwards. To illustrate your flaw - suppose the list skips over r in [a,b], and you find an x which generates a list which includes r... well, I can another $r' \in [a,b]$ which is not in this list... so you find a new, smaller x... so on and so on ad infinitum! What you should be proving is there exists an x $\Longrightarrow$ all $r \in \left[ a,b\right]$ are contained in the list generated by x, whereas what you proved is that for all $r \in \left[ a,b\right]$ there exists an x... see the difference?

 

[bomba923]

It looks like your question hinges on this: can all rational numbers in [a,b] be expressed as $$a+\lim_{x\rightarrow \infty}n \frac{b-a}{x}, \, n\in \mathbb{N}$$

*First of all, where the hell do you get $x \to \infty$ when I specifically stated:
$${0 < x < \left( {b - a} \right)}$$

*Secondly, where do you get off eliminating $x$ from
$${\mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } }$$
to getting

$$\lim_{x\rightarrow \infty}n \frac{b-a}{x}$$

?
---------------------------------
And what is this?

For your purposes, your finite lists (L1...Ln) are no different from the corresponding sequences
(L1...Ln, 0, 0...). The important feature is elementwise convergence - in each index, that element converges to zero.

The hell? (--Finite lists?) Since when does (the upper limit of $\left\{ n \right\}$--a.k.a #of terms in the "list")
$$\mathop {\lim }\limits_{x \to 0} \left\lfloor {\frac{{b - a}}{x}} \right\rfloor \ne \infty \;?$$

The important feature is elementwise convergence - in each index, that element converges to zero.

Mmm? If you don't incorrectly eliminate $x$, you would see that for
$$\mathop {\lim }\limits_{x \to 0} x \cdot \left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } \right\}$$
the last term (correctly multiplied by $x$) is
$$\mathop {\lim }\limits_{x \to 0} x \cdot \left\lfloor {\frac{{b - a}}<br /> {x}} \right\rfloor = b - a$$
If you correctly add $a$, as dictated by
$${a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } }$$
...you would see that
$$a + \left( {b - a} \right) = b$$

Therefore, the last term--aka your "convergence" in question, in the infinite "list" is $b$, which is what we want!.
-----------------------------------------------

the definition of the limit applies to sequences/lists, like
$${a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } }$$
in specific ways - in particular, you take the limit of each index element individually (if they converge). In your example, the elements all go to zero indexwise, so your 'limit' sequence does not have the properties you are trying to force from it.

The definition of a limit of a sequence:
"$A_n$ has limit $L$ if $$\exists \,\varepsilon > 0$$ such that $\forall n > {\text{N}}$, $$0 < \left| {A_n - L} \right| < \varepsilon$$

I suppose you're saying:
$$\forall n \in \mathbb{N},\;n\mathop {\lim }\limits_{x \to 0} x = 0$$
Alright-->The very limit in my expression is equal to $\left\{ {0,0, \ldots } \right\}$

*!BUT~: $$0 < \left| {A_n - L} \right| < \varepsilon$$, for all $n > N$, which in my expression is whatever natural value is assigned to ${\left\lfloor {\frac{{b - a}}{x}} \right\rfloor }$.
*And, consistently, for ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$,
$$\forall x > 0,\;\left( {x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } } \right) \cap \left\{ {0,0,0, \ldots } \right\} = \emptyset$$
(*except for the very first term, which a common trivial zero)
---------------------------------------------------------------------
|Thus, because the very limit is, indeed, equal to $\left\{ {0,0,0, \ldots } \right\}$, I'll rephrase my question:
"For ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$ (where $x \ne 0$), will the expression
$${a + x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } }$$
List out all the rationals with $$\left[ {a,b} \right]$$ As $x \to 0$ ?

 

[rachmaninoff]

|Thus, because the very limit is, indeed, equal to $\left\{ {0,0,0, \ldots } \right\}$, I'll rephrase my question:
"For ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$ (where $x \ne 0$), will the expression
$${a + x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } }$$
List out all the rationals with $$\left[ {a,b} \right]$$ As $x \to 0$ ?

This is a meaningless question. Define "As $x \to 0$" and we can talk.

Therefore, the last term--aka your "convergence" in question, in the infinite "list" is b, which is what we want!.

There is no 'last term' in an infinite sequence. This is why your abusive use of the concept 'limit' doesn't give you anything useful.

 

[bomba923]

- particularly those rational numbers in [a,b] whose integral denominators are divisble by 1/x. Hence x = 0.1 yields only rational multiples of 1/10, such as {3.0, 3.1, 3.2... 5.0}.

Eh? I never said
$$x \ne \frac{{93}}{{347}}$$.
But that's not important, because all rationals are divisible (with integral quotients) by 1/denominator. E.g., $$\frac{{93}}{{347}}$$ is divisible (with integral quotients) by $$\frac{{1}}{{347}}$$.

so you find a new, smaller x... so on and so on ad infinitum

Essentially the idea when somebody finds some "$r$" I skipped over!

Dead wrong. All x > 0 yield only a finite number of rational numebers in [a,b]

True:
$$\forall x > 0,\;\left\lfloor {\frac{{b - a}}{x}} \right\rfloor = {\text{finite}}$$
!But That is why I've rephrased the question with simply:
"$$x \to 0$$"
(without actually $x = 0$)---*|You with me here?

 

[rachmaninoff]

*!BUT~: $$0 < \left| {A_n - L} \right| < \varepsilon$$, for all $n > N$, which in my expression is whatever natural value is assigned to ${\left\lfloor {\frac{{b - a}}{x}} \right\rfloor }$.
*And, consistently, for ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$,
$$\forall x > 0,\;\left( {x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } } \right) \cap \left\{ {0,0,0, \ldots } \right\} = \emptyset$$
(*except for the very first term, which a common trivial zero)
-

This behavior does not persist in the limiting case. Think of this analogy: $f_n(x)=x^n$ is continuous on R for all finite n, but the limit diverges everywhere except on (-1,1). $f_n(x)=x^n$ has every real number in (0,1) in its range, but the limiting case does not.

 

[bomba923]

(*Btw--what exactly happened to TenaliRaman and EnumaElish? How come no further posts/feedback from them??)

This behavior does not persist in the limiting case. Think of this analogy: $f_n(x)=x^n$ is continuous on R for all finite n, but the limit diverges everywhere except on (-1,1). $f_n(x)=x^n$ has every real number in (0,1) in its range, but the limiting case does not.

Of course, that's why I removed the "limiting" case -->
No "limit" (and for a good reason!)--
and rephrased my question & expression for that matter:
|Now,
"For ${\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}}$, will the expression
$${a + x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}<br /> {x}} \right\rfloor } }$$
->List all the rationals within $$\left[ {a,b} \right]$$ as $x \to 0$ ?

 

[rachmaninoff]

!But That is why I've rephrased the question with simply: "$$x \to 0$$"
(without actually x=0 )---*|You with me here?

I am well aware of the difference. Let me rephrase your question in several different ways:

I) Let A(r) represent the number of rational numbers (in [a,b]) in the list generated by r. For instance, with [0,1], A(0.2) = 6.
Then $\lim_{r \rightarrow 0} A(r)$ does not exist - for the pathological reason that some rs are irrational and do not divide rational numbers.

II) Let's restrict r to rational numbers of the form 1/n, where n is a natural number. Then $\lim_{n \rightarrow + \infty } A(1/n) = + \infty$.

III) Let B(r') be the maximum value of A(r) attained on $r \in (r', + \infty )$. Then trivially $\lim_{r' \rightarrow 0} B(r') = + \infty$.

Now Let 'x' be a fixed rational in [a,b], and let C(r) be a function which is
i) 0 if 'x' is not in the list generated by r, and
ii) 1 if 'x' is in the list generated by r.

Then
$\lim_{r \rightarrow 0} C(r)$ does not exist (you can get either '0' or '1' values arbitrarily close to r=0).

$\lim_{n \rightarrow + \infty} C(1\n)$ does not exist (also, you can get either '0' or '1' values arbitrarily close to r=0, thinking for a moment about divisibilty).

But, if D(r') = the maximum value of C(r) attained on all of $r \in (r', + \infty )$, then
$\lim_{r; \rightarrow 0 } D(r') = 1$, meaning that there is some r > 0 such that 'x' is in the list generated by C(r).

----------------------

Let's try something new. Define S(r) as the set of rationals in [a,b] contained in the 'list' generated by r (in your definition). Then,

$$\bigcup_{r \in (0, + \infty ) } S(r) = \mathbb{Q} \cap [a,b]$$,

which is a type of 'limiting behavior' from which you generate the rationals on [a,b]. More cleanly,

$$\lim_{n \rightarrow + \infty} \bigcup_{m \in \{1,2...n\} } S(1/m) = \mathbb{Q} \cap [a,b]$$.

The 'union' set operator is essential here. Clearly,
$$\lim_{n \rightarrow + \infty } S(1/n)$$
is undefined.

 

[EnumaElish]

Dead wrong. ... You got this totally backwards. To illustrate your flaw - suppose the list skips over r in [a,b], and you find an x which generates a list which includes r... well, I can another $r' \in [a,b]$ which is not in this list... so you find a new, smaller x... so on and so on ad infinitum! What you should be proving is there exists an x $\Longrightarrow$ all $r \in \left[ a,b\right]$ are contained in the list generated by x, whereas what you proved is that for all $r \in \left[ a,b\right]$ there exists an x... see the difference?

I cannot say I don't see the difference between your use of "lim" and the usage in the original post, but bomba acknowledged it was being used loosely. I think I understand bomba's thought process, and it works for me. I am not saying you are being pedantic, rachmaninoff, you have just taken a more detail-oriented approach than I have, and I am not going to quarrel with your approach here. Good luck, you two.

 

[EnumaElish]

rachmaninoff, curious here: if q is rational, and u is a suitably defined upper bound, will the expression $$a + \left\{ n/q \right\}_{n=0}^u$$ list all rationals in [a,b) as $$q\longrightarrow +\infty$$?

 

[rachmaninoff]

rachmaninoff, curious here: if q is rational, and u is a suitably defined upper bound, will the expression $$a + \left\{ n/q \right\}_{n=0}^u$$ list all rationals in [a,b) as $$q\longrightarrow +\infty$$?

I've answered this already; the notation is ambiguous, the question has no meaning! For instance, if you're looking at the limit of sequences

$$A_1=\left( a, b \right)$$
$$A_2=\left( a, a+\frac{1}{2}(b-a) , b \right)$$
$$A_3 = \left( a, a+\frac{1}{3}(b-a) , a+\frac{2}{3}(b-a), b \right)$$
$$A_4 = \left( a, a+\frac{1}{4}(b-a) , a+\frac{2}{4}(b-a), a+\frac{3}{4}(b-a), b \right)$$
$$A_x = \left( a, a+\frac{1}{x}(b-a), ..., a+\frac{x-1}{x}(b-a), b \right)$$

Then the limit we're looking at is for all purposes
$$\begin{align*} \lim_{q \rightarrow + \infty}A_x &= \lim_{q \rightarrow + \infty}\left( a, a+\frac{1}{q}(b-a), ..., a+\frac{q-1}{q}(b-a), b, 0, 0, 0, 0, 0... \right) \\<br /> &= (0,0,0,0,0...)\\<br /> &=\mbox{nada} \end{align}$$
(regardless of whether q is restricted to rationals or not).

If you're looking at the sets
$$S_q=\{ a+\frac{n}{q} (b-a) | n \in \mathbb{N}, 0 \leq n \leq q \}$$
then the limit does not exist, no matter how you attempt to define 'limits' of a sequence of sets (a tortured definition, no doubt). This is obvious if you consider that for any q and
$$S_q=\{ a+\frac{n}{q} (b-a) | n \in \mathbb{N}, 0 \leq n \leq q \}$$,
you can find any number of q' > q such that
$$S_{q'}= \{ a+\frac{n}{q'} (b-a) | n \in \mathbb{N}, 0 \leq n \leq q' \}$$
is a 'completely different set' (in qualitive terms); for instance, you can find a q' relatively prime to q; then the two sets would be almost disjoint (the intersection would be {a,b}). For example,
$$S_{4}=\{ a,a+\frac{1}{4} (b-a), a+\frac{2}{4} (b-a), a+\frac{3}{4} (b-a), b\}$$
$$S_{5}=\{a,a+\frac{1}{5} (b-a), a+\frac{2}{5} (b-a), a+\frac{3}{5} (b-a), a+\frac{4}{5} (b-a), b\}$$
$$S_{4} \cup S_{5} = \{ a,b\}$$
$$S_{400} \cup S_{401} = \{ a,b\}$$
etc.

There is no reasonable way to define 'convergence' to get the S's to converge to anything.

Or you could extend your definition with a set union over a sequence of rationals 1/n, like I did in my previous post. That would give you Q.

 

[TenaliRaman]

By the Cauchy definition of a limit of a sequence:
"$$A_n$$ has limit $L$ if $$\exists \,\varepsilon > 0$$ such that $\forall n > {\text{N}}$, $$\left| {A_n - L} \right| < \varepsilon$$ ."

I suppose your "extreme end values" are just values of $$A_n$$, for any $n > N$ ??

Yes.

Never seen this notation; what does it mean? (the set of reals exclusively between zero and one??)

Almost yes, it means union over all x in (0,1), one could specify "x in (0,1) and x in Q" but since you already declared x being in Q, i felt it unnecessary to repeat there.

(Just because it doesn't seem right does not make it wrong or any less right, unless you show mathematically why it is incorrect.)

I can see why you are using the limit notation and i believe its because, we are used to this language of speech when talking of limits (of a sequence say)
x approaches 0 (from right side) means x takes values of 0.5,0.1,0.01 and so on towards zero and we see that eventually all elements of the sequence get as close as we want to the limit.
This process however is an inituitive way of looking at limits, however its not the standard way of specifying values of x as you have done. Anyone looking at your notation, first hand, may not even realize what it means. Thats why my suggestion of the use of union.

Simply replace your "lim as x->0" with "union over all x in (0,1) and x rational" and what you are suggesting seems perfectly clear.

-- AI
P.S -> Apparently, i only had problems with your notation, but i do maintain that since x is specified as rational, your idea does generate all rationals (between a and b that is).

 

[rachmaninoff]

At least we agree:

Simply replace your "lim as x->0" with "union over all x in (0,1) and x rational" and what you are suggesting seems perfectly clear.

$$\lim_{n \rightarrow + \infty} \bigcup_{m \in \{1,2...n\} } S(1/m) = \mathbb{Q} \cap [a,b]$$.

The 'union' set operator is essential here. Clearly,

$$\lim_{n \rightarrow + \infty } S(1/n)$$

is undefined.

 

[EnumaElish]

From rachmaninoff:
$$\begin{align*} \lim_{q \rightarrow + \infty}A_x &= \lim_{q \rightarrow + \infty}\left( a, a+\frac{1}{q}(b-a), ..., a+\frac{q-1}{q}(b-a), b, 0, 0, 0, 0, 0... \right) \\&= (0,0,0,0,0...)\\&=\mbox{nada} \end{align}$$

In fact, with your (technically correct) usage of limits,
$$\begin{align*} \lim_{q \rightarrow + \infty}A_x &= \lim_{q \rightarrow + \infty}\left( a, a+\frac{1}{q}(b-a), ..., a+\frac{q-1}{q}(b-a), b\right) \\&= (a,a,...,a,b,...,b,b)\\&=\{a,b\} \end{align}$$

 

[rachmaninoff]

$$\begin{align*} \lim_{q \rightarrow + \infty}A_x &= \lim_{q \rightarrow + \infty}\left( a, a+\frac{1}{q}(b-a), ..., a+\frac{q-1}{q}(b-a), b\right) \\&= (a,a,...,a,b,...,b,b)\\&=\{a,b\} \end{align}$$

Actually it's not that, nothing converges to b anywhere. It's just
(a, a, a, a...) all the way through. Also, you can't interchange a sequence with a set, it's nonsensical.

-rachmaninoff

 

[EnumaElish]

Actually it's not that, nothing converges to b anywhere. It's just
(a, a, a, a...) all the way through. Also, you can't interchange a sequence with a set, it's nonsensical.

-rachmaninoff

$$\lim_{q \rightarrow + \infty}\left[a+\frac{q-1}{q}(b-a)\right]=a+(b-a)\lim_{q \rightarrow + \infty}\frac{q-1}{q}=a+(b-a)\times 1 = b,$$ n'est pas?

 

[TenaliRaman]

hmm?
Consider (q-q/2)/q = 1/2
a+(1/2)(b-a) = (a+b)/2, shouldn't this be in the list too
how abt (q-q/3)/q = 2/3
a+(2/3)(b-a) = (a+2b)/3, shouldn't this be in the list too
Some more bizarre results are in the offing too
Ofcourse we are dragging the idea of "limit of a sequence" rather too far out of track here.

At least we agree:

Oh yes we do.

-- AI

 

[rachmaninoff]

$$\lim_{q \rightarrow + \infty}\left[a+\frac{q-1}{q}(b-a)\right]=a+(b-a)\lim_{q \rightarrow + \infty}\frac{q-1}{q}=a+(b-a)\times 1 = b,$$ n'est pas?

That's exactly the problem - the limiting sequence does not contain things like (q-1)/q (as I've said repeatedly, it's ill-defined). That would be tantamount to an infinite sequence having a final element (the one after (q-1)/q)).

 

[EnumaElish]

That's exactly the problem - the limiting sequence does not contain things like (q-1)/q (as I've said repeatedly, it's ill-defined). That would be tantamount to an infinite sequence having a final element (the one after (q-1)/q)).

I know you are technically correct. But I cannot get rid of the feeling that your objection is like saying 1 $\ne$ 0.999... for the reason that "one would never get to the last 9." I know intuition can give you the wrong answer, this must be one of those times.