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 January 11th, 2017, 04:05 PM #121 Senior Member   Joined: Jun 2014 From: USA Posts: 363 Thanks: 26 I didn't see your above post when I did this, only your second to most recent. In the following I focused on making what I'm saying very clear and concise. The argument is fairly simple. It starts with defining, for any rational q, the set $A_q$: $A_q$ = { x : x + q is a rational in (0, 1] } = (-q, 1-q] $\cap$ $\mathbb{Q}$ As an example, if q = 0, then $A_0 = (0, 1] \cap \mathbb{Q}$. As q increases, each element of $A_q$ decreases. If q = 1, we see that there is no element of $A_1 = (-1, 0] \cap \mathbb{Q}$ that is greater than 0. If $0 \leq q < 1$, then $A_q$ will contain at least one rational number that is greater than 0. The following statement is therefore true: For all $q \in [0, 1) \cap \mathbb{Q}$, there exists at least one rational number in $A_q$ that is greater than 0. So, I asked the question: Is there a p such that p is in all $A_q$ such that $q \in [0, 1) \cap \mathbb{Q}$? We cannot compute what rational number p is actually in all $A_q$ such that $q \in [0, 1) \cap \mathbb{Q}$. For each computable p that we try, we’ll note that p > 1-q for an infinite number of $q \in [0, 1) \cap \mathbb{Q}$. Since p cannot be an element of $A_q$ if p > 1-q, we conclude that p is not an element of all $A_q$ such that $q \in [0, 1) \cap \mathbb{Q}$. This allows for another true statement: There is no computable p such that p is in all $A_q$ such that $q \in [0, 1) \cap \mathbb{Q}$. I don’t find a contradiction in recognizing that all relevant $A_q$ contain a rational greater than 0 but no computable rational greater than 0 is actually in all relevant $A_q$. This is just a consequence of the rationals being dense in the reals. … Now Maschke, here is where we appear to differ: … However, if all relevant $A_q$ contain a rational greater than 0 but no rational greater than 0 is actually in all relevant $A_q$, we do in fact have a contradiction.
January 11th, 2017, 04:19 PM   #122
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Quote:
 Originally Posted by AplanisTophet I didn't see your above post when I did this, only your second to most recent. In the following I focused on making what I'm saying very clear and concise. The argument is fairly simple. It starts with defining, for any rational q, the set $A_q$: $A_q$ = { x : x + q is a rational in (0, 1] } = (-q, 1-q] $\cap$ $\mathbb{Q}$ As an example, if q = 0, then $A_0 = (0, 1] \cap \mathbb{Q}$. As q increases, each element of $A_q$ decreases. If q = 1, we see that there is no element of $A_1 = (-1, 0] \cap \mathbb{Q}$ that is greater than 0. If $0 \leq q < 1$, then $A_q$ will contain at least one rational number that is greater than 0.
This "at least one rational number" terminology is killing me.

$0\leq q < 1 \Rightarrow A_q \text{ contains infinite rational numbers}$

there are infinite rationals in the tiniest non-degenerate interval.

January 11th, 2017, 04:28 PM   #123
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Quote:
 Originally Posted by AplanisTophet There is no computable p such that p is in all $A_q$ such that $q \in [0, 1) \cap \mathbb{Q}$.
I'm baffled that we are now talking about computable numbers, which are a completely different topic. In any event all rational numbers are computable so this is not good terminology to use here.

But you've already proven that there's no $p$ in the intersection of the $A_q$'s so I myself don't even have to make this point. You've already completed the conversation. There is no $p$, you proved it.

Quote:
 Originally Posted by AplanisTophet Now Maschke, here is where we appear to differ: … However, if all relevant $A_q$ contain a rational greater than 0 but no rational greater than 0 is actually in all relevant $A_q$, we do in fact have a contradiction.
Of course this is false. All cats have paws, but no paw belongs to all cats. That's not a contradiction, it's common sense.

You're simply having a hard time squaring your incorrect intuition with the math that shows you're wrong. When the math shows your intuition is wrong, believe the math.

Last edited by Maschke; January 11th, 2017 at 04:53 PM.

January 11th, 2017, 08:40 PM   #124
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Ok. I’ll assume my intuition has led me astray as you say.

If I do that, then I have one last question for you (sorry).

I assume I'm wrong, so there is no rational p such that p is an element of every set $A_q$ such that $q \in [0, 1) \cap \mathbb{Q}$ (see the copy of my definitions below if unfamiliar). From this, we know that 1 is not an element of:

B = $\{0\} \cup \{ y \in \mathbb{Q} : y > 0$ and $\exists p$ such that $p$ is an element of every set $A_q$ such that $q \in [0, y) \cap \mathbb{Q} \}$

For any rational y > 0, it is possible to test whether or not there exists a p such that p is an element of every set $A_q$ such that $q \in [0, y) \cap \mathbb{Q}$. For example, if y = 0.5, we know that 0.25 is a suitable p because 0.25 is in every set $A_q$ such that $q \in [0, 0.5)$, so 0.5 is in B.

For each rational y where 0 $\leq$ y < 1, we can compute a suitable p right? I believe that means B = $[0, 1) \cap \mathbb{Q}$, does it not... or wait, but it can't? If so, then by definition, isn't this a contradiction? What is B? EXPLAIN PLEASE!!!

Quote:
 Originally Posted by AplanisTophet The argument is fairly simple. It starts with defining, for any rational q, the set $A_q$: $A_q$ = { x : x + q is a rational in (0, 1] } = (-q, 1-q] $\cap$ $\mathbb{Q}$ As an example, if q = 0, then $A_0 = (0, 1] \cap \mathbb{Q}$. … I asked the question: Is there a p such that p is in all $A_q$ such that $q \in [0, 1) \cap \mathbb{Q}$? [Answer] …For each computable p that we try, we’ll note that p > 1-q for an infinite number of $q \in [0, 1) \cap \mathbb{Q}$. Since p cannot be an element of $A_q$ if p > 1-q, we conclude that p is not an element of all $A_q$ such that $q \in [0, 1) \cap \mathbb{Q}$. This allows for another true statement: There is no computable p such that p is in all $A_q$ such that $q \in [0, 1) \cap \mathbb{Q}$.

January 11th, 2017, 09:05 PM   #125
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Quote:
 Originally Posted by AplanisTophet I have one last question ...
Then you must be Lieutenant Columbo!

I will get to the details of your post tomorrow.

January 11th, 2017, 09:06 PM   #126
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Quote:
 Originally Posted by AplanisTophet For each rational y where 0 $\leq$ y ...
Again ???

Think you have a Set "A" of Apples, some Red and some Yellows, ...and now you wanna make a Set "B" including one Tomato {0}. You can do that, but no way to put in bijection the new Set "B" with "A", you agree ?

January 11th, 2017, 10:14 PM   #127
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 Originally Posted by complicatemodulus Again ??? Think you have a Set "A" of Apples, some Red and some Yellows, ...and now you wanna make a Set "B" including one Tomato {0}. You can do that, but no way to put in bijection the new Set "B" with "A", you agree ?
If $A$ and $B$ are infinite sets in bijection and we add a tomato to A then we can still find a bijection between them. This is the Hilbert Hotel idea. For example if $A = B = \mathbb N$ then of course they care in bijection. If I add a tomato to $A$ then it's still in bijection with $B$, simply by lining up $A$ as $\text{tomato}, 1, 2, 3, 4, 5, \dots$ and mapping those elements to $1, 2, 3, 4, \dots$ respectively.

This is also true in general, even if $A$ and $B$ are uncountable sets in bijection, but the proof I gave would have to be modified to cover that case.

 January 11th, 2017, 11:06 PM #128 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 In my opinion: be honest... going infimus ($\lim_{K\to\infty}1/K$ ) talk of function (or analitic extension of your Set problem), instead of Set: it differs from talking of infinite Set $\mathbb{Q}$, $\mathbb{N}$ or else (open on the right), since infimus add continuity property to the ?elements? so cannot be part of Set Theory because you cannot define the element $n$, than $n+1$, or in that case you can sometimes discard some $1/K^m$ with $m\in\mathbb{N} : m>1$ elements of the ?Set? in same operations... Set theory born when continuity die, of course your problem can (sometimes) lay on a continous function used to define the relation between Domain and Co-Domain. You agree ? Note: ?xx? means impossible element or impossible set Last edited by complicatemodulus; January 11th, 2017 at 11:09 PM.
January 12th, 2017, 02:17 PM   #129
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I'm going to respond to the argument you presented in #111, since it is the simplest and also the one you claimed was your final argument. It's perfectly clear and there's no reason to add anything else.

Quote:
 Originally Posted by AplanisTophet For any rational $q$, let $A_q = \{ x : (q+x) \in (0, 1] \cap \mathbb{Q} \}$. Let $A^* = \{ A_q : q \in [0, 1) \cap \mathbb{Q} \}$. Statement to prove: There exists at least one rational p such that p is an element of every set $A_q$ in $A^*$. Why attempt to prove this? It’s because then the set p + $[0, 1) \cap \mathbb{Q}$ would be a subset of $(0, 1] \cap \mathbb{Q}$.

Ok. Rewriting this, and using the convention that all intervals are implicitly intersected with the rationals (just a notational convenience) you are saying that

Let $A^* = \{ (-q, ~1 - q] : q \in [0, ~1)\}$.

Agreed?

Here is a picture (unfortunately not very precise) of a few of the $A_q$'s.

You ask if there can be some $p \in \bigcap_{q \in [0, ~1)} A_q$. Agreed?

Now it is obvious -- and what's more, you have already PROVEN YOURSELF,that $0 \notin A^*$ so that $p$ can't be $0$.

On the other hand if $p \in (0, ~1)$, then by choosing $1 - q < p$ we have that $p \notin [0, ~1 - q]$ so that $p \notin A_q$.

Therefore in fact $\bigcap_{q \in [0, ~1)} A_q = \emptyset$.

Agreed?

Ok. That's the math. That is literally the end of the story.

However, you have some intuitions. Which I agree are not actually bad intuitions. They're fairly normal. The thing is, they happen to be wrong.

You think that as $q$ gets very close to $1$, you'll "trap a single rational" in the interval $(0, ~1 - q)$. Of course you can see very clearly -- because you have actually worked it out for yourself -- that this is not so.

But you want to think there is some magic thingie, like a goat perhaps, that's infinitesimally stuck to $-1$ and just to the right of it in such a way that it's in every $A_q$.

Now this is a BELIEF, and an INTUITION, but you haven't got any math to show that it exists.

That's because mathematically, no such goat exists. Actually I like the word barnacle because it acts like a barnacle stuck to the hull of a ship. It's inseparable from the hull but not part of it. Any larger circle around the ship would include both the hull and the barnacle.

That's what you're THINKING.

But thinking doesn't make it so. If you had a PROOF that such a thing existed, you'd be right that there's in contradiction in math.

But you don't. You have two things:

1) A PROOF that no such barnacle exists, attached to and inseparable from and just to the right of $-1$; and

2) An INTUITION that it does.

When the proof contradicts the intuition, we must believe the proof. And over time, our intuitions get better!

So the sooner you update your intuition, the better.

What say you? And whatever it is, strive to make it clear.

I did not yet look at your "one last question" in #124 yet. I only want to make sure that we are in complete agreement about the math so far.

ps -- Or are you trying to trap the barnacle just to the right of $0$? I may be a little unclear on your argument. You'd still be wrong, but the details of the refutation would differ.

Last edited by Maschke; January 12th, 2017 at 02:37 PM.

January 12th, 2017, 06:14 PM   #130
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Quote:
 Originally Posted by AplanisTophet B = $\{0\} \cup \{ y \in \mathbb{Q} : y > 0$ and $\exists p$ such that $p$ is an element of every set $A_q$ such that $q \in [0, y) \cap \mathbb{Q} \}$ For any rational y > 0, it is possible to test whether or not there exists a p such that p is an element of every set $A_q$ such that $q \in [0, y) \cap \mathbb{Q}$. For example, if y = 0.5, we know that 0.25 is a suitable p because 0.25 is in every set $A_q$ such that $q \in [0, 0.5)$, so 0.5 is in B. For each rational y where 0 $\leq$ y < 1, we can compute a suitable p right? I believe that means B = $[0, 1) \cap \mathbb{Q}$, does it not... or wait, but it can't? If so, then by definition, isn't this a contradiction? What is B? EXPLAIN PLEASE!!!

I haven't understood this in detail yet, but I get the feeling you are showing that for every interval there's some point. But there's still no point that's in every interval.

Just like every cat has paws, but there is no single paw that belongs to every cat.

Is that about right? Or are you doing something different than that?

In other words you're reversing quantifiers. $\forall x \exists y P(x,y)$ is not the same as $\exists y \forall x P(x,y)$.

Last edited by Maschke; January 12th, 2017 at 06:45 PM.

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