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December 16th, 2016, 06:25 PM   #21
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Quote:
 Originally Posted by AplanisTophet I didn't. It was clear from the OP that r must be irrational in the diagram, but to make that clear see the one I've attached to this post. Now, look at the attached diagram. There is one and only one rational in (0, 1) for each rational in (r, 1+r) except 1. Therefore, we need to go from [0, 1) onto (r, 1+r) if we want a bijection via "shifting." Satisfy yourself of this, then let's continue. I will not concede this point because I am absolutely correct on this.
I'm gratified that we've found one specific thing to disagree on. I regard that as progress. I will take a look this evening.

Your exposition here is very clear and clean and concise. I understood what you wrote. I thank you for that.

Could you at least agree with me on two general things:

* That the rationals are dense in the reals.

* And that you can't map a half-closed-half-open interval of reals to an open interval both bijectively and continuously. (With the same function of course. You can certainly map them bijectively; and you can certainly map them continuously. Just not both).

Can you agree with these two points? Just yes or no, I just want to know your opinion, I don't need to debate them. Just want your opinion. Yes or no on each.

Last edited by Maschke; December 16th, 2016 at 06:33 PM.

December 16th, 2016, 06:35 PM   #22
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Quote:
 Originally Posted by Maschke Could you at least agree with me that the rationals are dense in the reals?
Absolutely.

Quote:
 Originally Posted by Maschke And that you can't map a half-closed-half-open interval of reals to an open interval both bijectively and continuously.
Yup, but I'm not sure why you're asking. In other words, we aren't going from $\mathbb{R}(0, 1)$ to $\mathbb{R}(r, 1+r)$. We are, in the case where $r$ is irrational, instead going from $\mathbb{R}[0, 1) \cap \mathbb{Q}$ to $\mathbb{R}(r, 1+r) \cap \mathbb{Q}$ or, in the case where $r$ is rational, we could go from $\mathbb{R}(0, 1) \cap \mathbb{Q}$ to $\mathbb{R}(r, 1+r) \cap \mathbb{Q}$.

December 16th, 2016, 06:44 PM   #23
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Quote:
 Originally Posted by AplanisTophet Absolutely. Yup, but I'm not sure why you're asking.
If you're not sure why I'm asking, then now I'm not sure why I'm asking!! My theory was that you would understand why I'm asking

Humor: Three logicians walk into a roadside diner. Waitress comes up to their table and asks, "Do y'all want coffee?" First logician says ... "I don't know." Second logician says ... "I don't know." Third logician says ... "Yes!!"

Quote:
 Originally Posted by AplanisTophet In other words, we aren't going from $\mathbb{R}(0, 1)$
Where did you find that notation? You write $\mathbb{R}(0,1)$ and I assume you mean simply $(0,1)$, the open unit interval. The set $\{x \in \mathbb R : 0 < x < 1\}$. You don't need the extra $\mathbb{R}$ in front.

Also I wanted to save you some keystrokes, this seems obvious but it was a great revelation at the time to me too. You don't have to write \mathbb {R}, you can write \mathbb R and it works just the same.

Quote:
 Originally Posted by AplanisTophet to $\mathbb{R}(r, 1+r)$. We are, in the case where $r$ is irrational, instead going from $\mathbb{R}[0, 1) \cap \mathbb{Q}$ to $\mathbb{R}(r, 1+r) \cap \mathbb{Q}$ or, in the case where $r$ is rational, we could go from $\mathbb{R}(0, 1) \cap \mathbb{Q}$ to $\mathbb{R}(r, 1+r) \cap \mathbb{Q}$.
Jeez man this stuff makes my eyes glaze over. Let me go cogitate on the simpler exposition in your earlier post. But still ... if you needed to qualify your answer to that question, that DOES confirm my suspicion that you are not grokking the topology.

December 16th, 2016, 07:02 PM   #24
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Quote:
 Originally Posted by Maschke Jeez man this stuff makes my eyes glaze over. Let me go cogitate on the simpler exposition in your earlier post. But still ... if you needed to qualify your answer to that question, that DOES confirm my suspicion that you are not grokking the topology.
I'm not qualifying anything. Before I noticed what is in the diagram, there is no way I would have suggested that we try going from the rationals in [0, 1) to the rationals in (r, r+1) as, without the explanation, that is "insane" as you suggest. You don't need to point that out, but the fact that you do so tells me that you should simply go back to the diagram (my most recent version clarifying that r is irrational), and then try rereading my most recent proof.

I appreciate the joke and I may not be correct (given what the proof is I know you'll be inclined to simply assert it isn't correct, like any rational mathematician would so I don't blame you, without actually getting too into it), but I do hope you take this seriously as yes, it is mind numbing. The thing is that I've been up and down this stuff 100 times before finally arriving at what I did, and since then I've been up and down this 100 times more. That doesn't mean I'm correct, but I've been taking this very seriously, and so while I know I'm asking a lot, please do also. You have my upmost appreciation.

December 17th, 2016, 10:48 AM   #25
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I would like to submit exhibit B, "Irrational Shift Diagram" (see attached .pdf file) which is very similar to my previous "Rational Shift Diagram." For this diagram, note specifically that adding k = q - r to each element of (r, r+1) will be a bijection between (r, r+1) and (q, q+1), but it will not be a bijection between the irrationals of (r, r+1) and the irrationals of (q, q+1) because for some irrationals, adding k will result in a rational number (just as r + k is rational).

In light of the above mentioned diagram:

In my latest proof, $C$ in line 6 should be $C = \mathbb{Q} \cap (0, 1]$ where it was previously defined as $C = \mathbb{Q} \cap (0, 1)$.

A conforming amendment is that line 4 of my latest proof should read "...that $k_r \in \mathbb{Q} \cap (0, 1] \land k_r + A = \{ k_r + a : a \in A \} = B$."

Taken together, these minor updates correct my latest proof to be what is intended. It should now read:

....

See the "Rational Shift Diagram" before reading this proof (I suggest the one clarifying that r must be irrational in the diagram as was evident from the OP).

....

Continuum Hypothesis Proof:

1) Let $A = \mathbb{Q} \cap [0, 1)$.

2) Let $r$ be any irrational number such that $r \in I = \mathbb{R}(0, 1) \setminus \mathbb{Q}$.

3) Let $B = \mathbb{Q} \cap (r, r+1)$.

4) Let $k_r = y - x$, where $x \in A \land y \in B \land \{ x - q : q \in A \} = \{ y - q : q \in B \}$. Note that $k_r$ is a rational number such that $k_r \in \mathbb{Q} \cap (0, 1] \land k_r + A = \{ k_r + a : a \in A \} = B$.

5) $r, s \in I \land r \neq s \Rightarrow k_r \neq k_s$. $r, s \in I \land k_r \neq k_s \Rightarrow r \neq s$. Therefore, $r, s \in I \land r \neq s \iff k_r \neq k_s$.

6) Let $q_1, q_2, q_3,$ … be an enumeration of $C = \mathbb{Q} \cap (0, 1]$.

7) Let $f:C \rightarrow I$ be a bijection: $f(c) = r$ if $c = k_r$. The set $I$ is therefore enumerable.

The continuum hypothesis states that “there is no set whose cardinality is strictly between that of the integers and the real numbers.” This statement is true where:

$|I| = |\mathbb{R}| = |\mathbb{N}| = |\mathbb{Z}|$
Attached Files
 Irrational Shift Diagram.pdf (19.5 KB, 6 views)

 December 17th, 2016, 11:07 AM #26 Senior Member     Joined: Sep 2015 From: USA Posts: 1,763 Thanks: 905 Just stop it already. Maschke has clearly disproven your central assertion yet you repeatedly ignore him. If you really believe the cardinality of the reals is equal to the cardinality of the integers then you have so little understanding of numbers as to suggest finding another area to occupy your time. Meanwhile, stop posting the same disproven drivel over and over.
December 17th, 2016, 11:21 AM   #27
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Quote:
 Originally Posted by romsek Just stop it already. Maschke has clearly disproven your central assertion yet you repeatedly ignore him. Meanwhile, stop posting the same disproven drivel over and over.
The last thing Maschke wrote regarding where we left off was:

Quote:
 Originally Posted by Maschke I'm gratified that we've found one specific thing to disagree on. I regard that as progress. I will take a look this evening. Your exposition here is very clear and clean and concise. I understood what you wrote. I thank you for that.
So no, he hasn't gotten back to me yet.

Why don't you read both "Rational Shift Diagram" (again I suggest the copy clarifying r is irrational in the diagram) and "Irrational Shift Diagram." Note that what Maschke wrote (prior to him understanding what I was doing) was something that I acknowledged myself in the OP (that there are an infinite number of rationals between any rational and an irrational), so seeing as how we both acknowledged that and we are now on the same page with me waiting for him to get back to me, please don't call what I've done "disproven drivel" unless you have a disproof yourself (in which case I'm all ears because I have not yet asserted myself that my proof is accurate!).

December 17th, 2016, 01:13 PM   #28
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Quote:
 Originally Posted by AplanisTophet So no, he hasn't gotten back to me yet.
Busy, haven't forgotten you. For the record your diagram proves nothing (nor frankly does it say anything comprehensible at all); and it's mathematically impossible to map a half-closed interval of reals to an open one both bijectively and continuously. I'll supply more detail when I get a chance, but in the meantime you'd do well to study the construction I gave for finding two rationals in your shifted interval whose difference is $k$. That construction falsifies your argument.

December 17th, 2016, 04:17 PM   #29
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Quote:
 Originally Posted by Maschke For the record your diagram proves nothing (nor frankly does it say anything comprehensible at all
Explain that. Be specific as I do here: Given a standard < ordering, the Rational Shift Diagram proves that the rationals of (r, r+1) contain one and only one rational that is not able to be covered by the rationals of (0, 1) if the two sets were to otherwise be aligned.

Quote:
 Originally Posted by Maschke and it's mathematically impossible to map a half-closed interval of reals to an open one both bijectively and continuously.
True. However, I never do that (I go from rationals to rationals in the "Rational Shift Diagram" or from irrationals to irrationals in the "Irrational Shift Diagram," and neither of those sets are continuous), so I think you'd be wasting your time trying to show me something I already agree with and that doesn't apply.

Quote:
 Originally Posted by Maschke I'll supply more detail when I get a chance, but in the meantime you'd do well to study the construction I gave for finding two rationals in your shifted interval whose difference is $k$. That construction falsifies your argument.
Why, when in the OP I already did better by simply saying:

Quote:
 Originally Posted by AplanisTophet 10. ...Clearly, something is wrong now because m is rational while r is irrational, and there are an infinite number of rationals between any rational and an irrational.
I suggest you prove that there cannot exist a $k_r = y - x$ from step 4 of my proof, because if there is a faulty statement, it will be that one (which should be obvious). Where the two sets $A$ and $B$ align perfectly given the standard < ordering, I don't think you will be able to show that statement is faulty. Then again, as I keep saying, maybe I'm wrong... You'll have to use pure logic to show it though, not simply some axioms describing the real numbers, for what should be obvious reasons (the same reasons behind your joke above where two said I don't know and one (me) said yes).

December 17th, 2016, 08:48 PM   #30
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Quote:
 Originally Posted by AplanisTophet Therefore, we need to go from [0, 1) onto (r, 1+r) if we want a bijection via "shifting."
I'm going to continue to focus on your post #20 since it's short and narrowly defined so it's a good place for us to attempt common understanding.

Can you explain this remark? It's clear that no shift operator can map $0$ anywhere, since a shift by $r$ maps $0$ to $r$ yet $r \notin (r, \ 1+r)$. If you map $0$ somewhere else then it's not a shift (obvious) and can't even be continuous if the map is a bijection (needs proof).

You put "shifting" into quotes which I assume means something to you but not to me. Can you explain this quoted remark please? What do you mean that you can map $[0,1)$ to $(r, \ 1+r)$ via a "shifting" (in your quotes) operation?

Last edited by Maschke; December 17th, 2016 at 08:55 PM.

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