December 16th, 2016, 05:25 PM  #21  
Senior Member Joined: Aug 2012 Posts: 1,565 Thanks: 378  Quote:
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 halfclosedhalfopen 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 05:33 PM.  
December 16th, 2016, 05:35 PM  #22  
Senior Member Joined: Jun 2014 From: USA Posts: 308 Thanks: 21  Quote:
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, 05:44 PM  #23  
Senior Member Joined: Aug 2012 Posts: 1,565 Thanks: 378  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!!" 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:
 
December 16th, 2016, 06:02 PM  #24  
Senior Member Joined: Jun 2014 From: USA Posts: 308 Thanks: 21  Quote:
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, 09:48 AM  #25 
Senior Member Joined: Jun 2014 From: USA Posts: 308 Thanks: 21 
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: $I = \mathbb{R} = \mathbb{N} = \mathbb{Z}$ 
December 17th, 2016, 10:07 AM  #26 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,490 Thanks: 749 
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, 10:21 AM  #27  
Senior Member Joined: Jun 2014 From: USA Posts: 308 Thanks: 21  Quote:
Quote:
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, 12:13 PM  #28 
Senior Member Joined: Aug 2012 Posts: 1,565 Thanks: 378  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 halfclosed 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, 03:17 PM  #29  
Senior Member Joined: Jun 2014 From: USA Posts: 308 Thanks: 21  Quote:
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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, 07:48 PM  #30  
Senior Member Joined: Aug 2012 Posts: 1,565 Thanks: 378  Quote:
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 07:55 PM.  

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