Maschke

You posted at the same time I did. Please read what I wrote.

v8archie

Except that $y$ and $z$ are supposed to be rational.

Maschke

Oh I see, thanks. I have to patch my argument. It's late so I'll fix this in the morning.

We are given $w$ and $k$. We have $0 < w < 1$ and $r < r + w < 1 + r$. There's some rational $t$ with $r < t < r + w$. The points $t$ and $t + k$ will satisfy the conditions. They're both rational, they're both within the shifted-by-$r$ interval, and their difference is still $k$.

Note that (for positive $r$) I nudged the right-shifted values to the left to make sure they stay within the right-shifted interval. If $r$ were negative we'd be left-shifting the original unit interval; and then I'd right-nudge by finding a rational between $x$ and $1$. Make sense?

The key here is that the rationals are dense in the reals. Between any two reals there's a rational. I think the OP is hoping there will be a tiny little interval of reals that contains no rational, but such a thing doesn't exist.

I'll check this tomorrow if I made a mistake but it's perfectly obvious that you have plenty of room find suitable rationals in the second (shifted) interval.

AplanisTophet

As far as that proof goes, I've got a much nicer way of doing it now too but it'll have to wait until tomorrow. Dividing the sum of the $k_{r_i}$'s by 2 obviously wasn't necessary.

And yes Maschke, I believe we can actually select that natural number uniformly at random after all.

AplanisTophet

Please see the attached .pdf file entitled "Rational Shift Diagram" before reading the first proof.


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:

$|I| = |\mathbb{R}| = |\mathbb{N}| = |\mathbb{Z}|$



Proof of the Selection of a Natural Number Uniformly at Random:

If we define the term “select an element of a set uniformly at random” to simply mean that each element of the set has an equal chance of being selected using a method that selects one and only one element of the set, then we are able to select a natural number uniformly at random in the same way we are able to select an element of $\mathbb{R}[0, 1]$ uniformly at random. The method by which mathematicians generally agree that it is possible to select an element of $\mathbb{R}[0, 1]$ is the theoretical flipping of a coin an infinite number of times.* Such a number can be related to an element of $\mathbb{N}$ by putting $\mathbb{N}$ into a bijection with $\mathbb{R}[0, 1]$ given the cardinality implications of the above proof.

*Where each dyadic rational has two binary expansions (e.g. 0.1 = 0.011111…), a trivial correction must be made to ensure that the number is selected uniformly at random.



Thank you for reading the above proofs. I look forward to your feedback. If these are truly acceptable (and I’m not making any assumptions until others have reviewed them), then please separate this post into its own thread in this forum where it can permanently stay. Thank you also to Maschke who worked with me on developing the second proof regarding the selection of a natural number uniformly at random. The Continuum Hypothesis Proof is my original work.

Sincerely,

B. J. K.

Attached Files

Rational Shift Diagram.pdf (17.2 KB, 8 views)

Maschke

This is my correction to my posts #9 and #13. My late-night error was to shift the left-hand subinterval to the left. Rather, we need to shift the right-hand subinterval to the left till it matches up with a pair of rationals.

I see in your diagram you are overlapping the two intervals to further confuse yourself. Although my diagram shows the intervals as disjoint, my mathematical argument does not rely on that assumption and works just as well for overlapping intervals. In other words nothing in my argument relied on $r > 1$ and in fact $0 < r < 1$ works exactly the same way via the same argument.



* Given $w, x$ rationals in $(0,1)$. Those are shown in blue.

* Let $k = w - x$. So $k$ is the rational distance between $w$ and $x$. $k$ is not shown on the diagram.

* Shifting the unit interval to the right by the irrational $r$ moves $w$ and $x$ to $w + r$ and $x + r$, respectively. Clearly the distance $k$ is preserved. The two shifted points are the rightmost two points in blue.

* Now the challenge is to find $q$ and $q'$ in $(r, 1+r)$ with $q' - q = k$ and both $q, q'$ rational.

* Since $r < w + r$, there exists a rational $q$ strictly between them. It is a fact that the rationals are dense in the reals. You can always find a rational between any two distinct reals. You seem in my opinion to be forgetting this.

* Now let $t = w + r - q$. In other words $t$ is exactly the distance by which we have to nudge $w + r$ to the left to make it rational.

* Now let $q' = x + r - t$. In other words we shift $x + r$ to the left by $t$. I claim that $q'$ is the desired second point.

* Certainly $q' - q = k$.

* Certainly both $q$ and $q'$ are in the $r$-shifted (right hand) interval.

* We know $q$ is rational. It remains for me to show that $q'$ is rational as well. Now $q' = x + r - t$. And $t = w + r - q$. Then

$q' = x + r - t = x + r - (w + r - q) = x + r - w - r + q$

$= x - w+ q = k + q \in \mathbb Q$.

Done.

Now I have refuted your original contention of this thread. If your original contention is material to the rest of your argument, I've refuted your argument.

Therefore you have to respond before proceeding further. Your "rational shift" idea is flat out wrong.

If you mean that I helped you to understand that there is no such uniform probability distribution on the naturals, you're welcome. If you were enlightened by my pointing you to the Kolmogorov axioms, you're welcome. If you mean that I shed light on your understanding of the real numbers, you are definitely very welcome.

If on the other hand you mean that you didn't read what I wrote, didn't understand what I wrote, or just flat out ignored what I wrote, I'm afraid I can't accept your thanks. Your rational shift argument is wrong. Your claim that there's a bijection between an uncountable set and a countable one is wrong and cranky. Your claiming to have solved CH is wrong and cranky.

To bravely struggle to comprehend the nature of the real numbers is commendable. We are all on that path. The real numbers are extremely mysterious; and the more you study them the more you realize that. But to confuse your naive misunderstanding with revolutionary discoveries that overthrow 140 years of mathematical understanding is crankery. You have asked to be informed when you veer into crankery. Consider yourself so informed.

Maschke

More ...

I'm going through your most recent post to try to get a handle on what you are doing. A couple of things popped out at me.

People who do that might not be inclined to read the rest of your post at all. You just have this diagram without explaining anything, then out of nowhere you talk about the "extra rational" and it is very dispiriting because it's not possible to understand what you are talking about.

Also as I mentioned previously, your overlapping the intervals is adding an element of confusion. Do you actually ever use the hypothesis that $0 < r < 1$? I haven't read that far.


Ah!! LOL. You are now starting with a half-closed interval. Yes of course that will have a point left over when you do an additive shift bijection with an open interval. Duh. Is that what you're doing? In which case yeah, you're right, but to what end exactly?

Have you used this notation before? Did you just slip it into this post? Or have I been seeing the left square bracket as a left paren all this time and misunderstanding your argument?

Yes, if you add a limit point to an open interval you are topologically adding a point. There are still bijections with an open interval, of course, but they are not going to be additive shifts. You are correct about that. Is that your point here?

I will keep reading.

AplanisTophet

Maschke,

If $r \in \mathbb{R}(0, 1) \cap \mathbb{Q}$, then we know that $\exists k : k + C = \{ k + q : q \in \mathbb{Q} \cap \mathbb{R}(0, 1) \} = \mathbb{Q} \cap \mathbb{R}(r, 1+r)$. As you can tell from the diagram, when $r$ instead is irrational meaning $r \in \mathbb{R}(0, 1) \setminus \mathbb{Q}$, no such $k : k + C = \{ k + q : q \in \mathbb{Q} \cap \mathbb{R}(0, 1) \} = \mathbb{Q} \cap \mathbb{R}(r, 1+r)$ can exist (yes, we agree on this!) because there will be one and only one element in $\mathbb{Q} \cap \mathbb{R}(r, 1+r)$ not covered where each of the two and only two possible $k$ where $k : k + C = \{ k + q : q \in \mathbb{Q} \cap \mathbb{R}(0, 1) \}$ cover every element of $\mathbb{Q} \cap \mathbb{R}(r, 1+r)$ except for one. So yes, to make sure there can exist one and only one $k$ for any given $r$ when $r \in \mathbb{R}(0, 1) \setminus \mathbb{Q}$, instead of trying to go from $\mathbb{Q} \cap \mathbb{R}(0, 1)$ onto $\mathbb{Q} \cap \mathbb{R}(r, 1+r)$ by adding $k$ to each element of $\mathbb{Q} \cap \mathbb{R}(0, 1)$ which cannot be done, we must instead add $k$ to each element of $\mathbb{Q} \cap \mathbb{R}[0, 1)$ which can be done (as $k = y - x$, where $x \in \mathbb{Q} \cap \mathbb{R}[0, 1) \land y \in \mathbb{Q} \cap \mathbb{R}(r, 1+r) \land \{ x - q : q \in \mathbb{Q} \cap \mathbb{R}[0, 1) \} = \{ y - q : q \in \mathbb{Q} \cap \mathbb{R}(r, 1+r) \}$).


So yes, I did just slip it (using a half-closed interval when $r$ is irrational) into this post.

No, when $r$ is irrational it will be 1-to-1 and onto with no points left over BECAUSE I'm using a half-closed interval as you can tell from the diagram where if the interval was instead open there would be one and only one element of $\mathbb{Q} \cap \mathbb{R}(r, 1+r)$ that is not covered as you can also see from the diagram. When $r$ is rational (which isn't part of the proof because we already know those are enumerable), an open interval is 1-to-1 and onto while a half-closed interval would leave "a point left over" as you say on either side (pick which as that is trivial).


Yes, please do, as (regardless of whether I'm right or wrong and I'm not asserting I'm necessarily right yet either) based on your comment you aren't quite understanding the proof yet.

PS - I do see one problem with the diagram, which is that it should explicitly state that $r$ is irrational (this was known from the OP, but if my above proof was separated from this thread it would not be clear).

Maschke

You are insane. I don't mean that in a hostile way at all. I regard this as a very civil discussion. But one in which I am allowed to speak my mind.

I just refuted your basic argument. You can not throw out a long paragraph of symbology that you claim "proves" something that simply cannot be.

I'm not going to read that paragraph. Please write something that's more clear and concise.

You can't continuously map a half-closed interval onto an open interval for topological reasons. An additive shift is a continuous function. Continuous functions map limit points to limit points. There is no continuous bijection between a half-closed and an open interval of real numbers. So you are simply hallucinating when you say you can create such a function.

Do you understand that between every two reals there's a rational? I am suspecting you aren't taking this into consideration. For example in earlier posts you've spoken of some "minimum" separation between cosets. This mental image is at the very heart of your misunderstanding. It's turtles all the way down. Or in this case, it's rationals all the way down. The rationals are extremely sparse in measure but they make up for it by getting into everything, like sand at the beach. No matter how tiny you make an interval of reals, there are plenty of rationals in there.




You haven't got a proof, you've got a misunderstanding.

Instead of just flinging back another convoluted version of your flawed idea, could you please respond directly to the points I've made? That would take us a long way toward mutual understanding.

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.

Attached Files

Rational Shift Diagram.pdf (17.2 KB, 4 views)