I Are Some Real Numbers Countable and Others Uncountable?

[FactChecker]

Some jokes fall more than flat.

I think we should give some "artistic license" to the terminology used and recognize that @Svein has made an important point.

 

[sysprog]

Some jokes fall more than flat.

When you check your pockets on leaving the house $-$ let's see, keys, wallet, phone, TI-83 (graphing calculator), ok, good to go $-$ is a too-flat $-$ hmm $-$ more than flat $-$ maybe a B♭ (B flat) clarinet that's more than too flat maybe an A♮ (A natural) clarion?

 

[dextercioby]

I think @Svein is referring to the fact that the power set of $\mathbb{R}$ has a greater cardinality than $\mathbb{R}$.

Is the cardinality of the powerset of $\mathbb{R}$ (or $\mathbb{R}^n$, with $n$ finite) aleph_2?
If not, what is aleph_2?

 

[stevendaryl]

Is this correct?

The cardinality of an interval in R is proportional to its length, and as it is uncountable, the cardinality of some arbitrarily small nonzero interval in R is greater than all of N

No, that's not correct. The cardinality of an interval is not affected by its length. An interval of length 1 has the same cardinality as the full set of real numbers.

Two sets have the same cardinality if you can map one set to the other in a one-to-one fashion. The mapping:

$x \mapsto log(\frac{x}{1-x})$

maps any $x$ in the range $(0,1)$ to a real number in the range $(-\infty, +\infty)$, showing that those two ranges have the same cardinality.

 

[BWV]

Gazing up into the darkness I saw myself as a creature driven and derided by vanity; and my eyes burned with anguish and anger.

No, that's not correct. The cardinality of an interval is not affected by its length. An interval of length 1 has the same cardinality as the full set of real numbers.

Two sets have the same cardinality if you can map one set to the other in a one-to-one fashion. The mapping:

$x \mapsto log(\frac{x}{1-x})$

maps any $x$ in the range $(0,1)$ to a real number in the range $(-\infty, +\infty)$, showing that those two ranges have the same cardinality.

So do (0,1) and (0,12) in R have the same cardinality because a function exists to map them (multiply/divide by 12) or do, by definition, all uncountable sets have the same cardinality?

by the above, do (0,1) and (0,12) in the rationals have the same cardinality?

 

[stevendaryl]

So do (0,1) and (0,12) in R have the same cardinality because a function exists to map them (multiply/divide by 12) or do, by definition, all uncountable sets have the same cardinality?

No, there are uncountable sets with cardinality larger than the cardinality of all real numbers.

The question of whether there are uncountable sets with cardinality smaller than the cardinality of all real numbers is an unsolved problem. And unsolvable. Given the usual tools of mathematics, there is no way to prove or disprove the claim that the set of all real numbers is the smallest uncountable cardinality.

But certainly all sets of reals that are going to come up in ordinary mathematics are in one of three categories:

1. Finite sets.
2. Countably infinite sets.
3. Uncountably infinite sets.

by the above, do (0,1) and (0,12) in the rationals have the same cardinality?

Yes, every infinite set of rationals has the same cardinality.

 

[Infrared]

Is the cardinality of the powerset of $\mathbb{R}$ (or $\mathbb{R}^n$, with $n$ finite) aleph_2?
If not, what is aleph_2?

Allowing for $n\geq 2$ doesn't change anything since $\mathbb{R}$ and $\mathbb{R}^n$ have the same cardinality.

This question is an example of the generalized continuum hypothesis, which is known to be independent from ZFC (https://en.wikipedia.org/wiki/Continuum_hypothesis#The_generalized_continuum_hypothesis).

 

[TeethWhitener]

Is the cardinality of the powerset of R (or Rn, with n finite) aleph_2?
If not, what is aleph_2?

This depends on whether your system accepts or rejects the continuum hypothesis. The cardinality of the natural numbers is $\aleph_0$ and the cardinality of the real numbers is $2^{\aleph_0}$, but the continuum hypothesis (the assertion that $2^{\aleph_0}=\aleph_1$) is independent of the axioms of ZFC.

 

[FactChecker]

do, by definition, all uncountable sets have the same cardinality?

This part is not right. There are higher cardinal numbers. There is a whole sequence of them of increasing size.

 

[sysprog]

Cantor proved that the power set of $\mathbb{R}$ has cardinality greater than that of $\mathbb{R}$. The continuum hypothesis is that nothing has cardinality in between those. Please if you find such a thing, or can prove that there is no such thing, let us know.

 

[TeethWhitener]

do, by definition, all uncountable sets have the same cardinality?

Just to pile on, Cantor's theorem is that, for any set $A$, given its powerset $\mathcal{P}(A)$--defined as the set of all subsets of $A$--the cardinality of the powerset will always be strictly greater than the cardinality of the set: $|\mathcal{P}(A)|>|A|$

 

[WWGD]

Your logic is like this.

1 - 1
2 - 2
3 - 3
...
10 - 10
11 - 11
...
123-123
...

Assume that this list contains all natural numbers. Now add 1 to the ones place of the first number.(1+1=2), add 1 to the tens place of the second number(0+1=1), add 1 to the hundreds place of the third number(0+1=1), and so on. (If you do it third times, you get 112.)

The number you get, if you concatenate all changed digits is a number that was not in your list, i.e. whichever list you choose, there is a number that is not listed. Hence the natural numbers are uncountable.

What's the image of $\pi$? Or any other Irrational?
Sorry @emptyboat , I thought I was replying to the OP.

 

[WWGD]

Cantor proved that the power set of $\mathbb{R}$ has cardinality greater than that of $\mathbb{R}$. The continuum hypothesis is that nothing has cardinality in between those. Please if you find such a thing, or can prove that there is no such thing, let us know.

Good point. It's true for every set, finite or infinite.

 

[Jarvis323]

So do (0,1) and (0,12) in R have the same cardinality because a function exists to map them (multiply/divide by 12) or do, by definition, all uncountable sets have the same cardinality?

by the above, do (0,1) and (0,12) in the rationals have the same cardinality?

There is also measure theory which is probably more along the lines of your original intuition.

https://en.wikipedia.org/wiki/Lebesgue_measure#:~:text=Any closed interval [a, b,b and has measure zero.&text=Moreover, every Borel set is Lebesgue-measurable.

 

[stevendaryl]

There is also measure theory which is probably more along the lines of your original intuition.

Except that standard measure theory gives 0 as the measure for any collection of rational numbers. So it doesn't support the intuition that there are twice as many rationals in (0,2) as in (0,1).

 

[mathwonk]

but it is consistent with it since twice zero is zero. ?

 

[WWGD]

So do (0,1) and (0,12) in R have the same cardinality because a function exists to map them (multiply/divide by 12) or do, by definition, all uncountable sets have the same cardinality?

by the above, do (0,1) and (0,12) in the rationals have the same cardinality?

Yes, two sets have the same cardinality iff there is a bijection between them.

 

[sysprog]

So do (0,1) and (0,12) in R have the same cardinality because a function exists to map them (multiply/divide by 12) or do, by definition, all uncountable sets have the same cardinality?

by the above, do (0,1) and (0,12) in the rationals have the same cardinality?

Yes, two sets have the same cardinality iff there is a bijection between them.

That is of course true; however, it is clearly not true that "all uncountable sets have the same cardinality", $-$ e.g. $|2^\mathbb{R}|>|\mathbb{R}|$ (i.e., the cardinality of the power set of $\mathbb{R}$ is strictly greater than that of $\mathbb{R}$), and both $2^\mathbb{R}$ and $\mathbb{R}$ are uncountable sets.

 

[WWGD]

That is of course true; however, it is clearly not true that "all uncountable sets have the same cardinality", $-$ e.g. $|2^\mathbb{R}|>|\mathbb{R}|$ (i.e., the cardinality of the power set of $\mathbb{R}$ is strictly greater than that of $\mathbb{R}$), and both $2^\mathbb{R}$ and $\mathbb{R}$ are uncountable sets.

I agree. The fact that for any set S $|P(S)|>|S|$ proves this. And also proves there is no set of all sets. Its powerset would be of cardinality larger than it.

 

[dextercioby]

Is $\mathbb{R}^{\infty}$ properly defined (as infinity is properly understood in classical analysis)? If so, does it have the same cardinality as $\mathbb{R}^{n}$ for arbitrary but finite $n$? What would be a proof of that?

 

[WWGD]

Is $\mathbb{R}^{\infty}$ properly defined (as infinity is properly understood in classical analysis)? If so, does it have the same cardinality as $\mathbb{R}^{n}$ for arbitrary but finite $n$? What would be a proof of that?

It depends if by $\infty$ you mean $| \mathbb N |$ or $| \mathbb R |$. First is ( or can be seen as; it's equivalent to ) the set of all sequences of Real numbers ( using $A^B$ as the set of all maps from A to B ), second is the set of all functions from the Reals to itself ). If you see R as a metric space, first is metrizable, second one is not. Cardinalities are different. Maybe someone can double-check my set theory/infinite arithmetic, which is pretty rusty. Assuming continuum hypothesis , $2^{\aleph_0}=\aleph_1$. Then $| |\mathbb R|^{|\mathbb N| } = 2^{ \aleph_0 \times \aleph_0}= 2^{\aleph_0^2}= 2^{ \aleph_0}$ while $|\mathbb R|^{| \mathbb R| }= 2^{ \aleph_0 \times \aleph_1}= 2^{ \aleph_1}$

 

[dextercioby]

Yes, sorry, not fully explicit. The $\infty$ in my post is the cardinality of $\mathbb N$. So you say that $\left| \mathbb{R}^{|\mathbb{N}|}\right| = \aleph_1 \equiv |\mathbb{R}|$.

 

[WWGD]

Yes, sorry, not fully explicit. The $\infty$ in my post is the cardinality of $\mathbb N$. So you say that $\left| \mathbb{R}^{|\mathbb{N}|}\right| = \aleph_1 \equiv |\mathbb{R}|$.

Yes, but please wait for someone to double-check my arithmetic. I haven't done any in a really long time.

 

[PAllen]

Yes, but please wait for someone to double-check my arithmetic. I haven't done any in a really long time.

Looks ok to me, but I am also rusty on this, so consider it a fractional vote of confidence.

 

[TeethWhitener]

It depends if by $\infty$ you mean $| \mathbb N |$ or $| \mathbb R |$. First is ( or can be seen as; it's equivalent to ) the set of all sequences of Real numbers ( using $A^B$ as the set of all maps from A to B ), second is the set of all functions from the Reals to itself ). If you see R as a metric space, first is metrizable, second one is not. Cardinalities are different. Maybe someone can double-check my set theory/infinite arithmetic, which is pretty rusty. Assuming continuum hypothesis , $2^{\aleph_0}=\aleph_1$. Then $| |\mathbb R|^{|\mathbb N| } = 2^{ \aleph_0 \times \aleph_0}= 2^{\aleph_0^2}= 2^{ \aleph_0}$ while $|\mathbb R|^{| \mathbb R| }= 2^{ \aleph_0 \times \aleph_1}= 2^{ \aleph_1}$

You don’t even need the continuum hypothesis, as $|\mathbb{R}|=2^{|\mathbb{N}|}$. The rest of the math looks fine to me.

 

[vortextor]

I think it is worthwhile to realize how easy it is to create irrational numbers. Just make sequences that never start repeating, like 0.101001000100001000000100000001000000001...
And for any such number, you can add it to a list of rational numbers to get a list of irrational numbers.
IMHO, that makes it easier to accept how the irrational numbers are such a huge set compared to the rationals (although it is not a proof by any means).
Cantor's proof is genius and is worth studying. Mathematicians of his time hated and resisted his conclusion that there are higher, uncountable, orders of infinity, but it was undeniable.

Premises:
The real numbers include all the rational numbers and all the irrational numbers.
An irrational number has a representation of infinite length that is not, from any point, an
indefinitely repeating sequence of finite length.
A rational number has a terminating sequence or has an indefinitely repeating sequence of finite length.
The set of rational numbers is countable, and the set of real numbers is uncountable.
My conclusion;
Since any irrational number can be transformed in an infinite set of rational numbers with
terminating sequence, the set of rational numbers must be more dense and uncountable ?
This is obviously invalid, however I can't see why?

 

[WWGD]

Yes, sorry, not fully explicit. The $\infty$ in my post is the cardinality of $\mathbb N$. So you say that $\left| \mathbb{R}^{|\mathbb{N}|}\right| = \aleph_1 \equiv |\mathbb{R}|$.

Correct. Sorry for delay in replying.

 

[jbriggs444]

Premises:
The real numbers include all the rational numbers and all the irrational numbers.
An irrational number has a representation of infinite length that is not, from any point, an
indefinitely repeating sequence of finite length.
A rational number has a terminating sequence or has an indefinitely repeating sequence of finite length.
The set of rational numbers is countable, and the set of real numbers is uncountable.
My conclusion;
Since any irrational number can be transformed in an infinite set of rational numbers with
terminating sequence, the set of rational numbers must be more dense and uncountable ?
This is obviously invalid, however I can't see why?

Can you justify the implicit assertion I have highlighted above? I do not even understand what it is saying.

Best guess: You can chop the decimal representation of an irrational number into an infinite number of pieces, each distinct from one another and each corresponding in some way to a unique rational number. So if there are infinitely many rational numbers for any single irrational number, how can there be more irrational numbers than rational numbers.

So far, so good. This is all correct. For one irrational number there is indeed a deterministic generating procedure we can use to generate an associated countably infinite set of rational numbers.

So what? Will the countably infinite set of rational numbers generated for the next irrational you pick re-use any of the rational numbers you found for the first irrational. If so, the union of all generated rational numbers for all irrational numbers can remain a countable set despite being the union of an uncountable number of countable sets.

 

[TeethWhitener]

Since any irrational number can be transformed in an infinite set of rational numbers with
terminating sequence,

But you can generate an infinite set of irrational numbers from an irrational number as well:
$$\pi=3.14159\ldots$$
$$\pi-3=0.14159\ldots$$
$$\pi-3.1=0.04159\ldots$$
etc., each of which is a new irrational. So your argument just tells you that rationals are dense in the reals, but nothing about how numerous the rationals vs. reals are.

 

[TeethWhitener]

Can you justify the implicit assertion I have highlighted above? I do not even understand what it is saying.

I think he’s referring to the sequence $\{3,3.1,3.14,3.141,\ldots\}$ of rationals converging on the irrational $\pi$. I could be wrong though.