Cardinality of real #s

May 2015
516
32
Arlington, VA
What is the cardinality of the real numbers, if they include all possible number magnitudes?
 
Aug 2012
2,496
781
What is the cardinality of the real numbers, if they include all possible number magnitudes?
$2^{\aleph_0}$. The notation literally means the set of all functions from the natural numbers to the 2-element set {0,1}. You can think of it as a binary sequence with a binary point in front, representing a real number between 0 and 1.

Now the funny thing is that if you ask, well what number is $2^{\aleph_0}$, mathematicians have no idea. It might be $\aleph_1$ or $\aleph_2$ or $\aleph_{47}$ or some transfinite cardinal much larger than that. Nobody knows. More accurately, the answer is independent of the standard axioms of set theory, and nobody's found a plausible, universally acceptable axiom that settles the matter one way or another.
 
Last edited:
  • Like
Reactions: Loren
May 2015
516
32
Arlington, VA
Isn't Aleph-naught a minimum of some sort? Is there a simple derivation for it?

Some great new information.
 
Aug 2012
2,496
781
Isn't Aleph-naught a minimum of some sort? Is there a simple derivation for it?
$\aleph_0$ is the cardinality of the set of natural numbers $\{0, 1, 2, 3, \dots \}$. It's provably the smallest possible infinite cardinality.
 
  • Like
Reactions: idontknow

mathman

Forum Staff
May 2007
6,932
774
$\aleph_0$ is defined as the cardinality of the set of integers.
 
May 2015
516
32
Arlington, VA
Is there some simple significance of the base 2 with power Aleph-naught, or could a base approaching 1 from above yield a lesser minimum cardinality?
 
Aug 2012
2,496
781
Is there some simple significance of the base 2 with power Aleph-naught, or could a base approaching 1 from above yield a lesser minimum cardinality?
Turns out that the cardinality of the set of ternary (base 3) sequences has the same cardinality as those of base 2. In fact even if you had a countable infinity of symbols, the set of functions from the naturals to that set would still have the cardinality of the reals.
 
Dec 2015
1,084
169
Earth
\(\displaystyle card \: (\mathbb{R}) > card \: (\mathbb{N}) =\infty \).

also an easy fast method is to consider the set of reals as \(\displaystyle \mathbb{R}:\{-\infty , ...,-3 ,-2 ,-1, 1,...,\infty \}\) and \(\displaystyle \mathbb{N}: \{ 1 , 2 , 3 ,...,\infty \}\). which can be seen that reals have larger cardinality .

or write reals as \(\displaystyle \dfrac{n}{10^m } \) and compare the number of values over \(\displaystyle n\).
 
Oct 2009
942
367
OK, this post is problematic in every part:

\(\displaystyle card \: (\mathbb{R}) > card \: (\mathbb{N}) =\infty \).
Cardinality of $\mathbb{N}$ is not $\infty$. Rather it is $\aleph_0$.

also an easy fast method is to consider the set of reals as \(\displaystyle \mathbb{R}:\{-\infty , ...,-3 ,-2 ,-1, 1,...,\infty \}\) and \(\displaystyle \mathbb{N}: \{ 1 , 2 , 3 ,...,\infty \}\). which can be seen that reals have larger cardinality .
You shouldn't write $\mathbb{R}$ this way in any case. For example, it makes it seem like 1/2 is not part of the set. You cannot write $\mathbb{R}$ using this notation.
Furthermore, you seem to imply that $\mathbb{N}$ is a subset of $\mathbb{R}$ and hence $\mathbb{R}$ has a larger cardinality. This is NOT a valid argument. $\mathbb{Q}$ and $\mathbb{N}$ have the SAME cardinality despite the latter being a subset of $\mathbb{Q}$.

or write reals as \(\displaystyle \dfrac{n}{10^m } \) and compare the number of values over \(\displaystyle n\).
Not all reals (in fact, almost no reals) can be written in the form $\frac{n}{10^m}$.