Cardinality of the Power Series of an Infinite Set

[jaketodd]

According to this page: https://en.wikipedia.org/wiki/Cantor's_theorem

It says: "Cantor's theorem is a fundamental result that states that, for any set A, the set of all subsets of A (the power set of A) has a strictly greater cardinality than A itself."

Furthermore, it says: "Cantor's argument applies for any set, including countable and uncountable infinite sets."

So my question is: Is the cardinality of the power set of an infinite set equal to: 2^∞ ?

Thanks,

Jake

 

[andrewkirk]

So my question is: Is the cardinality of the power set of an infinite set equal to: 2∞ ?

It would not be written like that, because the symbol $\infty$ does not denote a cardinal.

You could either give the simple version, which is that the cardinality of the power set of an infinite set is infinite, and strictly greater than that of the original set.

or you could give the more precise version, which is that the cardinality of a set with cardinality $c$ is $2^c$, including if $c$ is infinite.

So for instance the cardinality of the powerset of the integers, which have cardinality $\aleph_0$, is $2^{\aleph_0}$.

 

[jaketodd]

or you could give the more precise version, which is that the cardinality of a set with cardinality $c$ is $2^c$, including if $c$ is infinite.

Well, can't you just plug in $\infty$ for c?

So for instance the cardinality of the powerset of the integers, which have cardinality $\aleph_0$, is $2^{\aleph_0}$.

Isn't the cardinality of the power set of the integers = $\infty$ ? After all, it's an infinite set, right? And if so, then doesn't $\aleph_0$ = $\infty$ ?

Maybe I wrote the expression wrong in my original post. I was trying to ask if the power series of an infinite set has cardinality = 2^$\infty$

Please clarify..

Thank you!

Jake

 

[WWGD]

Well, can't you just plug in $\infty$ for c?
Isn't the cardinality of the power set of the integers = $\infty$ ? After all, it's an infinite set, right? And if so, then doesn't $\aleph_0$ = $\infty$ ?

Maybe I wrote the expression wrong in my original post. I was trying to ask if the power series of an infinite set has cardinality = 2^$\infty$

Please clarify..

Thank you!

Jake

$\infty$ is not a cardinal number, and the result applies to cardinal numbers ( and cardinality)

 

[andrewkirk]

Well, can't you just plug in ∞ for c?

I did a quick reflection on how the $\infty$ symbol is used in mathematics.

My tentative conclusion is that it's mostly only used in expressions like $\int_a^\infty f(x)dx$, $\sum_{k=1}^\infty s_k$, $\bigcup_{k=1}^\infty A_k$ and $\lim_{x\to\infty} f(x)$. All of those expressions are limits of one kind or another, and the $\infty$ symbol is part of telling us what kind of limit is meant.

The symbol is also sometimes used as a member of a set called the Extended Real Number Line.

I could not think of any other uses.

None of those uses has anything to do with the study of cardinality, which is arguably a branch of Number Theory. The symbol $\infty$ is avoided in studying cardinality, because we want to make distinctions between different infinite cardinalities, and $\infty$ does not make those distinctions.

 

[jaketodd]

Well how can the cardinality of an infinite set not be infinite??

 

[WWGD]

Well how can the cardinality of an infinite set not be infinite??

The symbol $\infty$ only states that the cardinality is not finite, but does not specify whether it is countable or uncountable ( in its many different ways). It is sort of like saying that " a lot" is a number; it is not precise/specific-enough to be a number.

 

[jaketodd]

The symbol $\infty$ only states that the cardinality is not finite, but does not specify whether it is countable or uncountable ( in its many different ways). It is sort of like saying that " a lot" is a number; it is not precise/specific-enough to be a number.

This stuff is just too strange. Is there any form of infinity (i.e. countable, uncountable, etc.?) that would be appropriate for the cardinality of an infinite set? If you're going to do away with infinity in set theory, then how can you even say a set is infinite? That uses the same concept you claim to be unsatisfactory. No offense my friend.

Cheers!

Jake

 

[WWGD]

This stuff is just too strange. Is there any form of infinity (i.e. countable, uncountable, etc.?) that would be appropriate for the cardinality of an infinite set? If you're going to do away with infinity in set theory, then how can you even say a set is infinite? That uses the same concept you claim to be unsatisfactory. No offense my friend.

Cheers!

Jake

You should specify the actual cardinality of the set: countably- , uncountably- , etc. infinite , having cardinality equal to that of a set whose cardinality is known, etc. "Infinite cardinality" does not narrow things enough; it is just saying , of a number, something along the lines that " it is a pretty big number". It is not specific-enough.

 

[jaketodd]

You should specify the actual cardinality of the set: countably- , uncountably- , etc. infinite , having cardinality equal to that of a set whose cardinality is known, etc. "Infinite cardinality" does not narrow things enough; it is just saying , of a number, something along the lines that " it is a pretty big number". It is not specific-enough.

Ok, so you are saying there are forms of infinity, which can be used as the cardinality of a set? Would you be so kind as to talk about those for a bit? Thanks

 

[WWGD]

Ok, so you are saying there are forms of infinity, which can be used as the cardinality of a set? Would you be so kind as to talk about those for a bit? Thanks

See, e.g. https://duckduckgo.com/?q=cardinality+of+an+infinite+set&t=h_&ia=web and let us know if you have any specific question; the topic is too broad otherwise.

 

[WWGD]

According to this page: https://en.wikipedia.org/wiki/Cantor's_theorem

It says: "Cantor's theorem is a fundamental result that states that, for any set A, the set of all subsets of A (the power set of A) has a strictly greater cardinality than A itself."

Furthermore, it says: "Cantor's argument applies for any set, including countable and uncountable infinite sets."

So my question is: Is the cardinality of the power set of an infinite set equal to: 2^∞ ?

Thanks,

Jake

If a set has cardinality , say $\aleph_0$ , then its powerset has cardinality $2^{\aleph_0}$ . Similar for any set, infinite or otherwise. But the properties of infinite cardinals are different from those of finite cardinals. Then you also have the continuum hypothesis, etc..

 

[jaketodd]

Is it possible to have a cardinality of 2^$\infty$ ?

Thanks,

Jake

 

[TeethWhitener]

Is it possible to have a cardinality of 2^$\infty$ ?

Thanks,

Jake

No, as WWGD said, this is like having a cardinality of $2^{several}$. The symbol $\infty$ is not well-defined enough to make the expression meaningful.

 

[jaketodd]

No, as WWGD said, this is like having a cardinality of $2^{several}$. The symbol $\infty$ is not well-defined enough to make the expression meaningful.

If $\infty$ isn't good enough for cardinality, then how can it be good enough for anything, such as an "infinite set," which is, from what I can gather, a very common thing in set theory?

 

[TeethWhitener]

If ∞∞\infty isn't good enough for cardinality, then how can it be good enough for anything, such as an "infinite set," which is, from what I can gather, a very common thing in set theory?

As andrewkirk pointed out, the symbol $\infty$ really isn't used in set theory. The symbol is used in analysis to denote a limiting procedure (for instance, $\lim_{x\rightarrow\infty} f(x)=L$ means (somewhat imprecisely) that for any number $a$, when $x>a$, we can get $f(x)$ arbitrarily close to $L$, regardless of how big $a$ gets) or as a "point at infinity" added to the real numbers to make them homeomorphic to the circle. Set theory uses specific cardinalities, for instance $\aleph_0$ to denote sets with countably infinite cardinality, or $c$ to denote sets with cardinality equal to that of the real numbers.

 

[jaketodd]

Does set theory replace calculus? If not, then how can it reject $\infty$?

 

[fresh_42]

No offense, but your "concrete example" doesn't involve infinite sets. And even if it did, then how would you ever map every element of an infinite set to another, without doing it an infinite number of times -- oh but set theory doesn't use that infinity. And the infinity that calculus uses doesn't require continuity; it works for discrete models as well.

You can define a bijective function $f\, : \, 2\cdot \mathbb{Z} \longrightarrow \mathbb{N}$ which numerates all even numbers. So both sets are equally big. One doesn't need to map all single elements an infinite number of times, as the rule
$$f(2k) = \begin{cases}
2k & \text{if } k \geq 0 \\
2|k|+1 & \text{if } k < 0
\end{cases}$$
tells what to do in every single case. It isn't actually done an infinite number of times.

I think it might help you to read the two Wikipedia articles about ordinal numbers and [URL="https://www.physicsforums.com/insights/informal-introduction-to-cardinal-numbers/"]cardinal numbers[/URL]. These should tell you about the two different concepts and what sorts of "infinity" there are. The symbol itself is always only used, if the context is obvious, as for limits or infinite sums. As a standalone symbol it's not of much value, as it neither can be used for arithmetic nor as an equivalent term for infinity, as long as the context isn't set. I would prefer to not use it at all, but in integrals, limits and sums it is a nice shortcut for what is really meant.

 

[TeethWhitener]

Please, keep it simple.

Given an arbitrary computer program with an arbitrary input, we can never tell if that program will loop endlessly or whether it will eventually halt. Never, that is, on a digital computer. If we could build a sufficiently refined analog computer, we could solve this problem (google hypercomputation). The ultimate reason for this is that the set of natural numbers and the set of real numbers have different cardinalities.

Is this knowledge useful? Probably it's not immediately applicable. Maybe it never will be. But we wouldn't even know this possibility existed were it not for the study of transfinite numbers.

Edit: one common theme here is insisting on an immediate application. This is unwise: G. H. Hardy famously announced that his academic endeavors would never lead to anything that could be usefully applied. The fields of study he mentioned specifically were number theory and relativity. This was in 1940. Project ULTRA, using number theory to crack Nazi codes, started in 1941, and the Manhattan project, using relativity to develop nuclear weapons, started in 1942.

 

[PeterDonis]

Does set theory replace calculus? If not, then how can it reject $\infty$?

You don't appear to get the basic point of the question you asked. You asked a question about infinite sets. The $\infty$ symbol in calculus does not denote an infinite set; it denotes other uses of the general concept of "infinity" that are irrelevant to the question you asked to start this thread. So the answer to the question you asked, as you asked it...

Is the cardinality of the power set of an infinite set equal to: $2^\infty$ ?

...is "no, because $\infty$ does not name any infinite set". That is what @andrewkirk meant when he said that $\infty$ is not a cardinal.

A number of post that are off topic, per the above, have been deleted. The OP question has been answered and the thread is closed.