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Does ZFC permit or acknowledge the existence of infinite sets that are uncountable?

By saying infinite sets that are uncountable, I mean that the cardinality of the sets is uncountable (that is $\ge\aleph_1$)


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ummm... Yes? Uncountable sets exist in ZFC. – Dustan Levenstein Feb 11 '12 at 19:10
@user24796: Easily. The axiom of infinity tells us there is an infinite set $X$, the powerset axiom tells us it has a powerset $\mathscr{P} X$, and Cantor's diagonal argument tells us that $\mathscr{P} X$ has cardinality strictly greater than $X$. – Zhen Lin Feb 11 '12 at 19:13
"Uncountable" means bigger than $\aleph_0$, not bigger than $\aleph_1$. The latter number is by definition the cardinality of the set of all countable ordinals, which is an uncountable set. – Michael Hardy Feb 11 '12 at 20:47
@Michael Hardy: The question was bigger than or equal to $\aleph_1$. This is the definition of uncountable in ZFC (where as in ZF it only means "not smaller or equal than $\aleph_0$"). – Asaf Karagila Feb 11 '12 at 23:11
up vote 6 down vote accepted

Cantor's theorem tells us that if $X$ is a countably infinite set then $P(X)$ is not countable.

In ZFC we assert that if $x$ is a set then $P(x)$ is a set as well, so if you assume the axiom of infinity which asserts that indeed an infinite set exists then there exists an uncountable set as well.

There is more to that as well: How do we know an $ \aleph_1 $ exists at all?

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but according to Wikipedia, it says that real numbers cannot be characterized in first-order logic alone. [] and I am little confused... – user24796 Feb 12 '12 at 14:36
The real numbers cannot be characterized in FOL in the language of rings (or fields), however within set theory you can define high-order structures for other languages. Furthermore, even if the real numbers cannot be characterized in first order logic, they are still a model of the theory. – Asaf Karagila Feb 12 '12 at 15:05
@user24796: To say that the real numbers cannot be characterized in FOL is to say that you cannot have a theory that every model which satisfies this theory is exactly the real numbers (namely complete and Archimedean field). – Asaf Karagila Feb 12 '12 at 15:08

The set of real numbers $\mathbb{R}$ spring to mind. Another example is the power set of the natural numbers (which exists by the power set axiom of ZFC), and that is uncountable by Cantor's theorem.

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$\mathbb{R}$ and $\mathbb{C}$ are constucted within ZF (so ZFC)

the answer is yes

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