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In the ZFC set theory the "Axiom schema of specification" states that given an infinite set $z$, for any formula $\phi$ the subset $\{x\in z:\phi(x)\}$ always exists. Starting with a countable infinite set $z$ it is possible to construct only a countable infinite number of subsets this way, all of them being countable. My question is: How can we assure the existence of a set which is uncountable?

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    We should note that there are countable models of ZFC, and also pointwise definable models of ZFC (models in which every set can be definable without parameters). In such a model, there are no truly "uncountable" sets, but in that model, there are also no bijections from $\omega$ to $\mathcal{P}(\omega)$. – 2017-02-10
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    @Athar: This is interesting. What is your opinion about the Skolem's paradox which is somehow related? – 2017-02-10
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    It's not much of a paradox, you just need to think about the distinction between "countable" and "countable according to M" where M is some particular model of set theory. – 2017-02-10

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The power set axiom ensures that there is a set which cannot be put in bijection with a set of integers. Without the power set axiom it is consistent that every set is countable.

The thing to understand is that we can prove that there is a set which cannot be put into bijection with any subset of the natural numbers of that model. Even if we live in a model which is pointwise definable, in which every set can be defined (such model is countable, of course, but the model itself is "unaware of that").

This is Skolem's paradox in disguise, that first-order logic cannot properly handle the whole "infinite" thing. But it's not really a paradox, it's something we learn to live with and use to our advantage.

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    &Asaf:The Axiom of Power Set states that for any set _z_, there is a set _y_ that contains every subset of _z_. But how to assure that there are really an uncountable number of such subsets? I understand your position, and clearly we want all the conceivable collections of the elements of _z_ to be subsets. However the Axiom of Power Set works with existing sets as subsets. What makes a collection of elements to be a set, if it cannot be specified using any formula ϕ? (I don't know whether it is clear what my problem is.) – 2017-02-10
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    Because ZF proves Cantor's theorem, there is no bijection between a set and its power set. – 2017-02-10
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    I know Cantor's argument. His aim was to specify a subset which is distinct of any subset on a countable list, using the diagonal method. If this specification really works than it must have been done using a ϕ formula. But if all the countable infinite subsets which are specifiable with a ϕ formula are on the list, then Cantor's construction itself must be already on the list, This is a contradiction as Cantor's construction is different from all the subsets on the list. This yields that the diagonal construction cannot be done using a ϕ formula. – 2017-02-10
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    This goes back to the fact that specification is allowed to use parameters, which I pointed out in comments on your previous question. – 2017-02-10
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    A ϕ formula which is being used in a specification, can contain any number of free variables, however there exist still a countable number of ϕ's – 2017-02-10
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    I think that the only way you'll make peace with these issues is by actually studying logic and set theory for a while. Maybe a year or two. – 2017-02-10
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    I have already studied a few textbooks on set theory and logic. Surely I am far from being an expert of the area. My questions and doubts may seem naive, however up to now, I could not find anybody who could give relevant and soundly grounded answers on my problems. – 2017-02-10
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    *This* is not the place to dispel your confusion. I don't know what you know or not know. How familiar you are with the distinction of theory and meta-theory; how well-versed you are with the subtle issues related to Goedel's theorems; how much you understand the "usual caveats" when it comes to models of ZF, and how much you even know about basic model theory which is relevant for understanding here. – 2017-02-10
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    If you are well-versed in all these topics, I could probably write a quick explanation for you. But if you *are* versed in them, you should already know the answer yourself. I can only conclude from our interaction that there is a gap which I cannot, and will not, fill over this website. – 2017-02-10
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    I am awfully sorry for I have posed a question which made you upset. Still I feel frustrated, not getting any useful answer from an expert like you. – 2017-02-10
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    Thank you for drawing my attention to Skolem's paradox, which I was not aware of before. I have just begun to study it and all the theoretical facts related. Maybe we could have an interesting discussion about it later. – 2017-02-10
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    I am not upset, it's fine. I wrote about Skolem's paradox in a few answers on this website. You might want to make sure that you search for them first. – 2017-02-10
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    Surely, I will search for them. Thank you once again. :-) – 2017-02-10
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What this boils down to is Cantor's diagonalisation argument that shows that any infinite sequence $\langle A_n\colon n\in\mathbb{N}\rangle$ of subsets $A_n\subseteq \mathbb{N}$ must "miss" some subset $X$ of $\mathbb{N}$, namely the one specified by setting $X=\{x\in\mathbb{N} \colon\; x\not\in A_n\}$. The definition of $X$ implies that for each $n$, we have $n\in X\cup A_n$ but $n\not\in X\cup A_n$. In particular $X\not=A_n$ for each $n$.

To summarize, no infinite sequence indexed by $\mathbb{N}$ can list all of the members of the powerset of $\mathbb N$. This is what mathematicians mean when they say that the power set of $\mathbb{N}$ is uncountable. To formulate this one certainly needs to be able to talk about the powerset (otherwise there is nothing to be declared uncountable), so the powerset axiom is a good candidate for an answer to your question.

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Hartogs' thereom asserts that for any set $X$ there is an ordinal number $\alpha$ such that there is no injective function $f\colon \alpha \to X$. For $X$ define $H(X)$ to be the smallest such ordinal.

Consequently, $H(\omega)$ exists - we call it $\omega_1$; this is the smallest uncountable ordinal number.