I am a beginner in set theory, so if this question is strange, I apologise. As well known, from the consequence about the continuum hypothesis it is provable in ZFC that "If ${\rm Con(ZFC)}$, then it is independent of ZFC that whether there exists a set $\Omega%$ satisfying $\aleph_0<{\rm card}(\Omega)<2^{\aleph_0}$." In other words, there are both models of ZFC where there exists a set $\Omega%$ satisfying $\aleph_0<{\rm card}(\Omega)<2^{\aleph_0}$ and models where there does not exist a set $\Omega%$ satisfying it. On this basis, when I think about a set $\{ \Omega \mid \aleph_0<{\rm card}(\Omega)<2^{\aleph_0} \}$, it goes over my head. As I understand it the following statements are provable in ZFC, e.g. there exists this set, (if ZFC is consistent then) the statement that the cardinality of this set is not $0$ is independent of ZFC, and the cardinality of this set is greater than or equal to $0$. Is this correct? I'd appreciate if you could answer the question.
Does there exist a set $\{ \Omega\mid \aleph_0<{\rm card}(\Omega)<2^{\aleph_0} \}$
set-theory
Related Solutions
$\omega$ and $\aleph_0$ are the same set (the natural numbers $\{0,1,2,3,\ldots\}$) by different names. $\omega$ stresses the fact that we see it as an ordinal number (the smallest one that is infinite) while $\aleph_0$ is a name that says it is the first (i.e. of index $0$) in the transfinite sequence of cardinal numbers. All cardinals are special ordinal numbers, which in turn are transitive sets well-ordered by $\in$. All this is standard modern set theory (assuming ZFC standard axioms, including well-foundedness). As to the continuum $\mathfrak{c}$ we can say that $\mathfrak{c} \ge \aleph_1$ (it's uncountable) but not much more. Also, there is a theorem that $\text{cf}(\mathfrak{c})$ (the cofinality) cannot be $\omega$, so that $\mathfrak{c}=\aleph_{\omega}$ is not possible; but other than that restriction it can be any uncountable aleph (consistently).
So it's more correct to say sets of cardinality $\aleph_0$ (not cardinality $\omega$, though one does see it occasionally). Ordinals are for "order things", cardinals for "sizes" of sets.
Yes, because the standard formalism for forcing that Kunen uses starts with a countable transitive model of $\sf ZFC$, rather than just any countable model, more care is needed. Whether there is a transitive model of $\sf ZFC$ is independent from $\sf ZFC + Con(ZFC),$ though once you have one, you can get a countable transitive model by Lowenheim-Skolem + Mostowski collapse.
On the other hand, the reflection schema allows you to construct, in $\sf ZFC$ alone, a countable transitive model of any finite subtheory of $\sf ZFC$. Analyzing the forcing formalism, if we have an arbitrary finite fragment $T$ of $\sf ZFC + \lnot CH,$ it's possible to find a finite fragment $T'$ of $\sf ZFC$ such that we can, working in $\sf ZFC,$ construct a countable transitive model $M$ of $T'$ and show the generic extension $M[G]$ satisfies $T.$
If $\sf ZFC + \lnot CH,$ were inconsistent, there would be some finite fragment witnessing this, and then the above procedure would produce a contradiction in $\sf ZFC$ alone. So, as a bonus, this is no mere infinitary proof of $\sf Con(ZFC)\to Con(ZFC+\lnot CH)$ using model theory within $\sf ZFC$, but rather a finitary syntactic proof of $\sf Con(ZFC)\to Con(ZFC+\lnot CH)$.
On a side note, it is possible to carry out an independence proof that starts out like you suggested. Note that in the transitive approach, the construction of $M[G]$ uses the (external) well-foundedness of $M$ in an essential way. An analogous construction in the not-necessarily-well-founded case is possible, but more complicated$^*$. Between that complication and the fact that transitive models are nicer to work with for all sorts of reasons, the transitive models approach is generally favored as an introduction, despite the mild metamathematical hiccups it causes.
$^*$ One approach that works is to construct the relevant Boolean-valued model inside $M$ and then take a quotient of it by any ultrafilter. But then again, Boolean-valued models already produce a nice relative consistency proof on their own. See also this answer for a brief description, and JDH's answer to the same question for more on the BVM approach.
Best Answer
In a model of ZF(C) where the continuum hypothesis holds, the class $\{ \Omega \mid \aleph_0 < card(\Omega) < 2^{\aleph_0} \}$ is empty. Hence it is a set, namely the empty set.
On the other hand, in a model of ZF(C) where the continuum hypothesis does not hold (i.e. where it is false), the class $C := \{ \Omega \mid \aleph_0 < card(\Omega) < 2^{\aleph_0} \}$ is not empty. But now we can show that it is a proper class (as mentioned by Michael Carey in the comments). So $C$ is not a set in this case.
As a proof, I claim that if $C$ is non-empty, then $\bigcup C$ is the class of all sets. Let $x$ be any set and $y \in C$ (which exists by non-emptyness of $C$). If $x \in y$ then we are done, as $x \in \bigcup C$ follows directly. Otherwise $x \notin y$. Because $\aleph_0 < card(y)$, there must be some $z \in y$. Define $w := (y \setminus \{z\}) \cup \{x\}$. We can find a bijection between $y$ and $w$ by taking the identity on $y \setminus \{z\}$ and mapping $z \mapsto x$. Therefore $card(w)=card(y)$ and $w \in C$. By construction $x \in w$ so we obtain $x \in \bigcup C$ as claimed.
The same proof that $C$ is a proper class works, whenever $C$ is a non-empty class only defined in terms of the cardinality of its elements.