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The countable chain condition for forcing says that every antichain in a forcing order is a countable set. Equivalently,In this question the assertion is evaluated inside $M$. Although every subset of the countable set $M$ is externally countable, that observation does not establish the internal chain condition: $M$ need not contain an enumeration of an antichain.

Because $X=\mathbb{N}$ is a countable set and $\mu (X)=1$, some $x\in X$ satisfies $\mu (\{x\})>0$. Invariance and the inclusion $\{T^{n}x\}\subseteq T^{-1}\{T^{n+1}x\}$ giveIf the forward orbit contained infinitely many distinct points, their disjoint singletons would each have measure at least $\mu (\{x\})>0$, contradicting that $\mu$ is a probability measure. The orbit therefore eventually enters a finite periodic orbit $Y$.

Now $T(Y)=Y$, so $Y\subseteq T^{-1}Y$. Since $\mu$ is invariant,Thus $Y$ belongs to the invariant sigma-algebra modulo null sets. It has positive measure, so the fact that $T$ is an ergodic transformation forcesThis is the ergodic invariant probability on a countable state space has finite cyclic support.

A set is countable if it is finite or admits an injection into the natural numbers; equivalently, its elements can be listed by a finite list or a sequence.

The product of two countable sets is countable: after choosing enumerations, its elements are indexed by pairs $(m,n)\in {\mathbb{N}}^2$, and these pairs can be listed diagonally. Thus an injection $X\to A\times B$ followed by an injection $A\times B\to \mathbb{N}$ proves that $X$ is countable.

The integers are listed as $0,1,-1,2,-2,\dots$, so $\mathbb{Z}$ is countable. Every rational number is represented by a pair $(p,q)\in \mathbb{Z}\times {\mathbb{N}}_{>0}$, so $\mathbb{Q}$ is countable as the image of a countable set.

Finally, if $A_0,A_1,\dots$ are countable, choose listings $A_n=\{a_{n,0},a_{n,1},\dots \}$, allowing repetitions for finite sets. The diagonal listing of the pairs $(n,m)$ lists every $a_{n,m}$, proving that a countable union of countable sets is countable.

U. A witness in the smaller model that $x$ is a countable set remains a function in the larger model, so countability is described by an upward absolute formula. It is not downward absolute: the larger model may have a new enumeration of $x$ that is absent from the smaller model, as explained by upward absoluteness of countability.

Let $M$ be the given transitive model and suppose that its ordinal height of a model of set theory $\mathrm{Ord}\cap M$ were a countable set. For every $x\in M$, the internal Axiom of choice gives a bijection from $x$ to an ordinal of $M$; transitivity makes this an actual bijection, and the ordinal is externally countable. Thus every element of $M$ is externally countable.

For each $\alpha \in \mathrm{Ord}\cap M$, the internal rank $V_{\alpha }^M$ belongs to $M$ and is therefore countable. Every $x\in M$ lies in one of these ranks, so $M$ is a countable union of countable sets and is itself countable, contrary to the hypothesis. By uncountable transitive set model has uncountable ordinal height, $M$ contains uncountably many ordinals.

Suppose that the countable ordinal $\alpha$ satisfied $V_{\alpha }\models \mathsf{ZFC}$. The axioms force $\alpha$ to be a limit ordinal above $\omega$, so choose an externally countable cofinal function $n\mapsto {\alpha }_n$ into $\alpha$, with ${\alpha }_n+1<\alpha$. The internal Axiom of choice gives a bijection in $V_{\alpha }$ between each $V_{{\alpha }_n}$ and some ordinal below $\alpha$. Every such ordinal is externally a countable set, hence every $V_{{\alpha }_n}$ is externally countable. The countable union of countable sets is countable, sowould be countable. But $V_{\alpha }$ contains the full power set $\mathcal{P}(\omega )$, which is uncountable by Cantor theorem. This contradiction is the result Countable rank-initial segment cannot model ZFC, and therefore

The definable power set $\mathcal{D}(A)$ performs one definability step over the single structure $(A,\in )$, whereas the constructible power set ${\mathcal{P}}^L(A)$ contains subsets of $A$ created at arbitrarily late stages of the constructible hierarchy.

For the concrete case $A=\omega$, there are only countably many first-order formulas and finite tuples of natural-number parameters, so $\mathcal{D}(\omega )$ is a countable set. In contrast, the constructible universe satisfies ZFC, and Cantor theorem makes its full power set ${\mathcal{P}}^L(\omega )$ uncountable inside $L$. Consequentlyso the two notions do not agree in general.

Let $F=\bigcup G$, the generic union for the Finite-condition Lévy collapse, and fix an infinite cardinal number $\lambda <\kappa$ in $M$. For every $n\in \omega$, the set of conditions defining a value at $(\lambda ,n)$ is dense: extend a condition at that fresh coordinate with any value below $\lambda$. Hence $F_{\lambda }(n)=F(\lambda ,n)$ is a total function $\omega \to \lambda$.

For every $\beta <\lambda$, the set of conditions assigning the value $\beta$ at some fresh coordinate $(\lambda ,n)$ is also dense. Genericity therefore makes $F_{\lambda }$ surjective. Thusso $\lambda$ is a countable set in $M[G]$. Finite cardinals are already countable.

A set is countable when it is finite or admits a bijection with a subset of $\mathbb{N}$; equivalently, its elements can be listed in a sequence. The integers are countable, for example in the order $0,1,-1,2,-2,\dots$. Enumerating pairs by diagonals, or applying the Cantor pairing function, then shows that $\mathbb{N}\times \mathbb{Z}$ is countable.

Suppose instead that the power set $\mathcal{P}(\mathbb{N})$ had an enumeration $A_1,A_2,\dots$. The Cantor diagonal argument formsFor every $n$, the set $D$ differs from $A_n$ at the element $n$, so it is absent from the alleged enumeration. Therefore

A set is countable when it is finite or its elements can be put in one-to-one correspondence with a subset of $\mathbb{N}$. Equivalently, there is an injection from the set into $\mathbb{N}$.

True. If $\mathrm{TC}(\{x\})$ is countable, then each of its members is a subset of this transitive set and is therefore countable.

Conversely, if $x$ is reasonable, then $x$ itself and every set occurring at every finite membership depth below it are countable. Starting with the countable set $\{x\}$, each next level is a countable union of countable sets, hence countable. Their countable union is $\mathrm{TC}(\{x\})$, so the transitive closure is countable.

A set is countable set if it is finite or admits an enumeration by natural numbers. If $A_n$ are countable, choose enumerations $a_{n,m}$. Since ${\mathbb{N}}^2$ is countable by diagonal enumeration, the array $(n,m)\mapsto a_{n,m}$ enumerates ${\bigcup }_n{A}_n$ after repetitions are removed. This proves that a countable union of countable sets is countable.

The integers are countable, so $\mathbb{Z}\times {\mathbb{N}}_{>0}$ is countable. Mapping $(p,q)$ to $p/q$ surjects onto $\mathbb{Q}$, hence $\mathbb{Q}$ is countable. The same diagonal argument shows that $A\times B$ is countable whenever $A,B$ are.

Finally, if the reals in $(0,1)$ had decimal expansions $x_1,x_2,\dots$, choose a decimal whose $n$th digit differs from the $n$th digit of $x_n$, avoiding digits $0$ and $9$ to remove expansion ambiguity. This number differs from every listed number. The Cantor diagonal argument proves that $\mathbb{R}$ is uncountable.

A set is countable when it is finite or admits a bijection with a subset of $\mathbb{N}$. For each degree, the coefficient tuples for polynomials in $A[x]$ form a finite Cartesian power of the countable set $A$; a countable union over degrees is countable. In particular the algebraic numbers, being roots of countably many integer polynomials with finitely many roots each, are countable. Since $\mathbb{R}$ is uncountable, uncountably many real numbers are transcendental.

The set $Q$ is uncountable. Partition $N$ into four-element blocks and, independently on each block, choose either of two fixed-point-free permutations. Binary sequences then inject into $Q$.

Finally choose a line $L$ distinct from and nonparallel to every $L_i$, possible because only countably many directions are excluded. Each $L\cap L_i$ has at most one point, so the union covers only countably many points of the uncountable line $L$. It cannot cover the plane.

Monomials are indexed by ${\mathbb{N}}^n$, a countable set. A polynomial is a finite list of monomials with rational coefficients, so $P_n$ is a countable union of countable sets.

For a countable set $T\subset {\mathbb{R}}^2$, join two points outside $T$ through an auxiliary point avoiding the countably many lines through either endpoint and a point of $T$. Such an auxiliary point exists: choose a line with direction outside the countable list of forbidden directions, then avoid its countably many intersections with those lines. The two segments avoid $T$, giving a continuous path.

If a smaller transitive model has a function witnessing that $x$ is a countable set, the same witness exists in every larger transitive model. Downward absoluteness can fail because a larger model may contain a new enumeration of $x$.