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The integers are countable: $j(z)=2z+1$ for $z\ge 0$ and $j(z)=-2z$ for $z<0$ is an injection $\mathbb{Z}\to \mathbb{N}$. Hence $\mathbb{Z}\times \mathbb{N}$ is a countable set. The map $(a,b)\mapsto a/b$ is a surjection onto the rational numbers, so the image result already proved gives $\mathbb{Q}$ is countable. Unique reduced fractions are not required for this surjection argument.

For the final unheaded geometry request, suppose for contradiction that no two distinct circles in the collection intersect. Restrict to the subcollection ${\mathcal{C}}_0$ of circles tangent to the $x$-axis. Each such nondegenerate circle has a unique tangency point, and the tangency-point map ${\mathcal{C}}_0\to \mathbb{R}$ is a surjection by the hypothesis.

The closed disk of a circle tangent at $(a,0)$ meets the $x$-axis only at $(a,0)$. Two distinct members of ${\mathcal{C}}_0$ cannot have nested disks: if one disk were inside the other, its tangency point would belong to the larger disk, so both tangency points would be equal and both circle boundaries would pass through it. That already contradicts nonintersection. Circles with disjoint boundaries and no nesting have disjoint disks. Thus their open disks are pairwise disjoint.

By the density of the rational numbers, every nonempty open disk contains a point of ${\mathbb{Q}}^2$, which is a countable set. Fix an enumeration of ${\mathbb{Q}}^2$ and assign to each disk the first enumerated point inside it. Disjointness makes this assignment an injection. This is the countability of pairwise disjoint open disks, and proves that ${\mathcal{C}}_0$ is a countable set. Its surjection onto $\mathbb{R}$ would make the real numbers countable, contrary to the Cantor diagonal argument. ThereforeThe proof allows circles on either side of the axis and counts tangential contact as intersection; it does not assume that all circles in the original collection are tangent to the axis.

Suppose $f:S\to \mathcal{P}(S)$ were a surjective function. Form the subsetSince $D$ belongs to the power set, surjectivity would give $D=f(d)$ for some $d\in S$. But then $d\in D$ holds if and only if $d\notin f(d)=D$, a contradiction. Thus no such function is surjective, and in particular there is no bijection from $S$ to its power set. This is Cantor theorem; the argument also applies when $S$ is empty.

For infinite binary sequences, use the bijectionwith $\mathcal{P}(\mathbb{N})$, taking $\mathbb{N}=\{1,2,\dots \}$. Conversely, a subset determines its sequence of membership indicators. The power set is infinite, since it contains all singletons, and Cantor theorem rules out a bijection with $\mathbb{N}$. Hence it is an uncountable set, as is the set of infinite binary sequences. Equivalently, a proposed enumeration $(x_n^{(j)}{)}_{n\ge 1}$ is defeated by the sequence $y_j=1-x_j^{(j)}$, which differs from its $j$th listed sequence at position $j$. This is the Cantor diagonal argument.

A set is countable if it is finite or bijects with a subset of $\mathbb{N}$. For any $f:X\to \mathcal{P}(X)$, the Cantor diagonal argument set $D=\{x:x\notin f(x)\}$ differs from every $f(y)$ at $y$. Hence no such map is surjective. Binary sequences are indicator functions of subsets of $\mathbb{N}$, so $\{0,1{\}}^{\mathbb{N}}\cong \mathcal{P}(\mathbb{N})$ is uncountable.

Every member of $L$ corresponds to a unique permutation $(a_0,a_1,\dots )$ of $\mathbb{N}$, with $F_n=\{a_0,\dots ,a_{n-1}\}$. Let $p(j)$ be the position of $j$.

For $L_0$, $p(n)<n$ eventually. Choose $N$ after which this holds and then $m$ larger than all $p(0),\dots ,p(N-1)$. The $m$ positions $0,\dots ,m-1$ would contain all $m+1$ values $0,\dots ,m$, a contradiction. Thus $L_0$ is empty, hence countable.

For $L_1$, $p(n)\le n$ eventually. For every sufficiently large $m$, all values $0,\dots ,m$ occur in positions $0,\dots ,m$, so this initial segment is invariant. Comparing consecutive invariant initial segments gives $p(m)=m$ eventually. These are the finite-support permutations, of which there are countably many. Hence $L_1$ is 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

For cardinals, $\kappa <\lambda$ means that there is an injection $\kappa \to \lambda$ but no bijection. Under the axiom of choice, every set is well-orderable, so every infinite cardinal is the initial ordinal ${\aleph }_{\alpha }$ for a unique ordinal $\alpha$. Cantor theorem givesand minimality of the successor cardinal givesThe strict inequality ${\aleph }_{\alpha +1}<2^{{\aleph }_{\alpha }}$ printed in the conversion is not provable and can be false under the generalized continuum hypothesis; the intended symbol must be $\le$.

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.