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Every family of pairwise disjoint nonempty open disks in the plane is countable. The density of the rational numbers gives each disk a point of the countable set ${\mathbb{Q}}^2$. Fixing an enumeration and taking the first point in each disk yields an injection from disks to labels. This argument requires no countable choice of representatives and is useful when geometric separation would otherwise produce an uncountable family.

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.

The Martingale convergence theorem says that if a discrete-time martingale $(M_n)$ satisfies $C={\sup }_n\mathbb{E}|M_n|<\infty$, then there is an integrable random variable $M_{\infty }$ such thatThe hypothesis does not by itself ensure convergence in L1.

For each pair of rational numbers $a<b$, the Doob upcrossing inequality gives $\mathbb{E}{U}_N[a,b]\le (C+|a|)/(b-a)$. The monotone convergence theorem therefore gives a finite expected value for $U_{\infty }[a,b]={\lim }_N{U}_N[a,b]$, so this upcrossing count is finite almost surely. There are only countably many such pairs, hence all their upcrossing counts are simultaneously finite outside one event of probability zero.

If ${\mathrm{liminf}}_n{M}_n<{\mathrm{limsup}}_n{M}_n$, the density of the rational numbers supplies $a<b$ strictly between them, forcing infinitely many upcrossings. Consequently $M_n$ has a limit in the extended real numbers almost surely. By the Fatou lemma,A limit of either $+\infty$ or $-\infty$ would make this lower limit infinite, so the limit is finite almost surely, and the same inequality proves its integrability.

The sigma-algebras ${\mathcal{G}}_n=\sigma (S_n,S_{n+1},\dots )$ decrease with $n$, and (b) gives $S_n/n=\mathbb{E}[X_1\mid {\mathcal{G}}_n]$. The permitted reverse martingale convergence theorem therefore gives an integrable random variable $L$ such that $S_n/n\to L$ almost surely and with convergence in L1. In particular,It remains to prove that $L$ is constant. We must not assume that ${\bigcap }_n{\mathcal{G}}_n$ is the tail sigma-algebra of the $X_j$; instead use tail measurability of limits of sample averages directly. Define the random variable valued in the extended real numbers $L^{\ast }={\mathrm{limsup}}_n{S}_n/n$. For every fixed $N$,because $S_N/n\to 0$ pointwise. Thus $L^{\ast }$ is measurable with respect to $\sigma (X_{N+1},X_{N+2},\dots )$ for every $N$, and hence to the tail sigma-algebra. Also $L^{\ast }=L$ almost surely.

By the Kolmogorov zero-one law, each tail event $\{L^{\ast }\le q\}$, for a rational number $q$, has probability zero or one. Since $L^{\ast }$ is finite almost surely, its distribution function can only be that of a constant $c$: taking $c=\inf \{q\in \mathbb{Q}:\mathbb{P}(L^{\ast }\le q)=1\}$ and using the density of the rational numbers gives $L^{\ast }=c$ almost surely. Its expected value identifies $c=m$. This proves the strong law of large numbers:

For a increasing function $F$, define its one-sided limits byThey exist because of monotonicity. The function is discontinuous at $x$ only if the jump intervalis nonempty. By the density of the rational numbers, choose $q_x\in J_x\cap \mathbb{Q}$.

If $x<y$, monotonicity gives $F(x+)\le F(y-)$, so $J_x$ and $J_y$ are disjoint. Consequently $q_x\ne q_y$, and $x\mapsto q_x$ injects the set of discontinuities into $\mathbb{Q}$. Since the rationals are countable by part (ii), the set of discontinuities is countable.