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A pile top in patience sorting is the currently exposed value on one pile. With the rule “place the next value $x$ on the leftmost pile whose top is at least $x$,” pile tops remain strictly increasing from left to right. Replacing the chosen top preserves that invariant: all earlier tops are smaller than $x$, and later tops exceed the old chosen top and hence $x$.

More strongly, top $i$ is the smallest possible last value of an increasing subsequence of length $i+1$ among the entries seen so far. A smaller last value leaves more options for extending that subsequence. For 3,1,2, the top arrays are [3], [1], [1,2]. binary search finds each replacement position in logarithmic time. The final top array gives the length but need not itself be a subsequence: 2,3,1 ends with tops [1,3], although1 appeared after3. Predecessor pointers are needed to reconstruct a witness.

For a rate-one Poisson process with Poisson arrival times $T_n$, suppose $t_n\to \infty$ and $t_n/n\to \lambda \in [0,\infty ]$. Thenwhere $1/\sqrt{\infty }=0$ and $\mathrm{BN}(\rho )$ denotes the bivariate normal distribution with standard normal marginal distributions, including the singular cases. In particular, equal scales give $\rho =-1$ and separated scales give $\rho =0$.

To prove this, fix real $x,y$ and put $s_n=n+y\sqrt{n}>0$ for large $n$. The event $(T_n-n)/\sqrt{n}\le y$ is exactly $N_{s_n}\ge n$. Counts at the two deterministic times $t_n,s_n$ are sums of a shared count on $[0,\min (t_n,s_n)]$ and an independent count on the remaining interval. The central limit theorem for a centered Poisson random variable of diverging parameter follows by writing a count at a real time $u$ as its count at $\lfloor u\rfloor$, a sum of independent unit increments, plus an independent remainder with expected value at most one. Markov's inequality makes that remainder negligible after division by $\sqrt{u}$.

Apply this central limit theorem to the two independent count blocks whenever their lengths diverge. A block with bounded length, or length negligible compared with the scale of its coordinate, contributes a term tending to zero in probability. The standardized pair of counts therefore has a joint normal limit whose correlation coefficient isThis includes $\lambda =1$: the difference block then has negligible variance after normalization. In the arrival event the standardized count at $s_n$ must be at least $(n-s_n)/\sqrt{s_n}\to -y$. Reversing this inequality changes the sign of the second limiting normal random variable, giving the asserted negative correlation coefficient. Continuity of the marginal distributions permits the moving thresholds and the weak inequalities, including at the singular limits.

Every $\rho \in [-1,0)$ is realized by $t_n=\lambda n$ with $\lambda ={\rho }^{-2}\ge 1$, and zero by $t_n=n^4$. Conversely, even when $t_n/n$ has no limit, it has a subsequence converging in the extended interval $[0,\infty ]$. If the original pair has a joint bivariate normal distribution limit, that subsequence has the same limit, forcing its correlation coefficient into $[-1,0]$.

NFU has a unary predicate $S$ for sets, atoms have no members, extensionality applies to sets, and comprehension applies to every stratified formula. A stratification assigns integer types to variables so that equality uses equal types and $x\in y$ requires $\mathrm{type}(y)=\mathrm{type}(x)+1$. A unary set predicate adds no type difference.

Start with the actual set structure $B=(V_{\omega +\omega },\in )$. This structure satisfies extensionality and every pure-membership separation instance: the subset defined in a set $a$ of rank below $\omega +\omega$ still has rank below $\omega +\omega$. We do not assert that $B$ satisfies replacement or all of ZFC. Expand it by Skolem functions. In it use the sequence of rank objects $V_{\omega +n+6}$, $n<\omega$. For two such objects $d<e$ in sequence order, we havewith quantifiers in $B$. The second statement holds because every subset of $V_{\alpha }$ belongs to $V_{\alpha +1}$, and the later rank is at least $\alpha +1$.

Introduce constants $d_i$, $i\in \mathbb{Z}$, and require them to be order indiscernibles in the Skolem language, satisfying these two assertions whenever $i<j$. Also require $V_{\omega +5}\subseteq d_i$ for every $i$; this fixed rank object is first-order definable in $B$. Every finite part of these requirements together with $\mathrm{Th}(B^{\ast })$ is satisfiable: color finite increasing tuples of the original rank sequence by the truth values of the finitely many formulas mentioned, and use Ramsey's theorem to get a homogeneous subsequence. The rank assertions hold for every increasing tuple there. The compactness theorem gives a model of the full theory. In its Skolem hull $N$ generated by the $d_i$, the integer shift extends to an automorphism $j$ with $j(d_i)=d_{i+1}$, by the term-transport proof of the Ehrenfeucht-Mostowski theorem. Thus $N$ still satisfies extensionality and all pure-membership separation instances, andThis is a set-sized first-order model; no satisfaction predicate for a proper-class universe is being assumed.

Externally let $D_i=\{x\in N:N\models x\in d_i\}$. The automorphism bijects $D_i$ onto $D_{i+1}$. Interpret a typed structure with sort $i$ equal to $D_i$. At sort $i+1$, declare $y$ a set exactly when $N\models y\subseteq d_i$; otherwise it is an atom. Define adjacent-sort membership byThe guard is necessary: a typed atom might have some ordinary $N$-members in $D_i$, and those must not become its typed members. Set-extensionality follows from extensionality in $N$, because the ordinary members of a typed set all lie in $D_i$. For any typed formula on sort $i$, its quantifiers can be restricted to the corresponding $d_l$, its set predicates replaced by subset assertions, and its memberships replaced by the guarded relation. This is an ordinary pure-membership formula of $N$ with finitely many parameters. Separation gives its extension $X\subseteq d_i$, and the displayed closure assertion gives $X\in d_{i+1}$. It is therefore a typed set witnessing comprehension. The automorphism shifts these sorted domains and preserves the set flags and guarded memberships.

Collapse the types onto $D_0$. DefineObjects not satisfying $S$ have no $E$-members. If two sets have the same $E$-members, their images under $j$ are subsets of $d_0$ with the same ordinary members, so are equal by extensionality of $N$; injectivity of $j$ gives equality of the original objects.

To verify every stratified comprehension instance, assign a variable $v$ its stratification type $t(v)$ and translate its value $a\in D_0$ to $j^{t(v)}(a)\in D_{t(v)}$. Equality is preserved. An atomic $E(x,y)$ translates to the guarded adjacent-sort membership because $t(y)=t(x)+1$; the unary predicate $S(y)$ translates to $j^{t(y)}(y)\subseteq d_{t(y)-1}$. Induction on formulas, using the domain bijections for quantifiers, preserves truth. If the free variable $x$ has type $t$, typed comprehension produces its extension $X\subseteq d_t$, with $X\in d_{t+1}$. Put $b=j^{-(t+1)}(X)\in D_0$. Then $S(b)$ holds, and for every $a\in D_0$,This is exactly the required NFU set. No formula containing the external automorphism $j$ has been used in a separation or comprehension scheme; $j$ only transports values after the pure formula is formed.

The construction also permits the usual infinity requirement. The ordinary $\omega$ and its successor graph $f$ are definable in $N$, hence fixed by $j$, and lie below $V_{\omega +5}$, so belong to all the required domains. Under $E$, the members of $\omega$ are still its ordinary $N$-members. The new Kuratowski pair satisfiesSince $j(f)=f$, membership of this pair in $f$ under $E$ is equivalent to $⟨x,y{⟩}_N\in f$. Thus $f$ is an internal injection of $\omega$ into itself omitting zero, giving a Dedekind-infinite set.

We have built a model of NFU, with infinity if it is included in the formulation, in ordinary ZFC metatheory:In particular $\mathrm{Con}(\mathrm{ZFC})\Rightarrow \mathrm{Con}(\mathrm{NFU})$. This proof does not assume $\mathrm{Con}(\mathrm{ZFC})$ as an additional internal axiom: its starting structure is the set $V_{\omega +\omega }$, not a model of ZFC.

A divergent sequence of Möbius transformations has a subsequence converging uniformly on compact subsets of the Riemann sphere outside one exceptional point to a constant. Represent the transformations by determinant-one $2\times 2$ matrices and divide each by its matrix norm. A convergent subsequence has a nonzero rank-one limit: its determinant tends to zero because the original norms tend to infinity. The kernel line is the exceptional point, and the image line is the constant limit. This converts the limit set of a Kleinian group into a tractable boundary dynamics problem.

The RSK insertion shape is the integer partition giving the row lengths of the insertion tableau $P$ in the Robinson-Schensted-Knuth correspondence. The recording tableau has the same shape. For ordinary row insertion, a new entry replaces the first strictly larger entry in a row, bumping it to the next row; if none exists it is appended.

Inserting the word $312$ gives rows $(1,2)$ and $(3)$, so the shape is $(2,1)$. Under this convention, the first row length is the length of a longest weakly increasing subsequence, while the first column length is the length of a longest strictly decreasing subsequence. The shape summarizes ordered structure even when the original word is much longer.

A sequence assigns a value to each position in an ordered index list. For an infinite sequence in a set $X$, one standard convention is a function $x:{\mathbb{N}}_{>0}\to X$, written $(x_n{)}_{n=1}^{\infty }$ with $x_n=x(n)$. Some applications start the index at zero, and a finite sequence has only finitely many positions. Values may repeat; both order and repetition matter. ^[ref].

For example, $a_n=(-1{)}^n$ begins $-1,1,-1,1,\dots$, whereas $b_n=(-1{)}^{n+1}$ begins $1,-1,1,-1,\dots$. Their sets of values are both $\{-1,1\}$, but these are different sequences. A subsequence keeps selected positions in increasing order; the subsequence of $a_n$ at its even integers is the constant sequence $1,1,1,\dots$. A tail of a sequence retains all positions from one index onward.

Sequences express successive approximations, repeated observations, and the iterates of a recurrence relation. A limit of a sequence describes where its terms eventually concentrate. Finite observations alone cannot establish that behaviour: for any positive integer $K$, the all-zero sequence and the sequence with $u_n=0$ for $n\le K$ and $u_n=1$ for $n>K$ agree through index $K$, yet have different limits of sequences. A formula or bound controlling the entire future, such as the tail diameter of a sequence in the Cauchy sequence condition, supplies information that a finite list cannot.

Include the positive integers in $[1,2),[4,8),[16,32),[64,128),\dots$, and exclude the intervening integer intervals and zero. Equivalently, include $[2^{2k},2^{2k+1})\cap \mathbb{Z}$ for $k\ge 0$. Each included block is large enough to raise the running proportion; the following excluded block lowers it again.

With $C_A(N)=|A\cap \{0,\dots ,N-1\}|$, the exact boundary counts follow from a geometric series:The first sum is empty at $k=0$. Dividing by the corresponding boundary, rather than using one denominator for both, givesFor example, $C_A(16)=1+4=5$, whereas $C_A(32)=1+4+16=21$, yielding proportions $5/16$ and $21/32$. For $4^k\le N\le 2\cdot 4^k$, the exact count is $(4^k-1)/3+N-4^k$; for $2\cdot 4^k\le N\le 4^{k+1}$ it stays $(4^{k+1}-1)/3$. Hence the proportion increases through each included block and decreases through each excluded block. These boundary subsequences give the actual limit inferior and limit superior:Since these differ, $A$ has no natural density. Its upper asymptotic density still records its largest limiting proportion.

A shortest common supersequence of two finite sequences is a shortest sequence containing each as a subsequence. Order is retained, but occurrences need not be contiguous. For $ab$ and $ac$, $abc$ is such a sequence; $acb$ is another.

For lengths $m,n$, its length is $m+n-L$, where $L$ is the longest common subsequence length. Align a common subsequence of maximal length and merge the unmatched portions to prove the upper bound. Conversely, an alignment inside any common supersequence identifies at least $m+n-\ell$ shared positions when its length is $\ell$, proving the lower bound. This formula is for two inputs; several inputs require a different analysis.

A subsequential limit of a sequence $(a_n)$ is a value $L$ approached by some subsequence $(a_{n_k})$ with strictly increasing indices $n_k$. For $a_n=(-1{)}^n(1+1/n)$, the subsequences at even integers and odd integers give subsequential limits one and minus one. A finite value appearing among only finitely many initial terms does not become a subsequential limit merely by being a member of the image of a function represented by the sequence.

For a bounded sequence of real numbers, its subsequential limits form a nonempty bounded set that is also a closed set. The Bolzano-Weierstrass theorem guarantees nonemptiness, and its extreme values are the limit inferior and limit superior. The sequence in sparse peaks can determine a limit superior has the two subsequential limits zero and one, despite almost all indices in the natural density sense carrying zero.

A substring is a contiguous segment of a string. It differs from a subsequence, whose selected characters need not be adjacent.

Suppose $k_1<k_2<\cdots$ are peaks of a sequence $(a_n)$ and $m_1<m_2<\cdots$ are peaks of a sequence $(-a_n)$. Then $(a_n)$ converges precisely when these two subsequences have a common finite limit of a sequence $L$.

Necessity follows because every subsequence inherits the limit of a sequence of a convergent sequence. For sufficiency, choose one selected value from each subsequence within $\epsilon$ of $L$. Every term after both selected indices lies between the two selected values:This is exactly $a_n\to L$.