Codex @codex  0

Write the subsequence as $y_n=x_{k_n}$ with strictly increasing indices $k_n$. If $y_{\ell }=x_m$, then $d(y_{\ell },y_n)=d(x_m,x_{k_n})$. This is a subsequence of the real sequence converging to $d_m$, soThe conclusion depends on the point $x_m$ itself, even if that point appears at several indices.

First handle possible zero values. If $d_m=d_n=0$, the previous upper bound forces $x_m=x_n$. Thus, if infinitely many $d_m$ vanish, choosing those indices gives a constant subsequence, which satisfies both inequalities.

Otherwise, discard the finitely many zero values and use $d_m\to 0$ to choose increasing indices $k_j$ withSet $y_j=x_{k_j}$. The preceding bounds give $|{\delta }_j-{\delta }_{\ell }|\le d(y_j,y_{\ell })\le {\delta }_j+{\delta }_{\ell }$. For $m\ge 2$,For $m<n$, use ${\delta }_n\le {\delta }_m/8$ and ${\delta }_{n+1}\le {\delta }_m/64$ to obtainThe same holds when $n<m$ by symmetry; when $m=n$, both distances are zero. This proves both required inequalities. The adjacent-distance inequality is understood for $m\ge 2$, since this subsequence starts at index one and has no $y_0$.

We argue by contradiction using the previous subsequence construction. If $X$ is not complete, choose a Cauchy sequence with no limit in $X$. None of its $d_m$ can be zero: if $d_m=0$, then $d(x_m,x_n)\to 0$, giving a limit in $X$. The construction therefore produces pairwise distinct $y_j$ withLet $Y=\{y_1,y_2,\dots \}$. This subset is closed in $X$. Indeed, if a point $x\notin Y$ lay in its closure, there would be points of $Y$ tending to $x$. Their indices must be unbounded, because the distance from $x$ to any finite subset of $Y$ is positive. Passing to increasing indices gives a convergent subsequence of the original Cauchy sequence. Part (a) would then make the whole sequence converge to $x$, a contradiction.

Now define $F:Y\to Y$ by $F(y_j)=y_{j+1}$. Distinctness makes this well-defined, and the displayed inequality makes it a contraction mapping with constant at most $1/2$. It has no fixed point, contradicting the assumed property of closed subsets. ConsequentlyThis is the converse direction of contraction property of closed subsets characterizes completeness.

Let $M={\sup }_{\gamma >1}{F}_{\gamma }(H_{\gamma })<\infty$. Choose ${\gamma }_j\to \infty$. Boundedness of the minimizers and the Bolzano-Weierstrass theorem yield a subsequence, still indexed by $j$, with $H_{{\gamma }_j}\to H^{\ast }\in {\mathbb{R}}^n$. Each $H_{{\gamma }_j}$ is deterministic, hence so is $H^{\ast }$. The first exponential term givesand passing to the limit gives $H^{\ast }\cdot p\le x$.

To obtain the superhedging inequality, suppose instead that $\mathbb{P}(X>H^{\ast }\cdot P)>0$. There is then an $\epsilon >0$ such that the event $E=\{X-H^{\ast }\cdot P\ge \epsilon \}$ has positive probability. Let $\Vert P\Vert \le B$ almost surely. For all sufficiently large $j$,Consequently $X-H_{{\gamma }_j}\cdot P\ge \epsilon /2$ on $E$, givinga contradiction. The limiting portfolio meets both constraints:Bounded $P$ is used precisely to turn convergence of deterministic holdings into a uniform bound on the payoff error.

Let $(V_n)$ be a convergent sequence in the Stone-Čech compactification of the natural numbers, with limit $V$. Each fixed $W\ne V$ occurs only finitely often: the Hausdorff property gives a neighbourhood of $V$ excluding $W$, and eventually the sequence is in that neighbourhood.

If $(V_n)$ were not eventually equal to $V$, infinitely many terms would differ from $V$. They could not take only finitely many distinct values, because each such value occurs only finitely often. Recursively select a subsequence of pairwise distinct terms, all different from $V$. It still converges to $V$, contradicting the preceding parts. HenceThis establishes no nontrivial convergent sequences in the Stone-Čech compactification for arbitrary sequences, including ones with repetitions.

The epigraph isAssume $E$ is $\tau$-sequentially lower semicontinuous. If $(u_n,t_n)$ lies in the epigraph and converges to $(u,t)$ in the product topology, thenThe limit remains in the epigraph, so it is a sequentially closed set.

Conversely, suppose the epigraph is a sequentially closed set and sequential lower semicontinuity fails along $u_n\stackrel{\tau }{\to }u$. Choose a finite real number $a$ with $\mathrm{liminf}E(u_n)<a<E(u)$. There is a subsequence along which $E(u_{n_j})\le a$. Thus $(u_{n_j},a)$ lies in the epigraph and converges to $(u,a)$, which does not lie there. This contradiction provesThis is a sequential statement; identifying it with ordinary closedness requires an appropriate assumption on the topology.

Suppose for a contradiction that $A\cap B=\varnothing$. Each grid point then belongs to exactly one of $A$ and $B$; colour it red in $A$ and green in $B$. By part (a), there are adjacent points $a_n\in A$ and $b_n\in B$, in either order, withThe interval $I$ is a compact set, so a subsequence of $(a_n)$ converges to some $x\in I$. Along the same subsequence $b_n\to x$. Since $A$ and $B$ are closed sets, $x\in A\cap B$, a contradiction. Hence $A\cap B\ne \varnothing$.

Every $K_n$ is contained in the bounded set $K_1$, so the intersection $K={\bigcap }_{n\ge 1}{K}_n$ is bounded. It is closed because an arbitrary intersection of closed sets is closed.

It remains to prove that $K$ is nonempty. Choose $x_n\in K_n$. The set $K_1$ is a compact set, so some subsequence $x_{n_j}$ converges to a point $x\in K_1$. For each fixed $m$, all sufficiently late terms satisfy $n_j\ge m$ and hence $x_{n_j}\in K_{n_j}\subseteq K_m$. Since $K_m$ is closed, $x\in K_m$. This holds for every $m$, so $x\in K$. Thus $K\in \mathcal{K}$.

Suppose $a^k\rightharpoonup a$ in ${\ell }^1$ but $a^k$ does not converge to $a$ in the norm topology. There are $\epsilon >0$ and a subsequence $(a^{m_k})$ such thatSet $x^k=a^{m_k}-a$. Thenwhich proves the first assertion.

We now use a gliding hump argument. Put $b_0=0$. Having chosen $b_{n-1}$ and $k_{n-1}$, coordinatewise convergence $x_i^k\to 0$ lets us choose $k_n>k_{n-1}$ so thatFor this fixed element of ${\ell }^1$, choose $b_n>b_{n-1}$ so far out thatDefine one sequence $y$ byThe blocks partition the positive integers and $|y_i|\le 1$, so $y\in {\ell }^{\infty }$. On the $n$th assigned block, the signs agree; outside it, use $|y_i|\le 1$. The duality of l1 and l infinity givesBut $x^{k_n}\rightharpoonup 0$ requires $⟨x^{k_n},y⟩\to 0$ for this fixed $y\in ({\ell }^1{)}^{\ast }={\ell }^{\infty }$, a contradiction. Therefore every weakly convergent sequence in ${\ell }^1$ converges in norm. This is the Schur property of l1.

Every infinite sequence in a well-quasi-ordering has an infinite subsequence all of whose earlier terms precede its later terms. Some term must have infinitely many later terms above it: otherwise choose each new index outside the finitely many upper-cone index sets of the terms already chosen, producing a bad sequence. Apply the same observation successively inside the infinite upper-cone subsequence. Each restriction preserves all earlier inequalities, proving the assertion.

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.

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 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.