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Yes. Let $(x^{(k)}{)}_{k\ge 1}$ be any sequence in $A$, where $x^{(k)}=(x_m^{(k)}{)}_{m\ge 1}$. At coordinate $m$, the values lie in the compact interval $[-1/m,1/m]$. By part (a)(ii) we can successively extract a convergent subsequence for coordinate one, a further one for coordinate two, and so on. Taking the diagonal indices gives a single subsequence $(x^{(k_j)})$ for whichThe closed interval bound passes to the limit, so $|x_m|\le 1/m$. Hence $x_m\to 0$ and $x=(x_m)\in A$.

Coordinate convergence alone would not imply supremum norm convergence. Here the common tail bound supplies the missing estimate. Given $ϵ>0$, choose $M$ with $2/(M+1)<ϵ$. For $m>M$,For the finitely many $m\le M$, coordinate convergence makes all differences less than $ϵ$ once $j$ is large. Therefore $d(x^{(k_j)},x)<ϵ$, andThis is an instance of compactness from uniformly vanishing sequence tails. The common $1/m$ bound is exactly what the coordinate vectors in part (i) lack.

The source prints $m\ne 0$ in the zero case. Since the coordinates are indexed by positive integers, that overlaps its $m=n$ case and does not define a sequence. We use the intended coordinate vectors, with $e_m^{(n)}=1$ for $m=n$ and $e_m^{(n)}=0$ for $m\ne n$.

Every $e^{(n)}$ belongs to $X$, because its entries are bounded by one and its coordinate sequence eventually vanishes. If $n\ne k$, their supremum norm distance isIndeed, the difference is $1$ at coordinate $n$, $-1$ at coordinate $k$, and zero elsewhere. No subsequence is a Cauchy sequence, since arbitrarily late distinct terms remain distance one apart. Every convergent sequence is a Cauchy sequence, so no convergent subsequence exists. Therefore $\boxed{X\text{ is not sequentially compact}}$.

Choose $n\ge 1$ as in part (i). Since distances between iterates are nondecreasing, the triangle inequality givesLetting $ϵ\downarrow 0$ and combining with the given reverse bound proves that $\varphi$ is an isometry:For surjectivity, fix $x\in X$ and use part (i) with $y=x$ and $ϵ=1/k$ to choose $n_k\ge 1$ satisfying $d(x,{\varphi }^{(n_k)}x)<1/k$. Put $z_k={\varphi }^{(n_k-1)}x$, interpreting ${\varphi }^{(0)}$ as the identity. By sequential compactness, a convergent subsequence satisfies $z_{k_j}\to z\in X$. The assumed continuity yieldsThus every $x$ has a preimage, and $\boxed{\varphi \text{ is surjective}}$. If $X$ is empty, both conclusions hold vacuously. This proves the distance-increasing self-map theorem for sequentially compact metric spaces.

By the permitted sequential compactness of the product metric space $X\times X$, the sequence of pairs $({\varphi }^{(k)}x,{\varphi }^{(k)}y)$ has a convergent subsequence, and hence a Cauchy subsequence. Choose two of its indices $r<s$ sufficiently late thatThe distance-increasing assumption iterates: $d({\varphi }^{(r)}u,{\varphi }^{(r)}v)\ge d(u,v)$. With $n=s-r\ge 1$, use $u=x,v={\varphi }^{(n)}x$, and then $u=y,v={\varphi }^{(n)}y$, to obtainBoth summands are nonnegative, soThe use of paired iterates ensures that the same positive return time works for both points.

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set, $d<\infty$, and $f:U\to {\mathbb{R}}^d$ a continuous function. Through every $(t_0,y_0)\in U$ there is a local solution of a differential equation: a continuously differentiable function $y:I\to {\mathbb{R}}^d$ on an open interval containing $t_0$, satisfying the displayed initial value problem pointwise. No Lipschitz condition is assumed.^[ref]

Continuity of a proposed rate law therefore guarantees at least one trajectory for a short time. For a physical model, this is a first consistency check: can the stated rule and initial condition actually be followed? Prediction also needs uniqueness. The delayed solutions of a square-root differential equation below show how the same rule and initial condition can leave a departure time undetermined. The Picard-Lindelöf theorem supplies a sufficient additional hypothesis for uniqueness; failure of that hypothesis alone does not prove nonuniqueness.

For the initial value problem $y^{\prime }=2\sqrt{|y|}$, $y(0)=0$, a forward solution of a differential equation can stay zero until any $c\ge 0$, then follow $y=(t-c{)}^2$. The delayed solutions of a square-root differential equation verify this freedom exactly.

Local existence gives neither a global solution of a differential equation nor a bound on all future values. The continuous function $f(t,y)=y^2$ already permits finite-time blowup; the finite-dimensional continuation criterion explains the additional issue of extending a trajectory within its domain.

Proof outline: choose a compact set $K=[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$. By the extreme value theorem, $M={\max }_K\Vert f\Vert <\infty$. Take $0<h\le a$ with $Mh\le r$. Forward Euler method polygons remain in the closed ball and satisfy $\Vert y_n(t)-y_n(s)\Vert \le M|t-s|$. The Arzelà-Ascoli theorem yields a convergent subsequence with uniform convergence. Uniform continuity of $f$ on $K$ makes their integral equation defect tend uniformly to zero, so the limit satisfiesThe fundamental theorem of calculus gives a continuously differentiable function. Repeat backward and join at $t_0$, where both derivatives equal $f(t_0,y_0)$.^[ref]

The compact set here uses a finite-dimensional vector space; bounded closed balls in an arbitrary Banach space need not be compact. When $M=0$, the constant solution of a differential equation $y=y_0$ works on the entire time interval in $K$. In general the proof selects a subsequence, without asserting that all polygons converge to one unique trajectory.

For metric spaces $(X_1,d_1)$ and $(X_2,d_2)$, the sum metric on $X_1\times X_2$ is the displayed formula. Nonnegativity, symmetry and the triangle inequality follow by adding the respective inequalities in the two factors, and zero distance forces equality in both coordinates. It induces the finite product topology, as does the maximum product metric, since their distances satisfy $d_{\max }\le d_{\mathrm{sum}}\le 2d_{\max }$. A sequence converges for this metric exactly when both coordinate sequences converge. Two successive extractions of a convergent subsequence therefore show that the product of two sequentially compact spaces is sequentially compact.

Every continuous map from a nonempty compact convex set $K$ in a Banach space to itself has a fixed point. A useful equivalent form takes a nonempty closed bounded convex set $C$ and a continuous map $T:C\to C$ whose image is relatively compact. To obtain this version from the compact-domain version, restrict $T$ to $\overline{\mathrm{conv}}T(C)\subset C$ and use closed convex hull of a compact set.

For a proof on $K$, the Schauder approximation lemma gives continuous maps $T_{\epsilon }:K\to K\cap L_{\epsilon }$ with finite-dimensional $L_{\epsilon }$ and $\Vert T-T_{\epsilon }{\Vert }_{\infty }\le \epsilon$. The Brouwer fixed-point theorem on the nonempty compact convex set $K\cap L_{\epsilon }$ gives $x_{\epsilon }=T_{\epsilon }{x}_{\epsilon }$, so $\Vert T{x}_{\epsilon }-x_{\epsilon }\Vert \le \epsilon$. A convergent subsequence as $\epsilon \downarrow 0$ and the continuity of $T$ give an exact fixed point.

Every finite Markov chain has a stationary distribution, without requiring Markov chain irreducibility or aperiodicity. For a row-stochastic matrix $P$, put $B=P^T$ and choose a column probability vector $q$. Every $B^{j}q$ lies in the probability simplex. Their arithmetic averages $x_N=N^{-1}{\sum }_{j=0}^{N-1}{B}^{j}q$ therefore lie in this compact set, while telescoping givesA convergent subsequence has a limit $x$ in the probability simplex. Continuity of matrix multiplication gives $Bx=x$, so $\pi =x^T$ is a stationary distribution. The individual iterates need not converge: a deterministic two-state swap alternates its point masses forever.

Selecting a subsequence is a regular matrix summability method: every row has one entry equal to one, every fixed column eventually vanishes, and all row sums equal one. Every bounded sequence of real or complex numbers has a convergent subsequence by the Bolzano-Weierstrass theorem, so each bounded sequence is summable by some sequence-dependent regular method. The order of the quantifiers matters: no single regular matrix sums all bounded sequences.

Topological analysis groups methods that use properties of topological spaces to establish results in analysis. Compactness turns bounded families with suitable regularity into convergent subsequences; connectedness underlies intermediate-value arguments; and fixed-point theorems can prove that a nonlinear equation has a solution without producing a closed formula.

For instance, if a continuous function maps $[0,1]$ into itself, $g(x)=f(x)-x$ satisfies $g(0)\ge 0$ and $g(1)\le 0$. The intermediate value theorem therefore gives $g(c)=0$, so $f(c)=c$. The Brouwer fixed-point theorem extends this existence mechanism to finite-dimensional convex compact sets. Such arguments provide existence, while uniqueness and numerical convergence need further information.