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Let $(x_n)$ be any sequence in the compact metric space $X$. There must be an $x\in X$ such that every open ball about $x$ contains $x_n$ for infinitely many indices $n$. Otherwise, for each $x\in X$ choose a ball $B(x,r_x)$ containing only finitely many terms, counted by indices. These balls are an open cover. A finite subcover would imply that only finitely many indices occur anywhere in $X$, a contradiction.

Now choose successively $n_k>n_{k-1}$ with $x_{n_k}\in B(x,1/k)$; infinitely many available indices make this possible. Then $d(x_{n_k},x)<1/k\to 0$, so $(x_{n_k})$ is a convergent subsequence. ThusThe proof counts indices rather than distinct points, so it also handles a value repeated infinitely often.

A metric space $X$ is a compact space if every open cover $X={\bigcup }_{\alpha \in I}{U}_{\alpha }$, with all $U_{\alpha }$ open in $X$, has a finite subcover: finitely many indices ${\alpha }_1,\dots ,{\alpha }_r$ cover $X$.

It is a sequentially compact space if every sequence $(x_n)$ in $X$ has a convergent subsequence $(x_{n_k})$ whose limit of a sequence belongs to $X$. In formulas,A finite subcover is the defining test for compactness; a convergent subsequence is the defining test for sequential compactness.

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 natural-number successor chain embeds into the integer successor chain by ordinary inclusion. Thus $\mathbb{Z}$ witnesses the existential joint satisfiability in the first assertion. The two chains are not elementarily equivalent: every integer has a predecessor, whereas zero has none in the natural-number chain.

For the universal assertion, construct an elementary extension of the natural-number chain containing a two-sided successor chain. To its elementary diagram add distinct constants $d_j$ for $j\in \mathbb{Z}$ with $s(d_j)=d_{j+1}$ and require them to differ from all named natural numbers. Every finite subset is realized in the natural-number chain: place the finitely many required consecutive blocks sufficiently far to the right, away from the finitely many forbidden named Elements. Compactness yields an elementary extension $A^{\ast }$ with the required two-sided chain. That chain is a substructure isomorphic to the integer successor structure and hence is a model of its complete theory.

Consequently both joint theories in the two assertions are satisfiable, though the integer chain cannot embed into the natural-number chain itself: repeated predecessors would give an infinite descending sequence of natural numbers. The second assertion correctly uses an elementarily equivalent ambient structure rather than the original one.

Against Tit for tat, let $a_t$ be an arbitrary deviator's action at round $t$, and set $a_{-1}=C$ to encode the opponent's initial cooperation. The opponent then plays $a_{t-1}$. Define $h(C)=0$ and $h(D)=3$. The four possible transitions giveThus every stage payoff is at most $3+h(a_t)-h(a_{t-1})$. The telescoping sum yields the uniform payoff bound against Tit for tat:After division by $T$, no long-run average payoff can exceed three. The bound applies to every action sequence, so in particular it covers every finite-state machine representing a unilateral deviation. Mutual Tit for tat produces $C,C$ in every round and gives each player three. Neither player can improve, proving $(\mathrm{TFT},\mathrm{TFT})$ is a Nash equilibrium.

Another pair is $(\mathrm{TFT},\text{grim trigger})$. Here grim trigger initially cooperates and continues cooperating until the opponent first defects, after which it defects forever. Both strategies are two-state finite-state machines: Tit for tat records the opponent's previous action, while grim trigger records whether defection has ever occurred. Their prescribed play again consists entirely of cooperation and gives $(3,3)$.

The player using grim trigger faces Tit for tat, so every unilateral deviation is bounded by the same telescoping sum and cannot improve on three. The player using Tit for tat faces grim trigger. If that player never defects, the payoff remains three. If the first defection occurs at a finite round $\tau$, it gives a one-time payoff six; thereafter the opponent always defects, against which each subsequent payoff is at most one. Any finite initial prefix has zero contribution to the long-run average payoff, so such a unilateral deviation has limiting upper average at most one. It cannot improve on three either. Both players are therefore best responses, proving the second claimed Nash equilibrium.

For deterministic finite-state machines, the joint state path is eventually periodic, so the displayed averages have a limit. The first bound also controls the limiting upper average for arbitrary action sequences, making the no-improvement argument stronger than an enumeration of finite-state deviations. Notice that equality in the bound permits alternating exploitation to average three: equilibrium requires absence of a strict improvement, not uniqueness of a best response.

Construct one infinite binary sequence $u$ by concatenating every binary word of length one, then every word of length two, then every word of length three, and so on, in lexicographic order within each length. Every finite binary word occurs somewhere in $u$.

Given any target sequence $s$ and any $\eta >0$, choose $N$ such that $2^{-N}<\eta$. The word $s_1\cdots s_N$ occurs in $u$, beginning after some $k$ symbols. Thus ${\sigma }^{k}u$ agrees with $s$ in its first $N$ positions, and the prescribed metric givesEvery open open ball about every point therefore meets the forward orbit of $u$. This forward orbit is dense in $\Sigma$. The argument identifies finite-prefix cylinder sets as arbitrarily small neighborhoods in this metric; no binary-expansion convention for real numbers is needed.

The space $\Sigma$ of all infinite binary sequences is uncountable by the Cantor diagonal argument: given any list $s^{(1)},s^{(2)},\dots$, the sequence $t_i=1-s_i^{(i)}$ differs from its $i$th entry at position $i$, so it is absent from the list. Part (i) shows that only countably many sequences are periodic. Their complement is therefore uncountable.

Each forward orbit $\{{\sigma }^n(s):n\ge 0\}$ is countable. If there were only countably many nonperiodic forward orbits, their union together with the periodic points would be countable, contradicting the preceding argument. Hence there are uncountably many nonperiodic orbits, rather than merely uncountably many nonperiodic starting points.

One can also exclude eventual periodicity. An eventually periodic sequence is specified by a finite initial word and a finite repeating word, so only countably many exist. Removing all of them still leaves an uncountable set. This proves the assertion even if “nonperiodic” is understood to exclude orbits that eventually enter a periodic orbit.

For the left shift, ${\sigma }^p(s)=s$ precisely when the binary sequence is obtained by repeating its first $p$ symbols. There are exactly $2^p$ such sequences, although some have smaller least period. Hence the set of all periodic sequences is a countable union of finite sets and is at most countable. Passing from periodic points to periodic orbits cannot increase this cardinality.

To prove infinitude, for every integer $p\ge 1$ take the repeated word $(0^{p-1}1{)}^{\infty }$. Its least period is $p$, because its ones are separated by exactly $p$ positions; a smaller positive period would carry a one to a zero. Distinct least periods give distinct periodic orbits. Thus there are countably infinitely many periodic orbits.

The set $I^n$ consists precisely of points whose first $n$ positions $x,f(x),\dots ,f^{n-1}(x)$ all lie in $L\cup R$. HenceDeleting the first position preserves this condition, proving $f(\Lambda )\subseteq \Lambda$. For the reverse inclusion, if $y\in \Lambda$, take $x=g_L(y)$. Then $x\in L$ and $f^k(x)=f^{k-1}(y)\in L\cup R$ for every $k\ge 1$. Thus $x\in \Lambda$ and $f(x)=y$. In fact both inverse branches give preimages in $\Lambda$. Thereforewhich proves full invariance, not merely forward invariance.

For any infinite symbolic sequence $s$, the itinerary cylinders $I_{s_1\cdots s_n}$ are nonempty compact nested intervals. The nested interval theorem gives at least one point in their intersection. Every $x\in \Lambda$ has a unique itinerary because $L\cap R=\varnothing$. To justify that the itinerary also determines a unique point, it suffices to show the interval lengths shrink. The fact that logistic inverse branches contract in an angular coordinate works for every $\mu >4$; a direct claim that $|g_L^{\prime }|,|g_R^{\prime }|<1$ in the original coordinate would fail for some $\mu$ just above four.

Use the homeomorphism $h(u)={\sin }^2(\pi u/2)$ of $[0,1]$. The transformed inverse branches areTheir derivatives satisfyThe inequality follows by squaring and using $4/\mu <1$. Thus every length-$n$ transformed itinerary cylinder has length at most ${\mu }^{-n/2}$. Since $|h^{\prime }|\le \pi /2$, its original interval has length at most $(\pi /2){\mu }^{-n/2}$ and shrinks to a point. This gives the claimed one-to-one identification with the one-sided full shift.

Finally part (ii) givesThus $f$ acts by the left shift. The identification is a topological conjugacy: matching long prefixes forces closeness by the shrinking estimate, while distinct finite itinerary cylinders are disjoint compact sets and so their inverse coding is continuous. In particular $f$ need not be one-to-one on $\Lambda$, just as a one-sided left shift has two possible first symbols before any given tail.

When $a=0$, the explicit amplification factor becomesA single Fourier mode has constant modulus in space and is not in the l2 sequence space. Instead choose a whole interval of frequencies on which there is definite growth. LetThe inverse Fourier transform here gives the real sequenceBy the Parseval identity, its squared norm is $\Vert U^0{\Vert }_{{\ell }_2}^2=\Delta x|E|/(2\pi )=\Delta x/2$, finite and strictly positive. These data are also compatible with sampling a bounded continuous function: useThe value at zero is the limit of the displayed quotient, and $u_0(j\Delta x)=U_j^0$.

On $E$ we have ${\sin }^2\theta \ge 1/2$. Therefore the transformed solution and the Parseval identity giveFor each fixed pair $\Delta x,\Delta t>0$, takeTaking square roots proves $\Vert U^m{\Vert }_{{\ell }_2}\ge (1+c{)}^m\Vert U^0{\Vert }_{{\ell }_2}$ for every $m\ge 1$. In particular the norm already increases at the first step, so no positive mesh pair gives practical stability in the contraction sense of (b).

The native PDF places the existence of $c>0$ before the universal quantifier over mesh pairs. If this means a single positive $c$ independent of $\Delta x,\Delta t$, that assertion is false for nonzero data. Indeed, for every datum the same Fourier symbol givesFor any proposed fixed $c>0$, choose $0<\nu <\sqrt{(1+c{)}^2-1}$. The upper bound then contradicts the requested lower bound at $m=1$. The correct quantifiers are $\forall \Delta x,\Delta t>0\exists c=c(\Delta x,\Delta t)>0$, with nonzero square-summable data as constructed above.

Finally, failure of this stronger practical stability of a finite difference scheme should not be confused with every possible refinement notion of numerical stability. If $\Delta t/(\Delta x{)}^2\le K$, then $m\Delta t\le T$ givesThus a finite-time mesh-uniform bound can hold under this severe refinement even though contraction fails on every positive mesh. This is the distinction recorded in centered-advection refinement-dependent stability.

A sequence $x_n$ converges to $x$ if every open neighbourhood of $x$ contains all sufficiently late terms. A Hausdorff space separates any two distinct points by disjoint open neighbourhoods.

If a sequence had distinct limits $x,y$ in a Hausdorff space, choose disjoint open neighbourhoods $U$ of $x$ and $V$ of $y$. Convergence to each limit implies that all sufficiently late terms lie in both $U$ and $V$, contradicting $U\cap V=\varnothing$. Hence a convergent sequence in a Hausdorff space has a unique limit.

Use three nested sequences:For fixed $n,m$, letting $k\to \infty$ produces the point $(1/n+1/(n^{2}m),0)$. Letting $m\to \infty$ produces $(1/n,0)$, and letting $n\to \infty$ produces $(0,0)$. Accordingly, putThe successive derived sets areTo see that no limits are missing, take any convergent sequence of distinct points from $A_3$. If $n$ is unbounded, a subsequence has $n\to \infty$ and converges to the origin. Otherwise fix $n$ on a subsequence. Unbounded $m$ then forces limit $(1/n,0)$; with $n,m$ fixed, distinctness forces unbounded $k$, giving the corresponding point of $B$. The same argument, with one index fewer at each stage, proves the next two equalities. Distinct $n$-clusters are separated away from the origin because their first coordinates lie in the disjoint intervals $[1/n,1/n+1/n^2]$.

The original set has positive second coordinates, the first derived set has isolated $B$-points on the axis, the second has infinitely many $C$-points, and the third is a singleton. All three required successive inequalities follow. This is a three-stage accumulation tower.

In a metric space,The sequence may equivalently be required to have pairwise distinct terms. The metric hypothesis makes a countable sequence sufficient; arbitrary topological spaces need not have this sequential characterization.

A nonzero polynomial solution of degree $n$ has $a_n\ne 0$ but $a_{n+2}=0$. The recurrence therefore forces $2n+1-ϵ=0$. Since $ϵ=2E/(\hslash \omega )$,Conversely, for this value of $ϵ$, choose only the parity sequence containing degree $n$ and set the other initial coefficient to zero. The recurrence terminates at $n$, producing a nonzero polynomial times the decaying Gaussian. Thus every displayed energy occurs under the supplied normalizability criterion.

Differentiate the power series twice and align the coefficient of ${\xi }^k$. The equation becomesThusEven coefficients are generated from $a_0$, and odd coefficients from $a_1$; the two parity sequences are independent.

For each $n$, let $N_n$ be the null set where $f_n\ne g_n$. Their union $N$ is null. At any $x\notin N$, the two sequences agree term by term, so their pointwise limits are equal: $f(x)=g(x)$. Therefore $f\sim g$. Taking the countable union first is essential: it produces one exceptional set outside which all equalities hold simultaneously.

No such sequence exists. By Fatou's lemma on $(0,1)$,But the proposed nonnegative measurable integrand is positive at every point of $(0,1)$, so its integral cannot be zero. Explicitly, the sets where it is at least $1/k$ cover $(0,1)$; at least one has positive measure, forcing a positive integral. This contradicts the displayed bound.

For each integer $m\ge 1$, list the $m^2$ functionsin increasing $j$, and concatenate the blocks in increasing $m$. Each function has $L^1$ norm $1/m$, so the resulting sequence has norm tending to zero. Every $x\in (0,1)$ lies in exactly one interval of each block, where the function takes value $m$. ThereforeThis unbounded typewriter sequence with vanishing L1 norm is a moving-spike construction: the support shrinks fast enough to beat the increasing height.

If $A$ is closed in the complete ambient metric space, any Cauchy sequence in $A$ converges in $X$ by completeness, and its limit belongs to $A$ by part (i). Thus $A$ is complete with its restricted metric.

Conversely, suppose $A$ is complete and a sequence $a_n\in A$ converges in $X$ to $x$. A convergent sequence is Cauchy, also in the restricted metric, so completeness of $A$ gives $a_n\to a$ for some $a\in A$. Uniqueness of a metric limit gives $x=a\in A$. Part (i) now makes $A$ closed. Hence

A metric space is complete when every Cauchy sequence in it converges to a point of that space. A sequence is Cauchy when, for each $ϵ>0$, there is $N$ such that $d(x_m,x_n)<ϵ$ whenever $m,n\ge N$. The requirement that the limit lie in the space distinguishes completeness from having limits only in a larger ambient space.