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If $A_n$ are countable, choose enumerations and map each element of ${\bigcup }_n{A}_n$ to the first pair $(n,k)$ at which it occurs. Since ${\mathbb{N}}^2$ is countable, the union is countable.

A periodic function of period $k$ is determined by its $k$ values on a complete residue system. Thus all periodic functions form a countable union over $k$ of the countable sets ${\mathbb{N}}^k$, and are countable.

The set of bijections $\mathbb{N}\to \mathbb{N}$ is uncountable. Indeed, each binary sequence determines a bijection that swaps $2j-1$ and $2j$ exactly when its $j$th bit is one. This is an injection from the uncountable set of binary sequences.

A Cauchy sequence $(x_n)$ in a metric space $(M,d)$ satisfies: for every $\epsilon >0$ there is $N$ such that $d(x_m,x_n)<\epsilon$ whenever $m,n\ge N$. A complete metric space is one in which every Cauchy sequence converges to a point of the space.

Every Cauchy sequence is bounded. Choose $N$ such that $d(x_n,x_N)<1$ for $n\ge N$, and putThen every term lies in the ball $B(x_N,R)$.

Now suppose $M$ is complete and $(F_n)$ is a decreasing sequence of nonempty closed sets with $\mathrm{diam}{F}_n\to 0$. Choose $x_n\in F_n$. Given $\epsilon >0$, choose $N$ with $\mathrm{diam}{F}_N<\epsilon$. For $m,n\ge N$, both points lie in $F_N$, so $d(x_m,x_n)<\epsilon$. Completeness gives $x_n\to x\in M$. For each fixed $N$, the tail lies in the closed set $F_N$, hence $x\in F_N$. Therefore

Conversely, assume the nested-set property and let $(x_n)$ be Cauchy. DefineThese sets are nonempty, closed, and decreasing. The Cauchy property implies $\mathrm{diam}{F}_n\to 0$; taking a closure does not change the diameter. Choose $x\in {\bigcap }_n{F}_n$. Since $x_n,x\in F_n$,Thus every Cauchy sequence converges and $M$ is complete. This proves the Cantor intersection theorem characterization.

The contraction mapping theorem states that a contraction of a nonempty complete metric space has a unique fixed point.

For each $\lambda \in \Lambda$, the map $T_{\lambda }(x)=T(\lambda ,x)$ is a contraction with the common constant $k$, so it has a unique fixed point $x^{\ast }(\lambda )$. This defines the required unique function. Fix ${\lambda }_0$. The fixed-point identities and the triangle inequality giveConsequentlyThe numerator tends to zero as $\lambda \to {\lambda }_0$ by the assumed continuity for the fixed point $x^{\ast }({\lambda }_0)$. Hence $x^{\ast }$ is continuous, an instance of continuous dependence of the fixed point of a uniform contraction.

A sequence $f_n:S\to \mathbb{R}$ converges uniformly to $f$ if, for every $\epsilon >0$, there is $N$ such thatfor every $x\in S$ and every $n\ge N$. A map $h:M\to N$ is uniformly continuous if, for every $\epsilon >0$, there is $\delta >0$ such thatfor all $x,y\in M$.

If $f\in C_0({\mathbb{R}}^d)$, choose a ball outside which $|f|<1$. On the closed ball, continuity gives boundedness, so $f$ is bounded everywhere.

Now let $(f_n)$ be Cauchy in the uniform metric. For each $x$, $(f_n(x))$ is Cauchy in $\mathbb{R}$; let its limit be $f(x)$. Passing to the pointwise limit in the uniform Cauchy estimate shows that $f_n\to f$ uniformly. Hence $f$ is continuous. Given $\epsilon >0$, choose $n$ with $\Vert f-f_n{\Vert }_{\infty }<\epsilon /2$, and then choose $K$ so that $|f_n(x)|<\epsilon /2$ for $\Vert x\Vert >K$. It follows that $f$ also vanishes at infinity. Thus $C_0({\mathbb{R}}^d)$ is complete, as in completeness of continuous functions vanishing at infinity.

Every $f\in C_0({\mathbb{R}}^d)$ is uniformly continuous. Given $\epsilon >0$, choose $R$ so that $|f(x)|<\epsilon /2$ outside the ball of radius $R$. On the compact ball of radius $R+1$, $f$ is uniformly continuous; choose the corresponding $\delta \le 1$. If two points at distance below $\delta$ are not both in that ball, then both lie outside the ball of radius $R$, and their function values differ by less than $\epsilon$.

For the final sequence, continuity of $\epsilon$ at zero gives, for each fixed $x$,Thus pointwise convergence is compulsory. Uniform convergence need not hold: with $\epsilon (t)=t^2$,which is unbounded as a function of $x$ for every fixed $n$.

Under the additional bound $\epsilon (t)\le M|t|$, however,for $x\ne 0$, and the difference is zero at $x=0$. Hence convergence is uniform, by the uniform square-root perturbation under linear growth estimate. The pointwise answer remains yes.

For the sequence in part (iii), part (ii) givesTaking reciprocals yields the factorial-over-power series limit

Every convergent sequence is bounded. Conversely, let $(x_n)$ be increasing and bounded above, and set $L=\sup \{x_n:n\ge 1\}$. For any $\epsilon >0$, $L-\epsilon$ is not an upper bound, so some $x_N>L-\epsilon$. Monotonicity then givesfor all $n\ge N$. Hence $|x_n-L|<\epsilon$, proving the monotone bounded sequence criterion.

For a nonempty set $S\subset \mathbb{R}$ bounded above, $\sup S$ is its least upper bound: every $s\in S$ satisfies $s\le \sup S$, and every $u<\sup S$ fails to be an upper bound.

A sequence $x_n$ converges to $x$ if for every $\epsilon >0$ there is $N$ such that $n\ge N$ implies $|x_n-x|<\epsilon$.

A metric space is complete when every Cauchy sequence in it converges to a point of the space. Let $(y_n)$ be Cauchy in a closed subspace $Y$ of a complete metric space $X$. It is also Cauchy in $X$, so $y_n\to x$ for some $x\in X$. Since $Y$ is closed and every $y_n$ lies in $Y$, one has $x\in Y$. Thus the closed-subspace completeness theorem proves that $Y$ is complete.

The independent identically distributed sequence $(X_n)$ is itself a Markov chain, since the conditional law of $X_{n+1}$ is its common marginal law and does not depend on the past.

Also,Because $X_{n+1}$ is independent of the past, the conditional law of $S_{n+1}$ depends on the history only through $S_n$. Thus the partial sums $(S_n)$ are a Markov chain.

The upper register index is the largest register number occurring in an instruction of $P$, with value zero if only register $0$ is used. A configuration records the current state and the contents of every register up to that index.

For input $\vec{w}$, let $C(0,M,\vec{w})$ be the initial configuration: the designated initial state, the words of $\vec{w}$ in the input registers, and the empty word in every remaining register. Recursively, if $C(t,M,\vec{w})$ is halting, keep it fixed; otherwise let $C(t+1,M,\vec{w})$ be the unique configuration into which $M$ transforms it. The resulting sequenceis the computation sequence.

Consider computations while a terminal block of $b$'s is removed from register $0$. Until that block has disappeared, the available instruction and next state depend on the current state and on the visible $b$, not on the untouched prefix $x$.

Take a block longer than $|Q|$. Record the state after each successive $b$ has been removed. Two records have the same state $q$, by the pigeonhole principle. If the corresponding removed tail lengths are $k\ne \ell$, the same finite instruction sequences work in front of every prefix $x$. Thus there are fixed times $t,t^{\prime }$ such thatfor every word $x$.

The closest point theorem in a Hilbert space states that every nonempty closed convex subset $C$ of a Hilbert space has a unique point nearest to any given $x$.

Let $d={\inf }_{y\in C}\Vert x-y\Vert$ and choose $y_n\in C$ with $\Vert x-y_n\Vert \to d$. The parallelogram identity and convexity giveso $(y_n)$ is Cauchy. Completeness and closedness give a limit $y\in C$ with $\Vert x-y\Vert =d$. If $y,z$ were both minimizers, the same identity with their midpoint would force $\Vert y-z\Vert =0$.

Apply the theorem to a closed subspace $F$. For $x\in H$, let $y\in F$ be closest and put $z=x-y$. For every $u\in F$, minimality of $\Vert z-tu{\Vert }^2$ at $t=0$ for real and imaginary $t$ gives $⟨z,u⟩=0$. HenceThe intersection is zero, so $H=F\oplus F^{\perp }$.

If $T$ is a shift with orthonormal basis $(e_n)$, it is an isometry,Conversely, suppose these three properties hold. Choose a unit vector $e_1$ spanning $(\mathrm{Im}T{)}^{\perp }$ and set $e_n=T^{n-1}{e}_1$. Isometry makes this sequence orthonormal. Iterating the orthogonal decompositiongivesA vector orthogonal to every $e_n$ lies in every $\mathrm{Im}(T^n)$ and is therefore zero. Thus $(e_n)$ is an orthonormal basis and $T{e}_n=e_{n+1}$, so $T$ is a shift. This is the wandering-vector characterization of a unilateral shift.

Expanding the square and using orthogonality of Fourier modes,Finite truncations justify the calculation directly, and passage to the ${\ell }^2$ limit proves Parseval identity for every square-summable sequence.

False. The interval $(0,1)$ with the Euclidean metric is uncountable and has no isolated points, but the Cauchy sequence $1/n$ has no limit in the space.

Sincewe haveThus $a_n=1$ for every $n\ge 0$.

Explicitly,For $n\ge 1$, the remainder after terms $0,\dots ,n-1$ isIts ratio to ${\varphi }_n$ tends to one, verifying every coefficient formula.

Because $e^{-x}=o(x^{-N})$ for every $N$, the exponentially small part is invisible to the power sequence ${\psi }_n=x^{-n}$. Hencewith coefficient one at ${\psi }_1$ and all other power coefficients zero.