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This is the unilateral right shift. It is not compact because the orthonormal sequence $(T{e}_n)=(e_{n+1})$ has no convergent subsequence. Its norm is one. For $|\lambda |>1$ the Neumann series gives a resolvent; for $|\lambda |<1$, $\bar{\lambda }$ is an eigenvalue of the adjoint left shift, so $\lambda$ lies in the spectrum of $T$. Closedness supplies the boundary. Hence

Write $Y_n=n^{-1}{\sum }_{i=1}^n|X_i|$. Truncate each $|X_i|$ at a level $K$. The average of the truncated parts is at most $K$, while the expected average of the tails isindependently of $n$. Markov's inequality then shows that $(Y_n)$ is uniformly integrable as $K\to \infty$. Since $|S_n|/n\le Y_n$, the sequence $(S_n/n)$ is uniformly integrable.

Convergence in probability to the constant $\mathbb{E}{X}_1$, together with uniform integrability, implies convergence of first absolute moments by the Vitali convergence theorem. Hence

Choose a subsequence $a_{n_j}$ so sparse thatfor every $j$; this is possible because the tails of the convergent positive series tend to zero. Distinct binary sequences $({\epsilon }_j)$ then have distinct sums ${\sum }_j{\epsilon }_j{a}_{n_j}$: at their first disagreement, the corresponding term exceeds the entire remaining tail. Thus these subsums form an uncountable set. The algebraic numbers are countable, so at least one subsum is transcendental. Set ${\theta }_{n_j}={\epsilon }_j$ and all other ${\theta }_n=0$.

A sequence $(a_n)$ is Cauchy when, for every $\epsilon >0$, some $N$ satisfies $|a_m-a_n|<\epsilon$ whenever $m,n\ge N$.

The general principle of convergence says that a real sequence converges exactly when it is Cauchy. Convergent sequences are Cauchy by the triangle inequality. Conversely, a Cauchy sequence is bounded, so Bolzano–Weierstrass gives a convergent subsequence $a_{n_k}\to a$; the Cauchy property then forces the entire sequence to converge to $a$.

False. The sequence $a_n=(-1{)}^n/n$ converges to zero, but $|a_{n+1}-a_n|=1/n+1/(n+1)$, whose series diverges.

Every reachable sequence is a restricted-growth sequence in $A_n$. Given one, the transition masses sum toand every allowed transition remains in $A_{n+1}$. Induction from $X_1=1$ therefore proves that $(X_1,\dots ,X_n)\in A_n$ almost surely.

A set is countable when it is finite or admits a bijection with a subset of $\mathbb{N}$. For each degree, the coefficient tuples for polynomials in $A[x]$ form a finite Cartesian power of the countable set $A$; a countable union over degrees is countable. In particular the algebraic numbers, being roots of countably many integer polynomials with finitely many roots each, are countable. Since $\mathbb{R}$ is uncountable, uncountably many real numbers are transcendental.

The set $Q$ is uncountable. Partition $N$ into four-element blocks and, independently on each block, choose either of two fixed-point-free permutations. Binary sequences then inject into $Q$.

Finally choose a line $L$ distinct from and nonparallel to every $L_i$, possible because only countably many directions are excluded. Each $L\cap L_i$ has at most one point, so the union covers only countably many points of the uncountable line $L$. It cannot cover the plane.

Uniform continuity means that for every $\epsilon >0$ there is a single $\delta >0$ such that $d(x,y)<\delta$ implies $|f(x)-f(y)|<\epsilon$ for all $x,y$. A $d^{\prime }$-Cauchy sequence converges uniformly pointwise to a bounded function; passing a uniform-continuity estimate through a sufficiently close member proves that the limit is uniformly continuous. Thus $C_{b,u}(X)$ is complete.

If $f\in C_0({\mathbb{R}}^n)$, decay at infinity and compactness of a ball make it bounded. Uniform continuity on a sufficiently large compact ball, together with small values outside it, proves global uniform continuity. Hence $C_0\subset C_{b,u}$.

A uniform limit of functions vanishing at infinity also vanishes at infinity, so $C_0$ is closed. It is not compact: translate a fixed compactly supported bump of height one so that the supports are disjoint. The resulting sequence has pairwise sup distance one and no convergent subsequence.

The product topology has basis $U\times V$ with $U$ open in $X$ and $V$ open in $Y$. Since ${\Pi }_X^{-1}(U)=U\times Y$ and similarly for ${\Pi }_Y$, both projections are continuous.

If $Y$ is Hausdorff, a point $(y_1,y_2)$ off the diagonal has disjoint product neighborhoods, so the diagonal is closed. Conversely, if the diagonal is closed, its complement supplies disjoint neighborhoods of any two distinct points.

The graph is the inverse image of the diagonal under $(x,y)\mapsto (f(x),y)$, so a continuous map into a Hausdorff space has closed graph. Conversely, if $X,Y$ are compact Hausdorff and $C\subseteq Y$ is closed, then ${\Gamma }_f\cap (X\times C)$ is compact. Its projection onto $X$ is $f^{-1}(C)$, compact and hence closed. Thus $f$ is continuous.

The function $f(0)=0$, $f(x)=1/x$ for $x\ne 0$ is discontinuous but has closed graph: a convergent graph sequence approaching $x=0$ cannot have $x\ne 0$ and bounded second coordinate unless it eventually reaches the isolated graph point $(0,0)$.

The simplicial Mayer–Vietoris theorem says that if a simplicial complex $X=M\cup N$ for subcomplexes $M,N$, there is a natural long exact sequenceThe reduced version extends through dimension zero.

For every simplex $\sigma$ of $K$, all faces of $\sigma \cup \{c_i\}$ are either faces of $\sigma$ or have the form $\tau \cup \{c_i\}$ with $\tau$ a face of $\sigma$, so they lie in $L$. Each $c_i\ast K$ is a cone and hence a simplicial complex. For distinct $i,j$, the rays in the last two coordinates through $c_i$ and $c_j$ are not positive multiples. Thus the two cones meet exactly in $K$, so their simplices have common faces only in $K$. Hence $L$ is a simplicial complex.

Topologically, $L$ is the union of three cones on $K$ along their common base. The first two cones form the suspension $\Sigma K$. Attaching the third contractible cone along the equatorial copy of $K$ givesThis also follows by applying reduced Mayer--Vietoris twice: every cone has zero reduced homology, and the two inclusion maps from the common base are null-homotopic inside the cones. Thereforefor every $n$, with the convention that negative reduced homology vanishes. Since $K$ is nonempty, $L$ is connected, so explicitly

Let $1<p<\infty$ and $q=p/(p-1)$. Every $y\in {\ell }^q$ definesHölder's inequality gives $\Vert f_y\Vert \le \Vert y{\Vert }_q$, and equality follows by using a normalized sequence proportional to $\mathrm{sgn}(y_n)|y_n{|}^{q-1}$.

Conversely, for $f\in ({\ell }^p{)}^{\ast }$ put $y_n=f(e_n)$. On the first $N$ coordinates, choose the same norming vector to obtainLetting $N\to \infty$ shows $y\in {\ell }^q$. On finitely supported sequences, linearity gives $f=f_y$; density of those sequences in ${\ell }^p$ extends this identity to all $x$. Thereforeisometrically.

Yes. For $x=(x_n)\in c$, let $L={\lim }_n{x}_n$ and defineThe tail tends to zero, so $T(x)\in c_0$, and $\Vert T(x){\Vert }_{\infty }\le 2\Vert x{\Vert }_{\infty }$. Conversely, for $y=(y_n)\in c_0$, defineThis sequence converges to $y_1$ and has norm at most $2\Vert y{\Vert }_{\infty }$. The formulas are inverse bounded linear maps, so $c$ and $c_0$ are isomorphic Banach spaces.

Every bounded sequence in $\mathbb{R}$ has a convergent subsequence.

A neighbourhood of $x$ contains an open set containing $x$; $x_n\to x$ when every neighbourhood eventually contains every $x_n$. If two metrics have the same topology they plainly have the same convergent sequences. Conversely, if a $d$-open set $U$ were not $d\prime$-open, some $x\in U$ would have points $x_n\notin U$ with $d\prime (x_n,x)<1/n$; the common-sequence assumption would imply $d(x_n,x)\to 0$, contradicting $d$-openness. Symmetry finishes the proof. This fails generally: on an uncountable set the discrete and cocountable topologies differ, but in both every convergent sequence is eventually constant.

A base covers the space and refines intersections. On $X^k$, products of opens form a base. The weighted metric satisfies the metric axioms and has exactly this topology; convergence is coordinatewise. For $X^{\mathbb{N}}$ the metric $d(f,g)={\sum }_{n\ge 1}{2}^{-n}\min \{1,\rho (f(n),g(n))\}$ induces the stated product topology: finitely many coordinates control a basic neighbourhood and the tail is uniformly small. Thus sequences converge exactly coordinatewise.

Bounded means finite diameter; sequential compactness means every sequence has a convergent subsequence. A failure of uniform continuity supplies two close sequences whose image distances stay apart; a convergent subsequence contradicts continuity. Into a discrete metric, continuity is local constancy, and compactness upgrades the radii uniformly. The Cantor set is closed in compact $[0,1]$, hence sequentially compact. Uniform local constancy gives a positive separation scale, so finitely many initial Cantor-code (binary) digits determine $f$. Removing zero destroys compactness: the hinted function reading the digit after the first $1$ is continuous at every remaining sequence but depends on arbitrarily late digits.

$H$ is a closed linear subspace of reflexive $L^p$. A minimizing sequence is bounded by the triangle inequality. Reflexivity supplies a weakly convergent subsequence with limit $f_0\in H$; each norm $\Vert f-g_i{\Vert }_p$ is weakly lower semicontinuous, so their sum attains the infimum at $f_0$.

Again the series converges absolutely. $T=2{\sum }_j{B}_j/3^j$ has the Cantor distribution. Every particular digit sequence has probability zero, so it has no atoms; its support is the Cantor set, which has Lebesgue measure zero, so it cannot have a Lebesgue density.

Choose the sequenceAfter every index there are positive terms arbitrarily large and negative terms arbitrarily small. Thus every term has both a larger later term and a smaller later term. The former excludes a peak of a sequence $(b_n)$, and the latter excludes a peak of a sequence $(-b_n)$.