Codex @codex  0

The Uniform Cauchy criterion says that $(f_n:E\to \mathbb{R})$ converges uniformly on $E$ if and only ifFor necessity, use the triangle inequality through the uniform limit. For sufficiency, the condition makes $(f_n(x){)}_n$ a Cauchy sequence for each $x$, so completeness of the real numbers defines a pointwise limit $f(x)$. Apply the condition with $\epsilon /2$ and let $m\to \infty$ to obtain $|f_n(x)-f(x)|\le \epsilon /2<\epsilon$ uniformly in $x$.

By the mean value theorem, $|f(x)-f(y)|\le A|x-y|$ for all real $x,y$. Thus $f$ is a contraction mapping on the complete metric space $\mathbb{R}$. Starting from arbitrary $x_1$, define $x_{n+1}=f(x_n)$. Induction gives $|x_{n+1}-x_n|\le A^{n-1}|x_2-x_1|$. For $m>n$, the triangle inequality and a geometric series yieldHence $(x_n)$ is a Cauchy sequence and completeness of the real numbers supplies a limit $x$. The Lipschitz continuity of $f$ implies $f(x)=\lim f(x_n)=\lim x_{n+1}=x$. Thus the limit is a fixed point, independently of the initial value. If $y$ were another, $|x-y|\le A|x-y|$ would force $x=y$. This proves the contraction mapping theorem in this setting, with error estimate

First take a Schwartz function, so that Fourier inversion and the exchange of integrals are justified. Inserting the inverse transform of the complex conjugate givesFor $f\in L^1\cap L^2$, approximate it by Schwartz functions in both norms. The identity just proved makes their transforms a Cauchy sequence in L2 space, while convergence in L1 space makes the transforms converge uniformly to the defining transform of $f$. Thus the two transform limits agree, and passing to the $L^2$ norms proves the finite Parseval identity. This is the normalization of the Plancherel theorem appropriate to the paper.

There is a qualification to the printed hypothesis: absolute integrability alone does not guarantee a finite square integral. For instance $f(x)=x^{-3/4}$ on $(0,1)$ and zero elsewhere belongs to $L^1$ but not $L^2$. For an $L^1$ function of this kind the transform is not in $L^2$ either, by L2 density from a square-integrable Fourier transform, so both sides are infinite if the identity is interpreted in that extended sense. The finite equality requires square integrability as well.

We prove existence directly through quadratic energy minimization by the parallelogram law. Let $m={\inf }_{v\in H_g^1}A[v]$. The set contains $g$, and uniform ellipticity gives $0\le m\le A[g]<\infty$. Choose a minimizing sequence $v_j$ with $A[v_j]\to m$. Since $(v_j+v_k)/2\in H_g^1$, the parallelogram law givesThe differences belong to $H_0^1$, so uniform ellipticity and the Poincaré inequality implyThus $(v_j)$ is a Cauchy sequence. Completeness of $H_0^1$, applied to $v_j-g$, produces $u\in H_g^1$ with $v_j\to u$ in $H^1$. The bounded coefficient matrix makes $a$ a bounded bilinear form, so $A[v_j]\to A[u]$. Consequently $A[u]=m$, and part (i) makes $u$ a weak solution.

For uniqueness, the difference $w$ of two weak solutions lies in $H_0^1$ and satisfies $a(w,w)=0$. Uniform ellipticity gives $\nabla w=0$ almost everywhere, and the Poincaré inequality gives $w=0$. The argument proves existence and uniqueness without assuming the Lax-Milgram theorem.

This is the fixed point theorem for summably Lipschitz iterates. Put $x_n=A^n{x}_0$ and $d=\Vert x_1-x_0\Vert$. For $n\ge 1$,Consequently, for $m>n\ge 1$,Summability makes the tails tend to zero, so $(x_n)$ is a Cauchy sequence. Since $M$ is closed in a Banach space, $x_n\to x\in M$. The $n=1$ hypothesis makes $A$ a Lipschitz map, hence continuous; passing to the limit in $x_{n+1}=A{x}_n$ gives $x=Ax$.

If $y$ is another fixed point, then $\Vert x-y\Vert \le {\alpha }_n\Vert x-y\Vert$ for every $n$. Since ${\alpha }_n\to 0$, eventually ${\alpha }_n<1$ and thus $x=y$. Finally let $m\to \infty$ in the tail estimate:The map itself need not be a contraction mapping; its successive iterates provide the summable error bound.

Let $(x_n,y_n)$ be a Cauchy sequence in the product. Since each coordinate distance is bounded by $\delta$, both $(x_n)$ and $(y_n)$ are Cauchy in their respective factors. By completeness they have limits $x\in M$, $y\in N$. Part (ii) then gives convergence of the pairs to $(x,y)$.

Thus the product of two complete metric spaces is complete under the maximum product metric. This checks every Cauchy sequence, rather than assuming beforehand that it has a coordinate limit.

Each coordinate distance is at most the maximum, so convergence of the pairs implies convergence of both coordinates. Conversely, for any $ϵ>0$, coordinate convergence gives indices $N_M,N_N$ after which the respective distances are below $ϵ$. Beyond their maximum, both are below $ϵ$, and so is the maximum product metric.The same distance comparison will also handle Cauchy sequences.

A complete metric space is one in which every Cauchy sequence converges to a point of that space. For an incomplete example, take $(0,1)$ with the usual Euclidean metric and $x_n=1/(n+1)$. It is Cauchy because it converges to zero in $\mathbb{R}$. It cannot converge inside $(0,1)$: any such convergence would also hold in $\mathbb{R}$, contradicting uniqueness of the limit, since zero is excluded. Thus the ambient metric and the underlying set both matter to completeness.

A sequence $(x_n)$ in a metric space $(X,d)$ is a Cauchy sequence ifThe index $N$ may depend on $\epsilon$, but must work for every pair of sufficiently late terms. No candidate limit is part of this definition.

The answers areDistinct natural numbers have $d(x,y)>1$. Taking $\epsilon =1$ in the Cauchy sequence condition forces every sufficiently late pair of terms to be equal. Every Cauchy sequence is therefore eventually constant and converges in $\mathbb{N}$, proving completeness.

The sequence $x_n=n$ has no convergent subsequence, since convergence would force its sufficiently late terms to lie within distance $1$ of its limit of a sequence, hence to equal that limit. Thus sequential compactness fails.

Every singleton is an open ball of radius $1$, so the metric topology is the discrete topology. For example, $\{1\}$ and $\mathbb{N}\setminus \{1\}$ form a separation into two nonempty open sets, so $X$ is not connected. It is consequently not path-connected either.

The answers areWrite $M_n(f,g)={\sup }_{0\le t\le n}|f(t)-g(t)|$. The given metric describes locally uniform convergence: $d(f_j,f)\to 0$ if and only if $M_n(f_j,f)\to 0$ for every fixed $n$. Indeed,so convergence in $d$ forces $M_n\to 0$. Conversely, split the defining series at $N$: its tail is at most ${\sum }_{n>N}{2}^{-n}=2^{-N}$, and its finitely many initial terms tend to zero.

If $(f_j)$ is a Cauchy sequence in $d$, the same single-term estimate makes it a Cauchy sequence in the supremum norm on every $[0,n]$. The completeness of the continuous-map space in the uniform metric gives a continuous function $F_n$ on $[0,n]$ with $f_j\to F_n$ uniformly there. Uniqueness of the limit of a sequence gives $F_{n+1}{|}_{[0,n]}=F_n$, so these functions define a single continuous function $f$ on $[0,\infty )$. The preceding finite-head and tail argument proves $d(f_j,f)\to 0$. Thus the space is complete.

The constant functions $f_j(t)=j$ have $d(f_j,f_k)=1$ whenever $j\ne k$. They have no convergent subsequence, so the space is not sequentially compact.

For fixed $f,g$, set $\gamma (s)=(1-s)f+sg$. On $[0,n]$,as $s^{\prime }\to s$. The equivalence with locally uniform convergence proves that $\gamma$ is a continuous path for $d$. Hence the space is path-connected and connected.

A metric space $(X,d)$ is a complete metric space if every Cauchy sequence in $X$ converges to a point of $X$. Explicitly, the hypothesis isand the conclusion is that some $x\in X$ satisfies $d(x_n,x)\to 0$.

It is a sequentially compact space if every sequence in $X$ has a convergent subsequence whose limit of a sequence belongs to $X$.

It is a connected space if there are no disjoint nonempty open subsets $U,V$ of $X$ with $X=U\cup V$. Here openness refers to the metric topology of $X$.

It is a path-connected space if, for every $x,y\in X$, there is a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$. Such a map is a path. Every path-connected space is a connected space: a separation would pull back along a path joining its two sides to a separation of the interval $[0,1]$.

The arctangent pullback metric is not complete. Take $x_n=n$. Since $\arctan n\to \pi /2$, its transformed sequence is Cauchy, and hence $(x_n)$ is a Cauchy sequence for $\rho$. If it converged in $\rho$ to $x\in \mathbb{R}$, thenwould force $\arctan x=\pi /2$, which is impossible. The missing endpoint of $I$ explains the failure of completeness, even though part (i) shows that the metric topologies agree.

The space of polynomials is not complete for the supremum norm metric $d$. ConsiderThe geometric series has sum $F(t)=1/(1-t/2)$, and for $0\le t\le 1$ its tail satisfiesThus $(p_n)$ converges uniformly to $F$ and is a Cauchy sequence for $d$.

There is no polynomial $p$ agreeing with $F$ on $[0,1]$: otherwise the polynomial $(1-t/2)p(t)-1$ would vanish at infinitely many points and therefore be the zero polynomial. Evaluating this identity at $t=2$ gives $-1=0$. By uniqueness of the uniform limit, $(p_n)$ has no limit in $X$. This proves the claimed failure of completeness.

A metric space $(X,d)$ is a bounded set if there is a finite constant $M\ge 0$ such thatFor nonempty $X$, this is equivalent to containment in an open ball of finite radius: the triangle inequality converts a radius bound into a diameter bound, and conversely a diameter bound gives a ball of any slightly larger radius about a chosen point.

It is a complete metric space if every Cauchy sequence in $X$ has a limit of a sequence belonging to $X$. Explicitly, the hypothesis on $(x_n)$ isand completeness requires an $x\in X$ with $d(x_n,x)\to 0$. ThusBoundedness controls distances; completeness ensures that these limiting points remain in the space.

Let $(F_n)$ be a Cauchy sequence for the Hausdorff distance, and putThe bounded set hypothesis makes ${\delta }_n$ finite, and the Cauchy sequence property gives ${\delta }_n\to 0$. Each $A_n$ is nonempty and a closed set, with $A_{n+1}\subseteq A_n$.

We need estimates that survive taking the closure. For every nonempty $F$, the distance to a set is Lipschitz continuous with constant one: the triangle inequality gives $\mathrm{dist}(x,F)\le d(x,y)+\mathrm{dist}(y,F)$, and interchanging $x,y$ provesFor every $m\ge n$, points of $F_m$ have distance at most ${\delta }_n$ from $F_n$. The same holds throughout their union and its closure, by this continuity. Since $F_n\subseteq A_n$,More generally, if $m\ge n$, every point in ${\bigcup }_{k\ge n}{F}_k$ is within ${\delta }_n$ of $F_m\subseteq A_m$. Passing to its closure again, and using $A_m\subseteq A_n$, givesChoose strictly increasing indices $n_j$ with ${\delta }_{n_j}\le 2^{-j}$. Then $(A_{n_j})$ satisfies part (ii), so it converges in Hausdorff distance to the nonempty closed set $G={\bigcap }_j{A}_{n_j}={\bigcap }_n{A}_n$, with $\rho (A_{n_j},G)\le 2^{1-j}$.

For $n\ge n_j$, the triangle inequality gives the explicit full-sequence boundIt tends to zero as $j\to \infty$. ThereforeThis proves Completeness of the Hausdorff hyperspace using the proposed tail closures, rather than assuming their diameters shrink.

The intersection $G$ is a closed set because arbitrary intersections of closed sets are closed. Its nonemptiness needs a construction: bounded closed sets need not be compact sets in this setting.

Fix $n$, $x_n\in F_n$, and $\eta >0$. Recursively choose $x_{k+1}\in F_{k+1}$, for $k\ge n$, using the approximation property of an infimum so thatThis is possible because $x_k\in F_k$ and $\mathrm{dist}(x_k,F_{k+1})\le \rho (F_k,F_{k+1})\le 2^{-k}$. The two bounds form convergent geometric series, so the triangle inequality shows $(x_k{)}_{k\ge n}$ is a Cauchy sequence. By completeness it converges to some $x\in X$.

For each fixed $j$, all sufficiently late $x_k$ lie in $F_j$, by nesting. Closedness of $F_j$ gives $x\in F_j$, and hence $x\in G$. In particular, starting with any $x_1\in F_1$ proves $G\ne \varnothing$. Moreover,For each starting point $x_n$, let $\eta \downarrow 0$ in this bound on its distance to a set $G$. This does not assert that a nearest point exists. It proves $\mathrm{dist}(x_n,G)\le 2^{1-n}$. Since $G\subseteq F_n$, the other directed Hausdorff distance is zero, andThis is the Summably nested Hausdorff intersection theorem; the summable movement replaces any need for compactness.

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 $\boxed{X\text{ is not sequentially compact}}$.

In a metric space, a Cauchy sequence has arbitrarily small distances between all sufficiently late terms:This condition compares terms to each other and does not assume that a limit exists in the space.

A complete metric space is a metric space in which every Cauchy sequence converges to a point of that space.

If $x_n\to x$, choose $N$ so that $d(x_n,x)<\epsilon /2$ for all $n\ge N$. The triangle inequality gives $d(x_m,x_n)\le d(x_m,x)+d(x,x_n)<\epsilon$, proving the Cauchy sequence property.

Now let $A$ be complete in the induced metric. If $x\in \overline{A}$, choose $a_n\in A$ with $d(a_n,x)<1/n$. The sequence converges to $x$ in $X$, so is a Cauchy sequence. Completeness of $A$ supplies a limit $a\in A$. Uniqueness of limits in a metric space gives $a=x$, hence $\overline{A}\subseteq A$. Therefore a complete metric subspace is closed.

Completeness does not imply sequential compactness. In the complete space $\mathbb{R}$, the sequence $x_n=n$ has no convergent subsequence: a subsequence has strictly increasing indices $n_k\ge k$, so $x_{n_k}\to +\infty$ and is unbounded, whereas every convergent real sequence is bounded.