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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.

The Baire category theorem states that every countable intersection of open dense subsets of a complete metric space is dense. Equivalently, no nonempty open subset of a complete metric space is a countable union of nowhere dense sets.

To prove the first form, let $G_1,G_2,\dots$ be open dense subsets of a complete metric space $X$, and let $V\subseteq X$ be nonempty and open. Choose a closed ballInductively, density and openness of $G_{n+1}$ allow a closed ballThe balls are nested and $d(x_m,x_n)<2^{-n}$ for $m>n$, so $(x_n)$ is Cauchy. Completeness gives $x_n\to x\in X$. For every $n$, all later centres lie in $\overline{B}(x_n,r_n)$; closedness gives $x\in \overline{B}(x_n,r_n)\subseteq G_n$. The first ball also lies in $V$, henceSince every nonempty open $V$ meets the intersection, that intersection is dense.

For $n\ge 0$, putEach $S_n$ is closed, the hypothesis gives $\mathbb{R}={\bigcup }_n{S}_n$, and $E_n\subset E_{n+1}$ because a function that vanishes on an open set has derivative zero there.

The open setis dense. Indeed, inside any nonempty open interval choose a nondegenerate closed subinterval $J$. The complete metric space $J$ is covered by the closed sets $J\cap S_n$, so the Baire's theorem makes one of them contain a relative open interval, which lies in some $E_n$.

On each connected component $I$ of $\Omega$, the function $F$ is one polynomial. To see this, any compact subinterval $K\subset I$ is covered by the increasing family $(E_n)$. A finite subcover therefore gives $K\subset E_N$ for some $N$, and $F^{(N)}=0$ on $K$. Hence $F$ is polynomial on $K$; overlapping compact intervals force these polynomials to agree throughout $I$.

Let $D=\mathbb{R}\setminus \Omega$. This closed set has no isolated points. Otherwise, on the two sides of an isolated point $x$, $F$ would equal polynomials. Smoothness makes all their one-sided derivatives agree at $x$, so the two polynomials are identical and extend across $x$. Some derivative would then vanish on a neighborhood of $x$, contrary to $x\in D$.

Suppose that $D$ is nonempty. It is complete andis a countable closed cover. A second application of Baire's theorem gives an index $k$ and an open interval $U_0$ such thatChoose $x_0\in D\cap U_0$ and a bounded open interval $U$ centred at $x_0$ whose closure lies in $U_0$.

Because $D$ has no isolated points, every $x\in D\cap U$ is approached by distinct points of $D\cap U$. Difference quotients first give $F^{(k+1)}(x)=0$, and induction gives

No component of $\Omega$ meeting $U$ can cross $x_0$, so it has an endpoint in $D\cap U$. If its polynomial had degree $d\ge k$, then its $d$th derivative would approach a nonzero constant at that endpoint, contradicting the preceding display. Its degree is therefore less than $k$, so $F^{(k)}$ also vanishes on the part of that component in $U$. We conclude that $F^{(k)}=0$ throughout $U$, which says $U\subset E_k\subset \Omega$ and contradicts $D\cap U\ne \varnothing$.

The pointwise inequality $\hat{g}(X,X)\ge g(X,X)$ impliesEvery $d_{\hat{g}}$-Cauchy sequence $(x_j)$ is therefore $d_g$-Cauchy. Since $g$ is geodesically complete, the Hopf-Rinow theorem makes $(M,d_g)$ a complete metric space, so $x_j\to x$ in $d_g$ for some $x\in M$.

On a coordinate neighbourhood with compact closure around $x$, smooth positive-definite Riemannian metrics are uniformly equivalent. Thus there is $C>0$ such thatthere. For all sufficiently large $j$, a short $g$-geodesic from $x$ to $x_j$ stays in this neighbourhood, and henceThus $(M,d_{\hat{g}})$ is complete. Another application of Hopf-Rinow shows that $\hat{g}$ is geodesically complete.

Let $T_N={\sum }_{n=1}^N|a_n|$ and $S_N={\sum }_{n=1}^N{a}_n$. If $(T_N)$ converges, it is a Cauchy sequence. Thus, for every $\epsilon >0$, some $N$ makeswhere the triangle inequality gives the bound. Therefore $(S_N)$ is a Cauchy sequence in the complex numbers. These form a complete metric space, so $(S_N)$ has a finite limit of a sequence. Equivalently, apply the Cauchy criterion for series. Hence absolute convergence implies convergence.

The polynomial factors as $p(x)=(3x-2{)}^2(3x+1)$, so $\alpha =2/3$ is a multiple root of multiplicity two. The Newton root-finding iteration isFor $x\ne 0,2/3$, cancelling a factor givesThis expression extends continuously to $g(2/3)=2/3$. Let $I=[1/3,1]$. On this closed interval,Thus $g$ is increasing, andBy the mean value theorem, $|g(x)-g(y)|\le (16/27)|x-y|$ for all $x,y\in I$. The contraction mapping theorem states that a contraction mapping from a nonempty complete metric space into itself has a unique fixed point, and every fixed-point iteration converges to it. The closed interval $I$ with ordinary Euclidean distance is complete, and $g(2/3)=2/3$, soIn fact $|x_k-2/3|\le (16/27{)}^k|x_0-2/3|$. The Newton iteration at a multiple root here has linear convergence, since $g^{\prime }(2/3)=1/2$. Thus the convergence assertion is true for the simplified fixed-point iteration, and for the usual Newton root-finding iteration that stops upon reaching an exact root of a polynomial. If one insists on applying the unsimplified quotient at every step, it is undefined for the allowed starting value $x_0=2/3$, since both $p$ and $p^{\prime }$ vanish there; that literal version of an infinite sequence requires the stated stopping convention or continuous extension.

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

Banach's fixed-point theorem states that a map $T:Y\to Y$ on a nonempty complete metric space with $d(Tx,Ty)\le qd(x,y)$ for a constant $0\le q<1$ has a unique fixed point, approached from every starting point by iteration.

Choose $x_0$ and let $x_{n+1}=T{x}_n$. Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. For $m>n$, the triangle inequality and geometric series giveThus the sequence is Cauchy and converges by completeness to some $x_{\ast }$. The contraction estimate implies continuity, so $T{x}_{\ast }=\lim T{x}_n=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ were another fixed point, $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$ would force equality of the points. Passing $m\to \infty$ also yields the quantitative error estimate

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.

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.

The answers, in the order completeness, sequential compactness, connectedness, and path connectedness, areDefine $F(0)=0$ and $F(x)=\sqrt{x}\sin (1/x)$ for $x>0$. The estimate $|F(x)|\le \sqrt{x}$ proves continuity at zero by the squeeze theorem; elsewhere it follows from the usual continuous functions. Thus $X$ is the graph of a function $F$ on the closed set $[0,\infty )$.

To check closedness directly, if $(x_n,F(x_n))\to (x,y)$ in the Euclidean plane, then $x\ge 0$ and continuity gives $y=F(x)$. Therefore $X$ is a closed subset of the complete metric space ${\mathbb{R}}^2$, hence is a complete metric space. The sequence $(n,F(n))$ has no convergent subsequence, since its first Cartesian coordinates tends to infinity, so sequential compactness fails.

For any two points $(u,F(u)),(v,F(v))$, the mapis a continuous path in $X$ joining them. Thus $X$ is path-connected, and consequently connected. The factor $\sqrt{x}$ is crucial: the oscillations shrink to the single point $(0,0)$.

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 $[0,1]$.

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.

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.

A metric space is complete if every Cauchy sequence in it converges to a point of that same space. A limit in a larger ambient completion is insufficient for this definition.

Picard iteration replaces a tentative trajectory by the accumulated integral of its proposed rate. For an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting function byHere $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A fixed point of $\mathcal{P}$ solves this integral equation, and the fundamental theorem of calculus gives the original ordinary differential equation.

For $y^{\prime }=y$, $y(0)=1$, successive integrals giveAt $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx 2.71828183$. The fourth polynomial misses by about $0.00994850$. On $0\le t\le 1$, every omitted term of the exponential series is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving uniform convergence to the exponential function on this interval.

The beauty of Picard iteration is that each proposed trajectory supplies a better rate history for the next one. For this example, one integration adds exactly the next term of the exponential series. In general, the integrals may be difficult to compute, but the construction still explains local existence and uniqueness.

The short contraction mapping argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$ on whichThe second bound is the uniform local Lipschitz condition. Set $J=[t_0-h,t_0+h]$ with $0<h\le a$, $Mh\le r$ and $Lh<1$. In the function space of continuous functions $J\to {\mathbb{R}}^d$, use the supremum norm $\Vert u{\Vert }_{\infty }={\sup }_{t\in J}\Vert u(t)\Vert$. On its closed ball $\Vert u-y_0{\Vert }_{\infty }\le r$, direct integral estimates giveThus the map preserves the closed ball and is a contraction mapping there. This closed ball is a complete metric space, so the contraction mapping theorem gives convergence to exactly one fixed point. Any other local solution of a differential equation stays in the same ball after shortening its interval, and hence agrees there.

For $y^{\prime }=y$, take $r=1$, so $|y|\le 2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx 0.64872127<1$. The plotted interval $[0,1]$ is longer than this simple guarantee: the explicitly computed exponential series converges there anyway. The condition $Lh<1$ is sufficient for this one-step contraction mapping argument, rather than a necessary lifespan bound.

The p-adic integers form a complete metric space and a discrete valuation ring. Every nonzero $x$ is uniquely $p^{k}u$ with $k=v_p(x)\ge 0$ and $u$ a p-adic unit. The prime $p$ is a uniformizer, and reduction modulo $p$ gives the residue field ${\mathbb{F}}_p$. The element $p$ has no inverse inside ${\mathbb{Z}}_p$, so the ring is not a field.

For a short ideal-classification outline, take a nonzero ideal and choose an element of smallest P-adic valuation $k$, which exists by the well-ordering principle. Its unit factor is invertible, so the ideal contains $p^k$. Every other element is divisible by $p^k$, proving that the ideal is $p^k{\mathbb{Z}}_p$. Completeness follows because ${\mathbb{Z}}_p$ is a closed subset of the complete metric space ${\mathbb{Q}}_p$. The zero ideal is separate from this nonzero-ideal classification.

A strictly distance-decreasing map decreases the distance between every pair of distinct points. Unlike a contraction mapping, it need not have a single Lipschitz constant below one. Two distinct fixed points would contradict strict decrease, so there is at most one. Existence can fail even on a complete metric space: on $[0,\infty )$, $T(x)=x+1/(1+x)$ has no fixed point, while the mean value theorem and $0\le T^{\prime }(x)<1$ show strict distance decrease.

Let $X$ be a complete metric space and let $F_n$ be nonempty closed bounded subsets with $F_{n+1}\subseteq F_n$ and Hausdorff distance $d_H(F_n,F_{n+1})\le a_n$, where $a_n\ge 0$ and ${\sum }_n{a}_n<\infty$. Then $G={\bigcap }_n{F}_n$ is nonempty and closed, and $d_H(F_n,G)\le {\sum }_{k=n}^{\infty }{a}_k\to 0$.

To prove this, start with any $x_n\in F_n$ and choose $x_{k+1}\in F_{k+1}$ with $d(x_k,x_{k+1})<a_k+\eta 2^{n-k-1}$. The triangle inequality makes the resulting sequence a Cauchy sequence. Its limit exists by completeness and belongs to every $F_j$ by nesting and closedness. Its distance from the starting point is at most ${\sum }_{k=n}^{\infty }{a}_k+\eta$. Let $\eta \downarrow 0$ and take the supremum over starting points. The reverse directed distance is zero since $G\subseteq F_n$. Unlike the usual shrinking-diameter form of Cantor's intersection theorem, the sets need not have diameters tending to zero.

A sequence of functions $f_n:E\to \mathbb{R}$ has a uniform limit if and only if, for every $\epsilon >0$, there is $N$ such that $|f_n(x)-f_m(x)|<\epsilon$ for every $m,n\ge N$ and every $x\in E$. Necessity follows from the triangle inequality. For sufficiency, completeness of the real numbers first gives a pointwise limit $f$. Letting $m\to \infty$ in the bound with tolerance $\epsilon /2$ gives $|f_n(x)-f(x)|\le \epsilon /2$ uniformly in $x$. The same proof works for a complete metric space as codomain; completeness is essential.