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For a metric space $(X,d)$, uniform continuity meansThe choice of $\delta$ is independent of the points. A Lipschitz continuous function satisfies $|f(x)-f(y)|\le Ld(x,y)$ for one finite constant $L\ge 0$. Choosing $\delta =\epsilon /(L+1)$ proves Lipschitz continuity implies uniform continuity.

Let $(x_n)$ be a Cauchy sequence. For a given $\epsilon >0$, uniform continuity supplies $\delta >0$, and the Cauchy property supplies $N$ with $d(x_m,x_n)<\delta$ for $m,n\ge N$. Then $|f(x_m)-f(x_n)|<\epsilon$, so $(f(x_n))$ is Cauchy in $\mathbb{R}$. By completeness of the real numbers, $(f(x_n))$ converges, even if the original sequence has no limit in $X$.

A Cauchy sequence is a sequence $(x_n)$ such that for every $\epsilon >0$ there is $N$ for whichThe general principle of convergence, or completeness of the real numbers, states that a real sequence converges if and only if it is Cauchy.

For the final claim, let $m=\lfloor n/2\rfloor$. Since $(a_n)$ is decreasing and positive,Because the series $\sum a_k$ converges, its tails tend to zero. Also $n-m\ge n/2$, soThe squeeze theorem gives

The Bolzano-Weierstrass theorem states: every bounded sequence of real numbers has a convergent subsequence.

By the monotone subsequence theorem proved in the preceding part, a bounded sequence $(a_n)$ has a monotone subsequence $(b_r)$. It inherits the bounds of $(a_n)$. For completeness, the bounded monotone sequence theorem follows directly from completeness of the real numbers. If $(b_r)$ is increasing, let $L=\sup \{b_r:r\ge 1\}$. For every $\epsilon >0$, the definition of supremum supplies $r_0$ with $L-\epsilon <b_{r_0}\le L$. Thenso $b_r\to L$. If $(b_r)$ is decreasing, apply the increasing case to $(-b_r)$, or use its infimum. Thus in either case

For real-valued functions, the Uniform Cauchy criterion applied to partial sums says that $\sum u_n$ has uniform convergence exactly whenSufficiency uses completeness of the real numbers.

Let $M$ be a common bound for the uniformly bounded family of functions $(g_n)$; increase it if necessary so that $M>0$. The assumed uniform absolute convergence of $\sum f_n$ gives $N$ such thatThe triangle inequality now yieldsThe Uniform Cauchy criterion therefore proves uniform convergence of $\sum f_n{g}_n$. The same estimate with $|f_n{g}_n|$ proves uniform absolute convergence as well.

The infimum of $A$ is its greatest lower bound: $\alpha =\inf A$ means that $\alpha \le x$ for every $x\in A$, and that every lower bound $l$ of $A$ satisfies $l\le \alpha$. The supremum is its least upper bound: $\beta =\sup A$ means that $x\le \beta$ for every $x\in A$, and every upper bound $u$ satisfies $\beta \le u$. The completeness of the real numbers ensures that both exist for a nonempty bounded set of real numbers.

A real or complex sequence $(a_n)$ is a Cauchy sequence ifFor $z_n=x_n+i{y}_n$, the complex modulus givesThus the same $N$ works for both coordinate Cauchy sequences. Conversely, if both real coordinate Cauchy sequences converge to $x$ and $y$, then $|z_n-(x+iy)|\le |x_n-x|+|y_n-y|\to 0$. This proves the needed completeness for complex numbers from the assumed completeness of the real numbers.

Now suppose ${\sum }_{n\ge 0}|z_n|$ converges. Its real partial sums form a Cauchy sequence. Given $\epsilon >0$, sufficiently late tails therefore satisfy ${\sum }_{k=n+1}^m|z_k|<\epsilon$. For $s_m={\sum }_{k=0}^m{z}_k$, the triangle inequality yieldsThe complex partial sums are a Cauchy sequence, hence converge by the coordinate argument. Therefore absolute convergence implies convergence for a complex series.

True. Put $\alpha =\sup A$, $\beta =\sup B$ and $u=\max \{\alpha ,\beta \}$. These suprema exist by completeness of the real numbers, since the sets are nonempty and bounded. Every set element of the union $A\cup B$ is at most $u$, so $u$ is an upper bound. Conversely, every upper bound of $A\cup B$ bounds both $A$ and $B$, and is therefore at least both suprema. Thus

Let $\epsilon >0$. By the definition of Cauchy sequence, choose an integer $K\ge 1$ such that $|x_j-x_k|<\epsilon /2$ whenever $j,k\ge K$. Apply the given inequality with $N=K+1$. It supplies $n>K+1$ for whichBoth $n$ and $K$ are in the Cauchy sequence estimate, soFor every $m\ge K$ the triangle inequality givesThis is the definition of convergence to zero: $\boxed{x_m\to 0}$. The proof uses only the Cauchy sequence estimate, positivity, and the stipulated inequality; it does not assume the general completeness of the real numbers theorem for Cauchy sequences. Taking $N=K+1\ge 2$ also avoids the empty initial set in a literal reading of the condition for very small $N$.

For a nonempty set $A\subseteq \mathbb{R}$, a supremum is a real number $s$ such that $a\le s$ for every $a\in A$, and every upper bound $u$ of $A$ satisfies $s\le u$. Equivalently, $s$ is an upper bound and, for every $\epsilon >0$, some $a\in A$ satisfies $a>s-\epsilon$. By completeness of the real numbers, it exists whenever $A$ is nonempty and bounded above.

An increasing monotone sequence satisfies $a_{n+1}\ge a_n$ for every $n$; a bounded sequence satisfies $|a_n|\le M$ for some finite $M$ and every $n$. Put $s=\sup \{a_n:n\ge 1\}$. Given $\epsilon >0$, the defining property of the supremum supplies $N$ with $a_N>s-\epsilon$. For every $n\ge N$, monotonicity givesThis is exactly the definition of convergent sequence, so the bounded monotone sequence theorem gives $\boxed{a_n\to s}$.

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

Real analysis studies limits, continuity, differentiation, integration, and convergence of sequences and functions taking values in the real numbers. Its central question is how finite approximations determine an exact answer: what does an arbitrarily small error mean, and which operations respect that error? A derivative is a limit of quotients, while a Riemann integral is obtained from increasingly fine Riemann sums. Both need conditions that make the answer independent of the approximations chosen.

Two organizing ideas are completeness of the real numbers, which ensures that suitable approximations have a destination, and uniform convergence, which controls approximation error over an entire domain. Pointwise convergence supplies separate guarantees at separate inputs; uniform convergence supplies one guarantee for all inputs. This distinction explains why replacing an approximating function by its limit can preserve continuity or integrals under appropriate hypotheses, yet fail without them.

For historical context, Cauchy's 1821 Cours d'analyse explicitly discusses convergence of series. Its Note VII treats double series and compares summing by rows and by columns under stated convergence assumptions: controlling an infinite process was already a concrete problem. This is historical context for questions about limits, rather than an attribution of the modern definitions below. ^[ref].

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.