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MathAnal

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泛函分析 近世代数,群论。 有限元。

Complete

Cauchy Sequence

Cauchy seq is a sequence $a_1, a_2, \ldots$ such that the metric $d(a_m,a_n) $ satisfies $\lim_(min(m,n)->\infty)d(a_m,a_n)=0$.

Cauchy sequences in the rationals do not necessarily converge, but they do converge in the reals.

Real numbers can be defined using either Dedekind cuts or Cauchy sequences.

If a real number $x$ is irrational, then the sequence ($x_n$), whose n-th term is the truncation to n decimal places of the decimal expansion of $x$, gives a Cauchy sequence of rational numbers with irrational limit $x$.

The sequence defined by $ x_0=1, x_{n+1}=\frac{x_n+\frac{2}{x_n}}{2}$ consists of rational numbers (1, 3/2, 17/12, ...) converges to $\sqrt{2}$. TODO: prove it.

The sequence $ x_n = F_n / F_{n-1}$ of ratios of consecutive Fibonacci numbers which, if it converges at all, converges to a limit $\phi$ satisfying $\phi^2 = \phi+1$, where $\varphi = (1+\sqrt5)/2$.

Properties of cauchy sequence

Every convergent sequence (with limit $s$, say) is a Cauchy sequence, since, given any real number ε > 0, beyond some fixed point, every term of the sequence is within distance ε/2 of s, so any two terms of the sequence are within distance ε of each other. Every Cauchy sequence of real (or complex) numbers is bounded (since for some N, all terms of the sequence from the N-th onwards are within distance 1 of each other, and if M is the largest absolute value of the terms up to and including the N-th, then no term of the sequence has absolute value greater than M+1). In any metric space, a Cauchy sequence which has a convergent subsequence with limit s is itself convergent (with the same limit), since, given any real number r > 0, beyond some fixed point in the original sequence, every term of the subsequence is within distance r/2 of s, and any two terms of the original sequence are within distance r/2 of each other, so every term of the original sequence is within distance r of s. These last two properties, together with a lemma used in the proof of the Bolzano–Weierstrass theorem, yield one standard proof of the completeness of the real numbers, closely related to both the Bolzano–Weierstrass theorem and the Heine–Borel theorem. The lemma in question states that every bounded sequence of real numbers has a convergent monotonic subsequence. Given this fact, every Cauchy sequence of real numbers is bounded, hence has a convergent subsequence, hence is itself convergent. It should be noted, though, that this proof of the completeness of the real numbers implicitly makes use of the least upper bound axiom. The alternative approach, mentioned above, of constructing the real numbers as the completion of the rational numbers, makes the completeness of the real numbers tautological.

The values of exp(x), sin(x), cos(x), are known to be irrational for any rational value of x≠0, but each can be defined as the limit of a rational Cauchy sequence, using, for instance, the Maclaurin series.

A metric space X in which every Cauchy sequence converges to an element of X is called complete. Rationals are not complete, but Reals are.

Compact, Closed, Bounded, Complete

Reference: http://www.physicsforums.com/showthread.php?t=113648

Examples (real line with usual topology).
Bounded not closed: 0<x<1
Closed not bounded or compact: 0<=x<inf.

In $\mathcal{R}^n$, compact is equivalent to closed and bounded. So a closed set is bounded iff compact, and a bounded set is closed iff compact.

In a metric space, compact is equivalent to complete and totally bounded.

In $\mathcal{R}^n$ which is itself complete, closed is equivalent to complete, and since every bounded set in R^n has compact closure, bounded is equivalent to totally bounded.

a totally bounded set is one in which every sequence has a cauchy subsequence, and a complete set is one in which every cauchy sequence converges.

The connection is that a compact metric space is one in which every sequence has a convergent subsequence. (i think. it has been a long time since i taught this course.)