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# 数学代写|拓扑学代写TOPOLOGY代考|Cauchy Sequences

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## 数学代写|拓扑学代写TOPOLOGY代考|Cauchy Sequences

Definition 6.24 A sequence $\left{a_n\right}$ in a metric space $(X, d)$ is a Cauchy sequence if for any $\varepsilon>0$ there exists an integer $N$ such that $d\left(a_n, a_m\right)<\varepsilon$ for any $n, m \geq N$. Every convergent sequence is a Cauchy sequence: in fact, if $p$ is the limit of $\left{a_n\right}$, then for every $\varepsilon>0$ there is an $N$ such that $a_n \in B(p, \varepsilon / 2)$ for any $n \geq N$; the triangle inequality implies $d\left(a_n, a_m\right)<\varepsilon$ for any $n, m \geq N$.

Lemma 6.25 A Cauchy sequence is convergent if and only if it has limit points. In particular, any Cauchy sequence in a sequentially compact metric space converges.
Proof Let $\left{a_n\right}$ be a Cauchy sequence in a metric space $(X, d)$, with limit point $p \in X$. We’ll show that for every $\varepsilon>0$ we can find $N \in \mathbb{N}$ such that $d\left(p, a_n\right)<\varepsilon$ for any $n \geq N$. The Cauchy property implies that there’s an $M \in \mathbb{N}$ such that $d\left(a_n, a_m\right)<\varepsilon / 2$ for any $n, m \geq M$; as $p$ is a limit point, there also is an $N \geq M$ such that $d\left(p, a_N\right)<\varepsilon / 2$. Now use the triangle inequality
$$d\left(p, a_n\right) \leq d\left(p, a_N\right)+d\left(a_N, a_n\right)<\varepsilon \text { for every } n \geq M$$
If a Cauchy sequence has a convergent subsequence, by Lemma 6.17 it has a limit point.

## 数学代写|拓扑学代写TOPOLOGY代考|Compact Metric Spaces

Definition 6.30 A metric space is said to be totally bounded if it can be covered by a finite number of open balls of radius $r$, for any positive real number $r$.

Each totally bounded metric space is bounded: if $X$ is the finite union of balls $B\left(x_1, 1\right), \ldots, B\left(x_n, 1\right)$ and $M$ is the maximum distance between any two centres $x_i$, by the triangle inequality the distance of any two points of $X$ is at most $M+2$. Any discrete and infinite space, with the distance of Example 3.34, is bounded but not totally bounded.
Lemma 6.31 Every sequentially compact metric space is totally bounded.
Proof Take $(X, d)$ sequentially compact and suppose, by contradiction, there is an $r>0$ such that $X$ cannot be covered by finitely many open balls of radius $r$. We construct recursively a sequence $\left{a_n\right}$ : choose $a_1 \in X$ arbitrarily, and for any $n>1$ let $a_n$ be some element in the non-empty closed set
$$X-\bigcup_{i=1}^{n-1} B\left(a_i, r\right)$$
Since $d\left(a_n, a_m\right) \geq r$ for every $n>m$, no subsequence of $\left{a_n\right}$ can converge.

Lemma 6.32 Every totally bounded metric space is second countable.
Proof Let $X$ be totally bounded. For any $n \in \mathbb{N}$ there’s a finite subset $E_n \subset X$ such that $X=\cup\left{B\left(e, 2^{-n}\right) \mid e \in E_n\right}$. The countable set $E=\cup E_n$ is thus dense, making $X$ separable. Invoking Lemma 6.7 allows to conclude.

## 数学代写|拓扑学代写TOPOLOGY代考|Cauchy Sequences

$$d\left(p, a_n\right) \leq d\left(p, a_N\right)+d\left(a_N, a_n\right)<\varepsilon \text { for every } n \geq M$$

## 数学代写|拓扑学代写TOPOLOGY代考|Compact Metric Spaces

$$X-\bigcup_{i=1}^{n-1} B\left(a_i, r\right)$$

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