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# 数学代写|线性代数代写Linear algebra代考|The Completion of a Metric Space

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## 数学代写|线性代数代写Linear algebra代考|The Completion of a Metric Space

While not all metric spaces are complete, any metric space can be embedded in a complete metric space. To be more specific, we have the following important theorem.

Theorem 12.8 Let $(M, d)$ be any metric space. Then there is a complete metric space $\left(\mathrm{M}^{\prime}, d^{\prime}\right)$ and an isometry $\tau: M \rightarrow \tau(\mathrm{M}) \subset \mathrm{M}^{\prime}$ for which $\tau(M)$ is dense in $M^{\prime}$. The metric space $\left(M^{\prime}, d^{\prime}\right)$ is called a completion of $(\mathrm{M}, d)$. Moreover, $\left(\mathrm{M}^{\prime}, d^{\prime}\right)$ is unique, up to bijective isometry.

Proof. The proof is a bit lengthy, so we divide it into various parts. We can simplify the notation considerably by thinking of sequences $\left(x_n\right)$ in $M$ as functions $f: N \rightarrow M$, where $f(n)=x_n$.
Cauchy Sequences in $M$
The basic idea is to let the elements of $M^{\prime}$ be equivalence classes of Cauchy sequences in M. So let $\operatorname{CS}(M)$ denote the set of all Cauchy sequences in $M$. If $f, g \in C S(M)$ then, intuitively speaking, the terms $f(n)$ get closer together as $n \rightarrow \infty$, and so do the terms $g(n)$. Therefore, it seems reasonable that $d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n})$ ) should approach a finite limit as $\mathrm{n} \rightarrow \infty$. Indeed, according to Exercise 2 ,
$$|d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n}))-d(\mathrm{f}(\mathrm{m}), \mathrm{g}(\mathrm{m}))| \leq d(\mathrm{f}(\mathrm{n}), \mathrm{f}(\mathrm{m}))+d(\mathrm{~g}(\mathrm{n}), \mathrm{g}(\mathrm{m})) \rightarrow 0$$
as $\mathrm{n}, \mathrm{m} \rightarrow \infty$, and so $d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n})$ ) is a Cauchy sequence of real numbers, which implies that
$$\lim _{\mathrm{n} \rightarrow \infty} d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n}))<\infty$$
(That is, the limit exists and is finite.)

## 数学代写|线性代数代写Linear algebra代考|Equivalence Classes of Cauchy Sequences in M

We would like to define a metric $d^{\prime}$ on the set $\operatorname{CS}(\mathrm{M})$ by
$$d^{\prime}(\mathrm{f}, \mathrm{g})=\lim {\mathrm{n} \rightarrow \infty} d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n}))$$ However, it is possible that $$\lim {\mathrm{n} \rightarrow \infty} d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n}))=0$$
for distinct sequences $\mathrm{f}$ and $\mathrm{g}$, so this does not define a metric. Thus, we are lead to define an equivalence relation on $\operatorname{CS}(\mathrm{M})$ by
$$\mathrm{f} \sim \mathrm{g} \Leftrightarrow \lim {\mathrm{n} \rightarrow \infty} d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n}))=0$$ Let $\overline{\operatorname{CS}(M)}$ be the set of all equivalence classes of Cauchy sequences, and define, for $\overline{\mathrm{f}}, \overline{\mathrm{g}} \in \overline{\mathrm{CS}(\mathrm{M})}$ $$d^{\prime}(\overline{\mathrm{f}}, \overline{\mathrm{g}})=\lim {\mathrm{n} \rightarrow \infty} d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n}))$$
where $\mathrm{f} \in \overline{\mathrm{f}}$ and $\mathrm{g} \in \overline{\mathrm{g}}$.
To see that $d^{\prime}$ is well-defined, suppose that $f^{\prime} \in \bar{f}$ and $g^{\prime} \in \bar{g}$. Then since $f^{\prime} \sim \mathrm{f}$ and $g^{\prime} \sim \mathrm{g}$, we have
$$\left|d\left(\mathrm{f}^{\prime}(\mathrm{n}), \mathrm{g}^{\prime}(\mathrm{n})\right)-d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n}))\right| \leq d\left(\mathrm{f}^{\prime}(\mathrm{n}), \mathrm{f}(\mathrm{n})\right)+d\left(\mathrm{~g}^{\prime}(\mathrm{n}), \mathrm{g}(\mathrm{n})\right) \rightarrow 0$$
as $\mathrm{n} \rightarrow \infty$. Thus,
\begin{aligned} \mathrm{f}^{\prime} \sim \mathrm{f} \text { and } \mathrm{g}^{\prime} \sim \mathrm{g} & \Rightarrow \lim {\mathrm{n} \rightarrow \infty} d\left(\mathrm{f}^{\prime}(\mathrm{n}), \mathrm{g}^{\prime}(\mathrm{n})\right)=\lim {\mathrm{n} \rightarrow \infty} d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n})) \ & \Rightarrow d^{\prime}\left(\mathrm{f}^{\prime}, \mathrm{g}^{\prime}\right)=d^{\prime}(\mathrm{f}, \mathrm{g}) \end{aligned}
which shows that $d^{\prime}$ is well-defined. To see that $d^{\prime}$ is a metric, we verify the triangle inequality, leaving the rest to the reader. If $f, g$ and $\mathrm{h}$ are Cauchy sequences, then
$$d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n})) \leq d(\mathrm{f}(\mathrm{n}), \mathrm{h}(\mathrm{n}))+d(\mathrm{~h}(\mathrm{n}), \mathrm{g}(\mathrm{n}))$$
Taking limits gives
$$\lim {\mathrm{n} \rightarrow \infty} d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n})) \leq \lim {\mathrm{n} \rightarrow \infty} d(\mathrm{f}(\mathrm{n}), \mathrm{h}(\mathrm{n}))+\lim _{\mathrm{n} \rightarrow \infty} d(\mathrm{~h}(\mathrm{n}), \mathrm{g}(\mathrm{n}))$$
and so
$$d^{\prime}(\overline{\mathrm{f}}, \overline{\mathrm{g}}) \leq d^{\prime}(\overline{\mathrm{f}}, \overline{\mathrm{h}})+d^{\prime}(\overline{\mathrm{h}}, \overline{\mathrm{g}})$$

## 数学代写|线性代数代写Linear algebra代考|The Completion of a Metric Space

$$|d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n}))-d(\mathrm{f}(\mathrm{m}), \mathrm{g}(\mathrm{m}))| \leq d(\mathrm{f}(\mathrm{n}), \mathrm{f}(\mathrm{m}))+d(\mathrm{~g}(\mathrm{n}), \mathrm{g}(\mathrm{m})) \rightarrow 0$$

$$\lim _{\mathrm{n} \rightarrow \infty} d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n}))<\infty$$
（也就是说，极限存在并且是有限的。）

## 数学代写|线性代数代写Linear algebra代考|Equivalence Classes of Cauchy Sequences in M

$$d^{\prime}(\mathrm{f}, \mathrm{g})=\lim \mathrm{n} \rightarrow \infty d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n}))$$

$$\lim n \rightarrow \infty d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n}))=0$$

$$\mathrm{f} \sim \mathrm{g} \Leftrightarrow \lim \mathrm{n} \rightarrow \infty d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n}))=0$$

$$d^{\prime}(\overline{\mathrm{f}}, \overline{\mathrm{g}})=\lim \mathrm{n} \rightarrow \infty d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n}))$$

$$\left|d\left(\mathrm{f}^{\prime}(\mathrm{n}), \mathrm{g}^{\prime}(\mathrm{n})\right)-d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n}))\right| \leq d\left(\mathrm{f}^{\prime}(\mathrm{n}), \mathrm{f}(\mathrm{n})\right)+d\left(\mathrm{~g}^{\prime}(\mathrm{n}), \mathrm{g}(\mathrm{n})\right) \rightarrow 0$$

$$\mathrm{f}^{\prime} \sim \mathrm{f} \text { and } \mathrm{g}^{\prime} \sim \mathrm{g} \Rightarrow \lim \mathrm{n} \rightarrow \infty d\left(\mathrm{f}^{\prime}(\mathrm{n}), \mathrm{g}^{\prime}(\mathrm{n})\right)=\lim \mathrm{n} \rightarrow \infty d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n})) \quad \Rightarrow d^{\prime}\left(\mathrm{f}^{\prime}, \mathrm{g}^{\prime}\right)=d^{\prime}(\mathrm{f}, \mathrm{g})$$

$$d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n})) \leq d(\mathrm{f}(\mathrm{n}), \mathrm{h}(\mathrm{n}))+d(\mathrm{~h}(\mathrm{n}), \mathrm{g}(\mathrm{n}))$$

$$\lim \mathrm{n} \rightarrow \infty d(\mathrm{f}(\mathrm{n}), \mathrm{g}(\mathrm{n})) \leq \lim \mathrm{n} \rightarrow \infty d(\mathrm{f}(\mathrm{n}), \mathrm{h}(\mathrm{n}))+\lim _{\mathrm{n} \rightarrow \infty} d(\mathrm{~h}(\mathrm{n}), \mathrm{g}(\mathrm{n}))$$

$$d^{\prime}(\overline{\mathrm{f}}, \overline{\mathrm{g}}) \leq d^{\prime}(\overline{\mathrm{f}}, \overline{\mathrm{h}})+d^{\prime}(\overline{\mathrm{h}}, \overline{\mathrm{g}})$$

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