Unravel the Secrets of Convergence: A Deep Dive into Sequences and Cauchy’s Theorem

Here is a persuasive advertisement for ghostwriting services for the topic of convergent and Cauchy sequences in metric spaces:

**Get Ahead in Math with Our Expert Ghostwriting Services!**

Are you struggling to grasp the concepts of convergent and Cauchy sequences in metric spaces? Do you need help with a math assignment or research paper on this topic? Look no further!

Our team of expert ghostwriters specializes in providing high-quality, customized writing services for math students like you. With our help, you can:

* Get a deep understanding of sequence convergence and Cauchy sequences
* Receive well-researched and accurately written content that meets your specific needs
* Save time and reduce stress with our fast turnaround and flexible deadlines
* Improve your grades and academic performance with our expertly crafted writing

Our ghostwriting services cover a wide range of topics related to convergent and Cauchy sequences, including:

* Definitions and notation for sequence convergence
* Properties and examples of Cauchy sequences
* Applications of convergent and Cauchy sequences in real-world problems

Our team of experienced writers holds advanced degrees in mathematics and has a proven track record of delivering high-quality content on time. We understand the importance of meeting academic standards and ensuring originality, so you can trust us to provide you with exceptional work.

Don’t let math anxiety hold you back. Let us help you achieve academic success with our expert ghostwriting services. Order now and get ahead in math!

**Contact us today to learn more and get a quote!**

1. We say a sequence \(\{x_{n}\}\)_converges_ in \((M,d)\) if there is a point \(x\in M\) such that for every \(\epsilon>0\) there is an \(N\in\mathbb{N}\) such that \[n\geq N\quad\text{implies}\quad d(x_{n},x)<\epsilon.\] In such case, we use the notation \(\{x_{n}\}\to x\) or simply write \(x_{n}\to x\). 2. We say \(\{x_{n}\}\) is _Cauchy_ if for every \(\epsilon>0\) there is an \(N\in\mathbb{N}\) such that \[n,m\geq N\quad\text{implies}\quad d(x_{n},x_{m})<\epsilon.\] 3. A sequence \(\{x_{n}\}\) is said to be _bounded_ if there is a constant \(K>0\) and a \(y\in M\) such that \[d(x_{n},y)\leq K\quad\text{for all}\quad n\in\mathbb{N}.\]

**Theorem 56**.: _Let \((M,d)\) be a metric space._

1. _If a sequence_ \(\{x_{n}\}\) _converges in_ \((M,d)\) _to a limit_ \(x\)_, then_ \(x\) _is unique._
2. _If a sequence_ \(\{x_{n}\}\to x\) _in_ \((M,d)\) _and_ \(\{x_{n_{k}}\}\) _is a subsequence of_ \(\{x_{n}\}\)_, then_ \(x_{n_{k}}\to x\) _(as_ \(k\to\infty\)_) as well._
3. _Every convergent sequence in_ \((M,d)\) _is bounded._
4. _Every convergent sequence in_ \((M,d)\) _is a Cauchy sequence._

Proof

: Modify the proofs of the similar theorems given in Chapter 1, Section 1.4 replacing \(|x-y|\) by the given metric \(d(x,y)\).

Now while several results about sequences in \(\mathbb{R}\) carry over to an abstract metric space \((M,d)\), there are exceptions. For example, Cauchy sequences need not converge to a point in \(M\). For example take \(M=(0,1]\) with the usual metric and the sequence \(\{x_{n}\}=\left\{\frac{1}{n}\right\}_{1}^{\infty}\subset(0,1]\). Then even though \(\{x_{n}\}\) is Cauchy, it does not converge to any point in \(M\).

### Completeness of Metric Spaces

**Definition 44**.: A metric space \((M,d)\) is called _complete_ if every Cauchy sequence of points of \(M\) converges to a point of \(M\).

Note that completeness requires two properties, one is every Cauchy sequence in \(M\) is convergent and the other condition is that the limit of every Cauchy sequence stays in \(M\). The Bolzano-Weierstrass theorem holds for \(\mathbb{R}^{n}\) and thus \(\mathbb{R}^{n}\) is complete for all \(n\in\mathbb{N}\) (you prove it). On the other hand \((\mathbb{Q},d)\) with \(d(x,y)=|x-y|\) is a metric space. Recalling the density of the rationals, we can choose a sequence \(\{x_{n}\}\) in \((\mathbb{Q},d)\), such that \(x_{n}\to\sqrt{2}\). Take

\[x_{1}=1.4,\,x_{2}=1.41,\,x_{3}=1.414,\ldots,x_{n},\ldots\]

where

\[x_{n}=(n+1)\text{-digits in the decimal expansion of }\sqrt{2}.\]

Then the sequence is Cauchy (why?), but the limit point \(\sqrt{2}\notin\mathbb{Q}\). Thus \((\mathbb{Q},d)\) is not a complete metric space.

**Remark 33**.: Cauchy sequences always converge in a discrete space. Because in a discrete metric space, a sequence \(\{x_{n}\}\) is Cauchy if and only if it is eventually constant, that is for sufficiently large \(N\) we have \(x_{n}=x\) for some fixed \(x\).

In the following we give two additional examples of metric spaces which are not complete.

**Example 61**.:
* Let \(\mathcal{P}\) be the set of all polynomials (of all degrees) defined on \([0,1]\). Define \(d(x,y)\) by \[d(x,y)=\max_{0\,\leq\,t\,\leq\,1}|x(t)-y(t)|.\] Then, \((\mathcal{P},d)\) is not a complete metric space. It is clear that \((\mathcal{P},d)\) is a metric space. To show \((\mathcal{P},d)\) is not complete, consider the following sequence: \[x_{n}(t)=\sum_{k=0}^{n}\left(\frac{t}{2}\right)^{k}=1+\frac{t}{2}+\cdots+ \frac{t^{n}}{2^{n}}\quad 0\leq t\leq 1.\] Clearly, \(x_{n}(t)\in\mathcal{P}\) for each \(n\in\mathbb{N}\). Next, we show that the sequence \(\{x_{n}\}\) is a Cauchy sequence. Taking \(m

Let \((M,d)\) be a metric space and \(S\) a nonempty subset of \(M\). As we defined earlier, by the restriction of \(d\) to \(S\) we mean the mapping

\[d_{S}:S\times S\to[0,\infty)\quad\text{such that}\quad d_{S}(x,y)=d(x,y)\]

for \(x,y\in S\). It is clear that \(d_{S}\) is a metric on \(S\) and we usually drop \(S\) and express this metric on \(S\) as \(d\) as well. The metric space \((S,d)\) is called a subspace of \((M,d)\) and a subspace \((S,d)\) of the metric space \((M,d)\) is said to be _closed_ if it contains the limits of all sequences in \(S\) which converge in \(M\). The next theorem guarantees the completeness of many metric spaces which arise as subspaces of complete metric spaces.

**Theorem 57**.: _Let \((M,d)\) be a complete metric space and let \((S,d)\) be a subspace of \(M\). Then \(S\) is complete if and only if it is closed._

Proof

: Assume that \(S\) is closed; we want to show that it is complete. Let \(\{x_{n}\}\) be a Cauchy sequence in \(S\). It is also a Cauchy sequence in \((M,d)\), and since \(M\) is complete \(\{x_{n}\}\) converges to a point \(x\) in \(M\); but since \(S\) is closed, the limit of the sequence must belong to \(S\), so \(S\) is complete. Now let us prove the converse. Suppose \(S\) is complete, let \(\{x_{n}\}\) be any sequence in \(S\) and suppose that as a sequence in \(M\) it converges with limit \(x\). Then \(\{x_{n}\}\) is a Cauchy sequence in \(M\) and hence also in \(S\). Since \(S\) is complete, we must have \(x\in S\), so \(S\) is closed.

Now we know that in a complete metric space, the notions of completeness and closedness for subspaces coincide. Another natural question in regards to the completeness of metric spaces is whether or not an arbitrary incomplete metric space can be somehow enlarged to a metric space which is complete. After all, \(\mathbb{Q}\) is an incomplete metric space but we can enlarge to the real line \(\mathbb{R}\) which is complete. This question has a positive answer, but we first need the following definition:

**Definition 45**.: Let \((M,d)\) and \((\tilde{M},\tilde{d})\) be metric spaces. A mapping \(T:M\to\tilde{M}\) is said to be _isometric_ if \(T\) preserves distances. That is for all \(x,y\in M\),

\[\tilde{d}(Tx,Ty)=d(x,y).\]

The space \(M\) is said to be _isometric_ with the space \(\tilde{M}\) if there is a bijective (one-to-one and onto) isometry of \(M\) onto \(\tilde{M}\). In this case the spaces \(M\) and \(\tilde{M}\) are called _isometric spaces_.

We can now state the following theorem without proof (for proof consult, e.g., [35], pp. 42-45):

**Theorem 58**.: _For a metric space \((M,d)\) there is a complete metric space \((\tilde{M},\tilde{d})\) (called the completion) which has the following properties:_

1. \((\tilde{M},\tilde{d})\) _has a subspace_ \(S\) _that is isometric to_ \((M,d)\)_._
2. \((M,d)\) _is dense in_ \((\tilde{M},\tilde{d})\)_._
3. \((\tilde{M},\tilde{d})\) _is unique except for isometries (i.e., if_ \(\hat{M}\) _is any complete metric space having a dense subspace_ \(\hat{S}\) _isometric with_ \(M\)_, then_ \(\hat{M}\) _and_ \(\tilde{M}\) _are isometric as well)._

### Completion and Ultrametric

The set of real numbers \(\mathbb{R}\) is the completion of \(\mathbb{Q}\) with respect to Euclidean norm where the induced metric by this norm is \(d(x,y)=|x-y|\). _The field of p-adic numbers_\(\mathbb{Q}_{p}\) defined for each prime \(p\) is constructed from the rational numbers \(\mathbb{Q}\) as its completion with respect to a certain norm which depends on the prime number \(p\). This norm is quite different from the Euclidean norm.

**Definition 46**.: Let \(F\) be a field. A _norm_ on \(F\) is a map denoted by

\[||\cdot||:F\to\mathbb{R}\geq 0\]

satisfying the properties:

1. \(||x||=0\) if and only if \(x=0\).
2. \(||x\,y||=||x|||y||\) for all \(x,y\in F\).
3. \(||x+y||\leq||x||+||y||\) for all \(x,y\in F\) (The Triangle Inequality).

This norm induces a metric (or a distance function) by \(d(x,y)=||x-y||\) and we can regard \((F,d)\) as a metric space.

**Definition 47**.: Let \(||\cdot||_{1}\) and \(||\cdot||_{2}\) be two norms on a field \(F\). We say that these two norms are _equivalent_ if and only if there exists a positive real number \(\alpha\) such that \(||x||_{2}=||x||_{1}^{\alpha}\) for every \(x\in F\).

This means that a sequence is Cauchy with respect to \(||\cdot||_{1}\) if and only if it is Cauchy with respect to \(||\cdot||_{2}\). We have already seen that the absolute value \(|\cdot|\) is a norm on \(\mathbb{Q}\). The proposition below describes all the norms on \(\mathbb{Q}\) which are equivalent to the absolute value norm.

**Proposition 21**.: _Let \(\alpha>0\). \(||x||=|x|^{\alpha}\) is a norm on \(\mathbb{Q}\) if and only if \(\alpha\leq 1\). In this case it is equivalent to the absolute value norm._

Proof

: First note that if \(\alpha>1\) then the triangle inequality fails since \(|1+1|^{\alpha}=2^{\alpha}>|1|^{\alpha}+|1|^{\alpha}=2\). Suppose \(\alpha\leq 1\). The first two properties of the norm are straightforward, so we need to check the triangle inequality. To show the triangle inequality holds, we first observe the behavior of \(t^{\alpha}\). Note that for \(t\geq 1\), \(t^{\alpha}\leq t\) and for \(0\leq t\leq 1\) we have \(t^{\alpha}\geq t\). Assume \(\dfrac{|y|}{|x|}\leq 1\). Then by setting \(t=1+\dfrac{|y|}{|x|}\), we have:

\[|x+y|^{\alpha}\leq(|x|+|y|)^{\alpha}=|x|^{\alpha}\left(1+\dfrac{|y|}{|x|} \right)^{\alpha}\leq|x|^{\alpha}\left(1+\dfrac{|y|}{|x|}\right)\leq|x|^{\alpha }\left(1+\dfrac{|y|^{\alpha}}{|x|^{\alpha}}\right)\]

and therefore

\[|x+y|^{\alpha}\leq|x|^{\alpha}+|y|^{\alpha}.\qed\]

There is a theorem due to Ostrowski [30], p. 43 which describes in a precise way all norms on \(\mathbb{Q}\) which are equivalent to the absolute value \(|\cdot|\). We will see this after we explain the \(p\)-adic absolute value \(|\cdot|_{p}\) on \(\mathbb{Q}\).

**Definition 48**.: A norm is called _non-Archimedian_ if it satisfies the non-Archimedian triangle inequality:

\[||x+y||\leq\max\{||x||,||y||\}.\]

The metric induced by a non-Archimedian norm is said to be an _ultrametric_. Note that instead of the triangle inequality for the usual metric

\[d(x,z)\leq d(x,y)+d(y,z),\]

an ultrametric satisfies the stronger triangle inequality:

\[d(x,z)\leq\max\{d(x,y),d(y,z)\}.\]

Metric spaces with ultrametrics are called _ultrametric spaces_.

**Example 62**.: We have already seen one example of an ultrametric space. The discrete metric is an ultrametric (why?). Perhaps the most important examples of ultrametric spaces come from the \(p\)-adic numbers; \(p\)-adic numbers form a complete ultrametric space. Before we define _p-adic numbers_ and _p-adic absolute value_, let us make one nice geometrical observation about ultrametric spaces.