Suppose $\{p_n\}$ converges to $p\in X.$ Let For some integer $N,$ when Therefore, $p_n\in B_{ϵ}(p)$ for all but the finitely many $p_n$ where $n\le N.$

Conversely, suppose every neighborhood of $p$ contains all but finitely many $p_n,$ i.e., for all but $N$ elements of $\{p_n\}.$ Let Then, $p_n\in B_{ϵ}(p)$ whenever $n\ge N,$ therefore,

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Suppose, for contradiction, that $p\ne p^{\prime }.$ Then, Let $\delta =ϵ/2$ and $B_{\delta }(p),B_{\delta }(p^{\prime })$ be balls around $p$ and $p^{\prime },$ respectively. This means that only finitely many points from $\{p_n\}$ are not in $B_{\delta }(p)$. However, since $B_{\delta }(p)$ and $B_{\delta }(p^{\prime })$ are disjoint by construction, this means only finitely many points of $\{p_n\}$ are in $B_{\delta }(p^{\prime }),$ a contradiction. Therefore, our assumption that $p\ne p^{\prime }$ is incorrect, so $p=p^{\prime }.$

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Let Only finitely many points in $\{p_n\}$ lie outside of $B_{ϵ}(p)$. That is, for some integer $N,$ only the points $p_n$ where $n\le N$ lie outside of $B_{ϵ}(p).$ Let $\delta =max\{ϵ,d(p,p_1),d(p,p_2),\dots ,d(p,p_n)\},n=1,2,\dots ,N.$ Then, for all $n=1,2,3,\dots .$

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For each $n=1,2,3\dots ,$ there is a point $p_n\in E$ such that . Let and pick $N$ so that Then, if so $\underset{n\to \mathrm{\infty }}{lim}{p}_n=p$.

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For (a), let pick $N_s$ such that when $n\ge N_s,$ and pick $N_t$ such that when $n\ge N_t.$ Then, let $N=max\{N_s,N_t\}.$ Then, when $n\ge N,$

For (b), If $c=0,$ let and then Otherwise,

Let For some $N,$ when $n\ge N,$ we have

For (c), let For some $N,$ when $n\ge N,$ we have

For (d), first note the identity

$$\begin{array}{cc}{s}_n{t}_n-st=(s_n-s)(t_n-t)+s(t_n-t)+t(s_n-s).& \text{(1)}\end{array}$$

Now, let Pick $N_s$ such that when $n\ge N_s,$ and pick $N_t$ such that when $n\ge N_t.$ Then, let $N=max\{N_s,N_t\}.$ Then, when $n\ge N,$

which means

$$\underset{n\to \mathrm{\infty }}{lim}(s_n-s)(t_n-t)=0.$$

Applying this, along with the results of (a) and (b) to (1) gives:

For (e), pick $m$ such that when so we have that

Now, let For some integer when $n\ge M,$ we have that

Thus, when $n\ge N,$

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For (a), assume ${\mathbf{x}}_n\to \mathbf{x}.$ Then, from the definition of the norm,

$$|{\alpha }_{j,n}-{\alpha }_j|\le |{\mathbf{x}}_n-\mathbf{x}|,$$

that is, the distance from ${\alpha }_{k,n}$ to ${\alpha }_n$ is always less than or equal to the distance from ${\mathbf{x}}_n$ to $\mathbf{x}.$ Therefore, for and we can pick $n$ to make this true for as small of $ϵ$ as we'd like. Therefore, $\underset{n\to \mathrm{\infty }}{lim}{\alpha }_{j,n}={\alpha }_j.$

Conversely, assume $\underset{n\to \mathrm{\infty }}{lim}{\alpha }_{j,n}={\alpha }_j.$ Let For some integer $N,$ when $n\ge N$ we have

$$|{\alpha }_{j,n}-{\alpha }_j|\le \frac{ϵ}{\sqrt{k}},(1\le j\le k).$$

Therefore, $n\ge N$ implies that

so ${\mathbf{x}}_n\to \mathbf{x}.$

Part (b) follows from part (a) and A sequence in $R^k$ converges iff its components converge.

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Suppose that $\{p_n\}$ converges to $p.$ Suppose some subsequence $\{p_{n_i}\}$ converges to $q,$ and suppose, for contradiction, that $q\ne p.$ Now, following an argument similar to the proof that limits of sequences are unique, we can see that if $p\ne q,$ arbitrary neighborhoods around both can't contain all but finitely many points, so we have a contradiction, and $p=q.$

Conversely, suppose every subsequence of $\{p_n\}$ converges to $p.$ Then, $\{p_n\}$ is a subsequence of itself, so it converges to $p.$

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Let $E$ be the range of $\{p_n\}.$ If $E$ is finite, then at least one point $p$ in $E$ must be repeated infinitely many times in $\{p_n\}.$ If we let $\{n_i\}$ be the indices of the occurrences of $p$ in $\{p_n\}:$

$$p_{n_1}=p_{n_2}=\cdots =p,$$

then the subsequence $\{p_{n_i}\}$ converges to $p.$

On the other hand, if $E$ is infinite, then $E$ has a limit point $p\in X$. Pick $n_1$ so that Now, after picking $n_1,\dots ,n_{i-1},$ we can pick such that , so $\{p_{n_i}\}$ converges to $p.$

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Note that any bounded sequence $\{p_n\}\subset R^k$ is a subset of some closed set, bounded and thus compact k-cell in $R^k.$ Therefore, $\{p_n\}$ is a sequence in a compact metric space, and has a convergent subsequence.

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Let $E^{\ast }$ be the set of all subsequential limits of $\{p_n\}$ and let $q$ be a limit point of $E^{\ast }.$ We want to show that $q\in E^{\ast }.$

First, note that if the range of $\{p_n\}$ is just $\{q\},$ then $q$ is the only subsequential limit of $\{p_n\}.$ In this case, $E\ast =\{q\}$ is a singleton and is closed set, as it vacuously contains all of its limit points. So, assume this is not the case.

Choose $n_1$ so that $p_{n_1}\ne q,$ and let $\delta =d(q,p_{n_1}).$ Suppose $n_1,\dots ,n_{i-1}$ are chosen. Since $q$ is a limit point of $E^{\ast },$ there is an $x\in E^{\ast }$ with Since $x\in E^{\ast }$ and is thus the limit (sequence) of some subsequence of $\{p_n\},$ there is an such that Now, via the triangle inequality,

This means that $\{p_{n_i}\}$ converges to $q,$ because we can find a $p_{n_i}$ as close as desired to $q.$ Therefore $q$ is a subsequential limit of $\{p_n\}$ and so $q\in E^{\ast },$ and $E^{\ast }$ is closed.

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Suppose $\{p_n\}$ is a Cauchy sequence. Then, let For some integer $N,$ when $n,m\ge N.$ Therefore, Since $ϵ$ was arbitrary, we can see that the sequence $\{\mathrm{diam}{E}_N\}$ converges to $0.$

Conversely, suppose $li{m}_{N\to \mathrm{\infty }}\mathrm{diam}{E}_N=0.$ Then, every neighborhood of $0$ contains $\{\mathrm{diam}{E}_N\}$ for all but finitely many N. Let Then, pick $N$ such that Letting $m,n\ge N,$ we have that

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Because $E\subset \bar{E},$ $\mathrm{diam}E\le \mathrm{diam}\bar{E}.$

Conversely, Let $p,q\in \bar{E},$ and $p^{\prime },q^{\prime }\in E,$ such that Therefore, by the triangle inequality,

Therefore, $\mathrm{diam}\bar{E}\le 2ϵ+\mathrm{diam}E,$ and since $ϵ$ was arbitrary, $\mathrm{diam}\bar{E}=\mathrm{diam}E.$

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Let $K=\bigcap _1^{\mathrm{\infty }}{K}_n.$ Then $K$ is not empty. Assume for the sake of contradiction that $K$ contains more than one point. Then, But, for each $n,K\subset K_n,$ so that $\mathrm{diam}{K}_n\ge \mathrm{diam}K.$ But, this contradicts our given that $lim+n\to \mathrm{\infty }\mathrm{diam}{K}_n=0,$ so our assumption that $K$ contains more than one point must be invalid, and thus $K$ contains exactly one point.

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Suppose $\{p_n\}$ is a convergent sequence in a metric space $X.$ Let Then for some when $n\ge N.$ Thus,

whenever $n,m\ge N,$ and so $\{p_n\}$ is Cauchy.

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Let $\{p_n\}$ be a cauchy sequence in the compact metric space $X.$ For $N=1,2,3,\dots ,$ let $E_n$ bet the set consisting of $p_N,p_{N+1},p_{N+2},\dots .$ Then

$$\underset{N\to \mathrm{\infty }}{lim}\mathrm{diam}\overline{E_N}=0,$$

by two theorems above. Each $\overline{E_n}$ is a closed subset of the compact space $X,$ and is thus compact. Also, $E_{N+1}\subset E_N,$ which implies that $\overline{E_{N+1}}\subset \overline{E_N}$. Now, we have that there is a unique $p\in X$ which lies in every $\overline{E_n}.$

Let Since $\underset{N\to \mathrm{\infty }}{lim}\mathrm{diam}\overline{E_N}=0,$ there is an integer $N_0$ such that if $N\ge N_0.$ Since $p\in \overline{E_n},$ we have that for every $q\in \overline{E_N}$ and thus for every $q\in E_n.$ That is, if $n\ge N_0,$ so $\{p_n\}$ converges to $p.$

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Let $\{{\mathbf{x}}_n\}$ be a cauchy sequence in $R^k.$ Define $E_N$ as in the proof above, but with ${\mathbf{x}}_i$ in place of $p_i.$ For some The range (sequence) of $\{{\mathbf{x}}_n\}$ is the union of $E_n$ and the finite set $\{{\mathbf{x}}_1,\dots ,{\mathbf{x}}_{N-1}\}.$ Hence, $\{{\mathbf{x}}_n\}$ is bounded (sequence) (since the finite set of points can be contained in some bounding box, and the remaining points can be contained in some ball with diameter 1.) Since $\stackrel{―}{\{{\mathbf{x}}_n\}}$ is bounded, it is compact, and thus $\{{\mathbf{x}}_n\}$ is a subset of a compact metric space (its closure), and so converges.

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Suppose $\{p_n\}\subset R^k$ converges. Then, because $R^k$ is a metric space, $\{p_n\}$ is cauchy.

Conversely, suppose $\{p_n\}\subset R^k$ is cauchy. Then, $\{p_n\}$ converges.

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Let $\{p_n\}\subset E$ be a cauchy sequence. Then, it converges to some point $p\in X,$ and actually $p\in E,$ because $E$ is closed. Therefore, $E$ is complete.

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