Giovanni Antonio Canal (Canaletto) (1697-1768) was an Italian painter.
By a finite sequence of \(n\) terms, we mean that a function \(s\) whose domain is the set of numbers \(\{1,2,\cdots ,n\}\). The range of \(s\) is the set \({s(1),s(2),\cdots ,s(n)}\), which is usually written as \(\{s_{1},s_{2},\cdots ,s_{n}\}\). By a infinite sequence, we mean that a function \(s\) whose domain is the set \(\{1,2,\cdots\}\) of all positive integers. The range of \(s\) is the set \(\{s(1),s(2),\cdots, s(n),\cdots\}\), which is also usually written as \(\{s_{1},s_{2},\cdots ,s_{n},\cdots\}\). We may also simply write \(s=\{s_{n}\}_{n=1}^{\infty}\).
Let \(k\) be another function whose domain is the set of positive integers and whose range is a subset of the positive integers satisfying \(k(m)<k(n)\) if \(m<n\). For \(n\geq 1\), the composition function \((s\circ k)(n)=s_{k(n)}\) is said to be a subsequence of \(s\). We also simply write \(\{s_{k_{n}}\}_{n=1}^{\infty}\) to denote a subsequence of \(\{s_{n}\}_{n=1}^{\infty}\). For example, let \(s=\{1/n\}\). We define \(k(n)=2^{n}\). Then \(s\circ k=\{1/2^{n}\}\) is a subsequence of \(s\). In \(\mathbb{R}\), we recall that the limit,
\[\lim_{n\rightarrow\infty}x_{n}=a\]
means that, given any \(\epsilon >0\), there is an integer \(N\) such that \(n\geq N\) implies \(|x_{n}-a|<\epsilon\). In what follows, we are going to study these concepts in metric space.
Definition. Let \(\{x_{n}\}_{n=1}^{\infty}\) be a sequence in metric space \((M,d)\). We say that the sequence \(\{x_{n}\}_{n=1}^{\infty}\) is convergent when there exists \(a\in M\) such that the following property is satisfied: given any \(\epsilon >0\), there exists an integer \(N\) such that \(n\geq N\) implies \(d(x_{n},a)<\epsilon\). In this case, we also say that the sequence \(\{x_{n}\}_{n=1}^{\infty}\) converges to \(a\) and is denoted by
\[\lim_{n\rightarrow\infty}x_{n}=a\mbox{ or }x_{n}\rightarrow a\mbox{ as }n\rightarrow\infty.\]
When there is no such \(a\in M\), we say that the sequence \(\{x_{n}\}_{n=1}^{\infty}\) is divergent.
\begin{equation}{\label{ma84}}\tag{1}\mbox{}\end{equation}
Remark \ref{ma84}. Let \(\{x_{n}\}_{n=1}^{\infty}\) be a sequence in \((M,d)\). It is clear to see that
\[\lim_{n\rightarrow\infty}x_{n}=a\mbox{ if and only if }\lim_{n\rightarrow\infty}d(x_{n},a)=0\]
where \(\{d(x_{n},a)\}_{n=1}^{\infty}\) is a sequence in \(\mathbb{R}\). \(\sharp\)
In \(\mathbb{R}\), it is easy to see
\[\lim_{n\rightarrow\infty}\frac{1}{n}=0.\]
Suppose that we take \(M=(0,1]\). Then, the sequence \(\{x_{n}=1/n\}_{n=1}^{\infty}\) is not convergent, since the only candidate for the limit is \(0\), which is not in \(M\). This says that the convergence and divergence in metric space \((M,d)\) depends on \(M\) itself.
Proposition. The limit of a sequence \(\{x_{n}\}_{n=1}^{\infty}\) in metric space \((M,d)\) is unique.
Proof. Suppose that
\[\lim_{n\rightarrow\infty}x_{n}=p\mbox{ and }\lim_{n\rightarrow\infty}x_{n}=q.\]
We shall prove \(p=q\). Remark \ref{ma84} says
\[\lim_{n\rightarrow\infty}d(x_{n},p)=0=\lim_{n\rightarrow\infty}d(x_{n},q).\]
Using the triangle inequality, we have
\[0\leq d(p,q)\leq d(p,x_{n})+d(x_{n},q),\]
which implies
\[0\leq d(p,q)\leq\lim_{n\rightarrow\infty}d(p,x_{n})+\lim_{n\rightarrow\infty}d(x_{n},q)=0.\]
Therefore, we obtain \(d(p,q)=0\), i.e., \(p=q\). This completes the proof. \(\blacksquare\)
The concept of adherent point can refer to the page point set topology in metric space.
Proposition. Suppose that the sequence \(T=\{x_{n}\}_{n=1}^{\infty}\) in metric space \((M,d)\) converges to \(p\). Then, the following statements hold true.
(i) \(T\) is bounded.
(ii) \(p\) is an adherent point of \(T\).
Proof. To prove part (i), since \(x_{n}\rightarrow p\), given \(\epsilon =1\), there exists an integer \(N\) satisfying \(x_{n}\in B(p;1)\) for each \(n\geq N\). We define
\[r=1+\max\left\{d(p,x_{1}),\cdots ,d(p,x_{N-1})\right\}.\]
Then, we have \(T\subseteq B(p,r)\), which says that \(T\) is bounded.
To prove part (ii), since every open ball \(B(p;\epsilon )\) contains a point of \(T\), this says that \(p\) is an adherent point of \(T\), and the proof is complete. \(\blacksquare\)
\begin{equation}{\label{ma85}}\tag{2}\mbox{}\end{equation}
Proposition \ref{ma85}. Let \((M,d)\) be a metric space, and let \(T\) be a subset of \(M\). Suppose that \(p\in M\) is an adherent point of \(T\). Then, there exists a sequence \(\{x_{n}\}_{n=1}^{\infty}\) in \(T\) that converges to \(p\).
Proof. Since \(p\) is an adherent point of \(T\), for each integer \(n\geq 1\), the open ball \(B_{M}(p;1/n)\) satisfying \(B_{M}(p;1/n)\cap T\neq\emptyset\). We take some point \(x_{n} \in B_{M}(p;1/n)\cap T\). Then, we have \(x_{n}\in T\) satisfying \(d(p,x_{n})\leq 1/n\). This shows
\[\lim_{n\rightarrow\infty}d(p,x_{n})=0.\]
Using Remark \ref{ma84}, we have
\[\lim_{n\rightarrow\infty}x_{n}=p\]
and the proof is complete. \(\blacksquare\)
Proposition. A sequence \(\{x_{n}\}_{n=1}^{\infty}\) in metric space \((M,d)\) converges to \(p\) if and only if every subsequence \(\{x_{k_{n}}\}_{n=1}^{\infty}\) converges to \(p\).
Proof. Suppose that \(x_{n}\rightarrow p\) as \(n\rightarrow\infty\). Let \(\{x_{k_{n}}\}_{n=1}^{\infty}\) be any subsequence of \(\{x_{n}\}_{n=1}^{\infty}\). Given any \(\epsilon >0\), there exists an integer \(N\) such that \(n\geq N\) implies \(d(x_{n},p)<\epsilon\). By the definition of subsequence, there exists an integer \(N^{*}\) satisfying \(k_{n}\geq N\) for \(n\geq N^{*}\). This says that \(n\geq N^{*}\) implies \(d(x_{k_{n}},p)<\epsilon\). Therefore, we obtain \(x_{k_{n}}\rightarrow p\) as \(n\rightarrow\infty\). The converse is obvious, since \(\{x_{n}\}_{n=1}^{\infty}\) is itself a subsequence, and the proof is complete. \(\blacksquare\)
Proposition. Let \(\{a_{n}\}_{n=1}^{\infty}\) be a convergent sequence, and let \({\sigma_{n}}_{n=1}^{\infty}\) a sequence of arithmetic means given by
\[\sigma_{n}=\frac{1}{n}\sum_{k=1}^{n}a_{n}.\]
Then, we have that
\[\lim_{n\rightarrow\infty}a_{n}=A\mbox{ implies }\lim_{n\rightarrow\infty}\sigma_{n}=A.\]
Proof. Let \(b_{n}=a_{n}-A\) and \(\tau_{n}=\sigma_{n}-A\). Then, we have
\[\tau_{n}=\frac{1}{n}\sum_{k=1}^{n}a_{k}-A=\frac{1}{n}\left (\sum_{k=1}^{n}a_{k}-nA\right )
=\frac{1}{n}\sum_{k=1}^{n}\left (a_{k}-A\right )=\frac{1}{n}\sum_{k=1}^{n}b_{k}.\]
Since \(\lim_{n\rightarrow\infty}b_{n}=0\), given any \(\epsilon>0\), there exists an integer \(N\) such that \(n\geq N\) implies \(|b_{n}|<\epsilon\). Let
\[M=\max\left\{\left |b_{1}\right |,\left |b_{2}\right |,\cdots,\left |b_{N}\right |\right\}.\]
Then, for \(n>N\), we have
\[|\tau_{n}|\leq\frac{1}{n}\sum_{k=1}^{N}\left |b_{k}\right |+\frac{1}{n}\sum_{k=N+1}^{n}
\left |b_{k}\right |\leq\frac{NM}{n}+\frac{n-N}{n}\epsilon<\frac{NM}{n}+\epsilon,\]
which implies
\[0\leq\lim_{n\rightarrow\infty}|\tau_{n}|\leq\epsilon,\]
Since \(\epsilon\) can be any positive number, it follows \(\lim_{n\rightarrow\infty}\tau_{n}=0\). This completes the proof. \(\blacksquare\)
\begin{equation}{\label{ma152}}\tag{3}\mbox{}\end{equation}
Example \ref{ma152}. Given a sequence \(\{I_{n}\}_{n=1}^{\infty}\) by
\[I_{n}=\int_{0}^{\pi/2}\cos^{n}xdx,\]
which is known as Wallis integral, it is clear to see
\[I_{0}=\frac{\pi}{2}\mbox{ and }I_{1}=1.\]
For \(n\geq 2\), using the integration by parts, we have
\begin{align*}I_{n+2} & =\int_{0}^{\pi/2}\cos^{n+1}x\cos xdx\\ & =\left [\cos^{n+1}x\sin x\right ]_{0}^{\pi/2}+(n+1)\int_{0}^{\pi/2}\cos^{n}x\sin^{2}xdx\\& =(n+1)\left (\int_{0}^{\pi/2}\cos^{n}xdx-\int_{0}^{\pi/2}\cos^{n+2}xdx\right )\\ & =(n+1)\left (I_{n}-I_{n+2}\right ),\end{align*}
which implies
\begin{equation}{\label{ma151}}\tag{4}I_{n+2}=\frac{n+1}{n+2}I_{n}.\end{equation}
Using induction, we can obtain
\begin{align*} I_{2n} & =\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdots\frac{1}{2}\cdot I_{0}\\ & =\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdots\frac{1}{2}\cdot\frac{\pi}{2}\\ & =\frac{(2n)!\pi}{2^{2n+1}(n!)^{2}}\end{align*}
and
\begin{align*} I_{2n+1} & =\frac{2n}{2n+1}\cdot\frac{2n-2}{2n-1}\cdots\frac{2}{3}\cdot I_{1}\\ & =\frac{2n}{2n+1}\cdot\frac{2n-2}{2n-1}\cdots\frac{2}{3}\\ & =\frac{2^{2n}(n!)^{2}}{(2n+1)!}.\end{align*}
Since
\[0\leq\cos^{n+1}x\leq\cos^{n}x\mbox{ for }0\leq x\leq\pi/2,\]
it follows
\[I_{n+2}\leq I_{n+1}\leq I_{n}.\]
Since \(I_{n}>0\), we obtain
\[1\leq\frac{I_{n+1}}{I_{n+2}}\leq\frac{I_{n}}{I_{n+2}}=\frac{n+2}{n+1},\]
which implies
\[\lim_{n\rightarrow\infty}\frac{I_{n+1}}{I_{n}}=1.\]
From (\ref{ma151}), we have
\[(n+2)I_{n+2}I_{n+1}=(n+1)I_{n+1}I_{n},\]
which implies
\begin{align*} (n+2)I_{n+2}I_{n+1} & =(n+1)I_{n+1}I_{n}=nI_{n}I_{n-1}\\ & =\cdots=I_{0}I_{1}=\frac{\pi}{2}.\end{align*}
Therefore, we obtain
\begin{align*} \lim_{n\rightarrow\infty}nI_{n}^{2} & =\lim_{n\rightarrow\infty}(n+1)I_{n+1}^{2}\\ & =\lim_{n\rightarrow\infty}(n+1)I_{n+1}I_{n}=\frac{\pi}{2},\end{align*}
which implies
\[\lim_{n\rightarrow\infty}2nI_{n}^{2}=\pi\mbox{ or }\lim_{n\rightarrow\infty}\sqrt{2n}I_{n}=\sqrt{\pi}.\]
Definition. A sequence \(\{x_{n}\}_{n=1}^{\infty}\) in metric space \((M,d)\) is called a Cauchy sequence when, given any \(\epsilon >0\), there exists an integer \(N\) such that \(n\geq N\) and \(m\geq N\) imply \(d(x_{n},x_{m})<\epsilon\). \(\sharp\)
\begin{equation}{\label{map430}}\tag{5}\mbox{}\end{equation}
Proposition \ref{map430}. Suppose that the sequence \(\{x_{n}\}_{n=1}^{\infty}\) in metric space \((M,d)\) is convergent. Then \(\{x_{n}\}_{n=1}^{\infty}\) is a Cauchy sequence.
Proof. Suppose that \(x_{n}\rightarrow p\) as \(n\rightarrow\infty\). Given any \(\epsilon >0\), there exists an integer \(N\) such that \(n\geq N\) implies \(d(x_{n},p)<\epsilon /2\). Therefore, we also have \(d(x_{m},p)<\epsilon /2\) for \(m\geq N\). For \(n\geq N\) and \(m\geq N\), by the triangle inequality, we obtain
\begin{align*} d(x_{n},x_{m}) & \leq d(x_{n},p)+d(p,x_{m})\\ & <\frac{\epsilon}{2}+\frac{\epsilon}{2}\\ & =\epsilon .\end{align*}
This completes the proof. \(\blacksquare\)
In general, the converse of Proposition \ref{map430} is not true. The counterexample is the sequence \({1/n}_{n=1}^{\infty}\) in the metric space \(((0,1],d)\), where \(M=(0,1]\). The sequence \({1/n}_{n=1}^{\infty}\) is a Cauchy sequence in \((0,1]\). However, it is not convergent. If \(M\) is taken to be the whole Euclidean space \(\mathbb{R}^{k}\), then the converse is true, which is shown below.
\begin{equation}{\label{mat432}}\tag{6}\mbox{}\end{equation}
Theorem \ref{mat432}. In the Euclidean space \(\mathbb{R}^{k}\), every Cauchy sequence is convergent.
Proof. Let \(T=\{{\bf x}_{1},{\bf x}_{2},\cdots ,{\bf x}{n},\cdots\}\) be a Cauchy sequence in \(\mathbb{R}^{k}\). We are going to show that \(T\) is bounded. Given \(\epsilon =1\), there exists an integer \(N\) satisfying \(\parallel {\bf x}_{n}-{\bf x}_{N}\parallel <1\) for each \(n\geq N\). This means that \({\bf x}_{n}\in B({\bf x}_{N};1)\) for each \(n\geq N\). We define
\[r=1+\max\left\{\parallel {\bf x}_{1}\parallel ,\cdots ,\parallel {\bf x}_{N}\parallel\right\}.\]
Then, we consider the following two cases
- For \(n\leq N\), we have \begin{align*} \parallel {\bf x}_{n}\parallel & \leq\max\left\{\parallel {\bf x}_{1}\parallel,\cdots ,\parallel {\bf x}_{N}\parallel\right\}\\ & <r.\end{align*}
- For \(n>N\), using the triangle inequality and the fact of \({\bf x}_{n}\in B({\bf x}_{N};1)\), we have \begin{align*} \parallel {\bf x}_{n}\parallel & \leq\parallel {\bf x}_{n}-{\bf x}_{N}\parallel + \parallel {\bf x}_{N}\parallel\\ & <1+\parallel {\bf x}_{N}\parallel\\ & \leq r.\end{align*}
Then, we have \(T\subseteq B({\bf 0},r)\), which says that \(T\) is bounded. Using the Bolzano-Weierstrass Theorem in page point set topology in Euclidean space, the set \(T\) has an accumulation point \({\bf p}\in\mathbb{R}^{k}\). Given any \(\epsilon >0\), Proposition \ref{ma85} says that \(\parallel {\bf x}_{m}-{\bf p}\parallel<\epsilon/2\) for sufficiently large \(m\). The Cauchy sequence says that we can take a sufficiently large integer \(N\) such that \(n\geq N\) and \(m\geq N\) implies \(\parallel {\bf x}_{n}-{\bf x}_{m}\parallel<\epsilon /2\) and \(\parallel {\bf x}_{m}-{\bf p}\parallel<\epsilon/2\). Therefore, if \(n\geq N\), we have
\begin{align*}\parallel {\bf x}_{n}-{\bf p}\parallel & \leq\parallel {\bf x}_{n}-{\bf x}_{m} \parallel +\parallel {\bf x}_{m}-{\bf p}\parallel\\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2}\\ & =\epsilon .\end{align*}
This shows \({\bf x}_{n}\rightarrow {\bf p}\), and the proof is complete. \(\blacksquare\)
Example. Given a sequence \(\{a_{n}\}_{n=1}^{\infty}\) in \(\mathbb{R}\) satisfying
\[\left |a_{n+2}-a_{n+1}\right |\leq\frac{1}{2}\left |a_{n+1}-a_{n}\right |\mbox{ for }n\geq 1,\]
we are going to show that this sequence is convergent without knowing its limit. Let \(b_{n}=|a_{n+1}-a_{n}|\). Then, we have
\[0\leq b_{n+1}\leq\frac{b_{n}}{2}.\]
By induction on \(n\), we can obtain
\[b_{n}\leq\frac{b_{1}}{2^{n}},\]
which implies
\[\lim_{n\rightarrow\infty}b_{n}=0.\]
For \(m>n\), we have
\[a_{m}-a_{n}=\sum_{k=n}^{m-1}\left (a_{k+1}-a_{k}\right ),\]
which implies
\begin{align*} \left |a_{m}-a_{n}\right | & \leq\sum_{k=n}^{m-1}b_{k}\leq b_{n}\left (1+\frac{1}{2}+\cdots +\frac{1}{2^{m-1-n}}\right )\\ & <2b_{n}\rightarrow 0\mbox{ as }n\rightarrow\infty.\end{align*}
It follows that \(\{a_{n}\}_{n=1}^{\infty}\) is a Cauchy sequence. Theorem \ref{mat432} says that the sequence \(\{a_{n}\}_{n=1}^{\infty}\) is convergent. \(\sharp\)
Example. Given a sequence \(\{x_{n}\}_{n=1}^{\infty}\) defined by
\[x_{n}=\int_{1}^{n}\frac{\cos t}{t^{2}}dt,\]
we are going to show that this sequence is convergent. Given any \(\epsilon >0\), there exists an integer \(n_{0}\) satisfying
\[n\geq n_{0}\mbox{ implies }\frac{1}{n}<\epsilon .\]
Therefore, for \(m>n\), we have
\begin{align*} \left |x_{n}-x_{m}\right | & =\left |\int_{n}^{m}\frac{\cos t}{t^{2}}dt\right |\\ & \leq\int_{n}^{m}\frac{|\cos t|}{t^{2}}\leq\int_{n}^{m}\frac{1}{t^{2}}dt\\ & =\frac{1}{n}-\frac{1}{m}\\ & <\frac{1}{n}<\epsilon.\end{align*}
This shows that \({a_{n}}_{n=1}^{\infty}\) is a Cauchy sequence. Theorem \ref{mat432} says that the sequence \({a_{n}}_{n=1}^{\infty}\) is convergent. \(\sharp\)
Definition. A metric space \((M,d)\) is called complete when every Cauchy sequence in \(M\) is convergent. A subset \(T\) of \(M\) is called complete when the metric subspace \((T,d)\) is complete. \(\sharp\)
Example. Theorem \ref{mat432} says that the Euclidean space \(\mathbb{R}^{k}\) is complete. In particular, the real number system \(\mathbb{R}\) is complete. However, the subspace \(T=(0,1]\) is not complete. On the other hand, we define the metric
\[d({\bf x},{\bf y})=\max_{i=1,\cdots ,k}\left |x_{i}-y_{i}\right |.\]
Then, we can show that \((\mathbb{R}^{k},d)\) is also complete. \(\sharp\)
Theorem. Every compact subset \(T\) of a metric space \((M,d)\) is complete.
Proof. Let \(A=\{x_{n}\}_{n=1}^{\infty}\) be a Cauchy sequence in \(T\). Theorem 5 in page point set topology in metric space says that the infinite subset \(A\) of \(T\) has an accumulation point \(p\in T\). Given any \(\epsilon >0\), Proposition \ref{ma85} says that \(d(x_{m},p)<\epsilon/2\) for sufficiently large \(m\). The Cauchy sequence says that we can take a sufficiently large integer \(N\) such that \(n\geq N\) and \(m\geq N\) implies \(d(x_{n},x_{m})<\epsilon /2\) and \(d(x_{m},p)<\epsilon/2\). Therefore, for \(n\geq N\), the triangle inequality says
\begin{align*} d(x_{n},p) & \leq d(x_{n},x_{m})+d(x_{m},p)\\ & <\frac{\epsilon}{2}+\frac{\epsilon}{2}\\ & =\epsilon .\end{align*}
This shows that \(x_{n}\rightarrow p\), and the proof is complete. \(\blacksquare\)
