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\[||x-x_{k}||<\epsilon\quad\text{whenever }k\geq N\]
proving \(x_{k}\to x\quad\text{in }\mathbb{R}^{n}\).
**Example 24**.: Consider the sequence \(\{x_{k}\}\) in \(\mathbb{R}^{2}\), where \(x_{k}=\left(\dfrac{1}{k^{2}},\dfrac{1}{k^{3}}\right)\). The components of this sequence are \(\dfrac{1}{k^{2}}\) and \(\dfrac{1}{k^{3}}\), and each converges to zero; then, by the above theorem the sequence \(x_{k}=\left(\dfrac{1}{k^{2}},\dfrac{1}{k^{3}}\right)\to(0,0)\) in \(\mathbb{R}^{2}\).
**Remark 11**.: To describe the convergence of a sequence \(\{x_{n}\}\) to \(x_{0}\) in \(\mathbb{R}^{2}\), one can consider the open \(\epsilon\)-neighborhood of the point \(x_{0}\). Recall that \(U_{\epsilon}(x_{0})\), an open neighborhood of \(x_{0}\) in \(\mathbb{R}^{2}\), is defined as
\[U_{\epsilon}(x_{0})=\{x\in\mathbb{R}^{2}:||x-x_{0}||<\epsilon\}.\]
By \(||\cdot||\) we mean the Euclidean distance in \(\mathbb{R}^{2}\). As one can see from the following Figure 1.18, \(U_{\epsilon}(x_{0})\) contains all but a finite number of terms of \(\{x_{n}\}\).
Much of our discussion of sequences in \(\mathbb{R}\) still makes sense in \(\mathbb{R}^{n}\), but some do not. For example, there is no natural order imposed on \(\mathbb{R}^{n}\), and so the discussion of monotone sequences does not apply in this context.
### Subsequences
**Definition 13**.: Given a sequence \(\{x_{n}\}\) of real numbers, consider a sequence \(\{n_{k}\}\) of positive integers such that \(n_{1} \[x_{n_{1}},x_{n_{2}},x_{n_{3}},\ldots\] is called a **subsequence** of \(\{x_{n}\}\) and is denoted by \(\{x_{n_{k}}\}\), where \(k\in\mathbb{N}\) indexes the subsequence. Any given sequence \(\{x_{n}\}\) has many subsequences, and note that the order of the terms in a subsequence is the same as in the original sequence and repetitions are not allowed. Subsequences may or may not behave like the original sequence. For example, if \(\{x_{n}\}=1,-1,1,-1,\ldots\), then \(\{x_{n}\}\) has no single limit; thus it is divergent, but the subsequences \[x_{1},x_{3},x_{5},\ldots=1,1,1,\ldots\qquad x_{2},x_{4},x_{6},\cdots=-1,-1,-1,\ldots\] converge to \(1\) and \(-1\), respectively. On the other hand, the sequence \(\{x_{n}\}=\left\{\frac{1}{10^{n}}\right\}\) converges to zero and so does every one of its subsequences. These examples lead us to the following: **Proposition 2**.: _Let \(\{x_{n}\}\) be a sequence and \(L\) be a real number._ * _If_ \(\{x_{n}\}\) _converges to_ \(L\)_, then every subsequence_ \(\{x_{n_{k}}\}\) _converges to_ \(L\) _too._ Figure 18: Convergence of a sequence in \(\mathbb{R}^{2}\) : To prove a) let \(\epsilon>0\) be given; since \(\{x_{n}\}\) converges to \(L\), there exists \(N\) such that \(|x_{n}-L|<\epsilon\) whenever \(n\geq N\). Note that for any subsequence, \(n_{k}\geq k\), so this very same \(N\) also works for the subsequence \(\{x_{n_{k}}\}\). If \(k>N\), then \[n_{k}\geq k>N,\quad\text{and}\quad|x_{n_{k}}-L|<\epsilon.\] The proof of part b) follows from part a). Indeed, \(\{x_{n}\}\to L\) if and only if all subsequences of \(\{x_{n}\}\) converge to \(L\). **Theorem 15** (The Bolzano-Weierstrass Theorem).: _Every bounded sequence of real numbers has a convergent subsequence._ : Suppose \(\{x_{n}\}\) is a bounded sequence in \(\mathbb{R}\) so that there exists \(M>0\) with \(-M\leq x_{n}\leq M\) for every \(n\). Bisect the interval \([-M,M]\) into two closed intervals \([-M,0]\) and \([0,M]\). At least one of these closed intervals must contain infinitely many \(x_{n}\). Select a half for which this is the case label that interval by \(I_{0}\), and select \(n_{0}\) for which \(x_{n_{0}}\in I_{0}\). Next, split \(I_{0}\) into closed intervals of equal length, and let \(I_{1}\) be the half that again contains infinitely many points of \(x_{n}\). As there are infinitely many \(x_{n}\)’s available, we can select \(n_{1}>n_{0}\) with the property that \(x_{n_{1}}\in I_{1}\) (Figure 1.19). Continue these processes to obtain subintervals, indices, and points. From this bisection processes we obtain the following: * \(I_{0}\supseteq I_{1}\supseteq I_{2}\supseteq\cdots\) (a nested set of closed and bounded intervals). Figure 1.19: The bisection process used in the proof of Bolzano–Weierstrass Theorem Next, we claim that this subsequence \(\{x_{n_{k}}\}\) is the convergent subsequence we have been looking for. However, we must first catch its limit \(x\). The Nested Intervals Property comes in handy here, which guarantees the existence of at least one point \(x\in\mathbb{R}\) contained in every \(I_{k}\), i.e., \(\bigcap_{n=1}^{\infty}I_{n}\neq\emptyset\). We claim \(x_{n_{k}}\to x\). If we label \(I_{k}=[a_{k},b_{k}]\) and consider the left hand of these intervals, we see a sequence \(a_{1},a_{2},a_{3},\dots\) Since \(I_{k+1}\subseteq I_{k}\subseteq[-M,M]\), we have \[-M\leq a_{k}\leq a_{k+1}\leq M\quad\text{for every $k$}.\] This sequence is monotone increasing and bounded, so by the Monotone Convergence Theorem it must converge to some number \(x\). Given \(\epsilon>0\), there is an integer \(K_{1}\) such that \(|a_{k}-x|<\epsilon\) whenever \(k\geq K_{1}\). Now observe that for each \(k\), we have \[|x_{n_{k}}-x|\leq|x_{n_{k}}-a_{k}|+|a_{k}-x|\leq\frac{M}{2^{k}}+|a_{k}-x|.\] Let \(\epsilon>0\). By construction the length of \(I_{k}=b_{k}-a_{k}=\frac{M}{2^{k}}\) and \(\frac{1}{2^{k}}\to 0\quad\text{as}\quad k\to\infty\). Therefore, there is an integer \(K_{2}\) such that \(\frac{M}{2^{k}}<\frac{\epsilon}{(2M)}\) whenever \(k\geq K_{2}\). Let \(K=\max\{K_{1},K_{2}\}\); thus if \(k\geq K\), we must have \[|x_{n_{k}}-x|\leq\frac{M}{2^{k}}+|a_{k}-x|<\frac{\epsilon}{2}+\frac{\epsilon}{ 2}=\epsilon.\qed\] ### Cauchy Sequences **Definition 14**.: A **Cauchy sequence** of real numbers is a sequence \(\{x_{n}\}\) with the property that for each \(\epsilon>0\) there is an integer \(N\) depending on \(\epsilon\) such that \[|x_{n}-x_{m}|<\epsilon\quad\text{for all}\quad n,m\geq N.\] Notice that this definition is similar to the definition of a convergent sequence, but there is no mention of the limit \(L\). Furthermore, the definition of a Cauchy sequence suggests that the terms of the sequence are “bunching up,” but it requires that **all** terms with large enough index are close to one another, not just consecutive terms. Sometimes in the definition the condition \(n>m\geq N\) is used. This does not change anything since \(n=m\) is the trivial case. Consider the sequence \(\left\{\frac{1}{n}\right\}\); clearly it is convergent, and we can guess that terms of this sequence “bunch up.” Let \(\epsilon>0\) be given, set \(N=\frac{1}{\epsilon}\), and since if \(n>m>N\), then \[|x_{n}-x_{m}|=\left|\frac{1}{m}-\frac{1}{n}\right|=\frac{1}{m}-\frac{1}{n}< \frac{1}{m}<\frac{1}{\frac{1}{\epsilon}}=\epsilon.\] There is nothing special about the convergent sequence \(\left\{\frac{1}{n}\right\}\), since, as shown in the next lemma, convergent sequences are Cauchy sequences. **Lemma 3**.: _Every convergent sequence is a Cauchy sequence._ : Assume \(\{x_{n}\}\) converges to \(L\). Then given \(\epsilon>0\), we can choose \(N\) such that \(|x_{n}-L|<\frac{\epsilon}{2}\) whenever \(n\geq N\). Now we apply the triangle inequality to \(|x_{n}-x_{m}|\) and observe that for \(n>m>N\), we have \[|x_{n}-x_{m}|\leq|x_{n}-L|+|x_{m}-L|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\] Thus \(\{x_{n}\}\) is a Cauchy sequence. As expected, if both \(x_{n}\) and \(x_{m}\) are close to a limit \(L\), then they are close to each other; thus every convergent sequence is Cauchy. The converse, however, is not so obvious. In fact for a Cauchy sequence to be convergent, we must have a limit \(L\) to approach. We will use the Bolzano-Weierstrass Theorem for the converse. For this purpose we first claim that Cauchy sequences are bounded. **Lemma 4**.: _Every Cauchy sequence is bounded._ : Choose \(\epsilon=1\), and refer to the definition of the Cauchy sequence to conclude that for some integer \(N\) the inequality \(|x_{m}-x_{n}|<1\) holds for all \(n,m\geq N\). Thus we must have \[|x_{n}|\leq|x_{N}|+|x_{n}-x_{N}|\leq|x_{N}|+1\] for all \(n\geq N\). But this implies that the sequence is bounded, since \[|x_{n}|\leq\max\{|x_{1}|,|x_{2}|,\dots,|x_{N}|,|x_{N}|+1\}\] for all \(n\in\mathbb{N}\). **Theorem 16** (Cauchy Criterion).: _Every Cauchy sequence of real numbers is convergent._
* _If_ \(\{x_{n}\}\) _has subsequences converging to different limits, then_ \(\{x_{n}\}\) _diverges._
Proof
Proof
* The length of \(I_{k}\) is \(\frac{M}{2^{k}}\).
* Increasing indices \(n_{0}
Proof
Proof
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