2.4: The Bolazno-Weierstrass Theorem
- Page ID
- 49102
The Bolzano-Weierstrass Theorem is at the foundation of many results in analysis. it is, in fact, equivalent to the completeness axiom of the real numbers.
Every bounded sequence \(\left\{a_{n}\right\}\) of real numbers has a convergent subsequence.
- Proof
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Suppose \(\left\{a_{n}\right\}\) is a bounded sequence. Define \(A=\left\{a_{n}: n \in \mathbb{N}\right\}\) (the set of values of the sequence \(\left\{a_{n}\right\}\)). If \(A\) is finite, then at least one of the elements of \(A\), say \(x\), must be equal to \(a_{n}\) for infinitely many choices \(n\). More precisely, \(B_{x}=\left\{n \in \mathbb{N}: a_{n}=x\right\}\) is infinite. We can then define a convergent subsequence as follows. Pick \(n_{1}\) such that \(a_{n_{1}}=x\). Now, since \(B_{x}\) is infinite, we can choose \(n_{2}>n_{1}\) such that \(a_{n_{2}}=x\). Continuing in this way, we can define a subsequence \(\left\{a_{n_{k}}\right\}\) which is constant, equal to \(x\) and, thus, converges to \(x\).
Suppose now that \(A\) is infinite. First observe there exist \(c, d \in \mathbb{R}\) such that \(c \leq a_{n} \leq d\) for all \(n \in \mathbb{N}\), that is, \(A \subset[c, d]]\).
We define a sequence of nonempty nested closed bounded intervals as follows. Set \(I_{1}=[c, d]\). Next consider the two subintervals \(\left[c, \frac{c+d}{2}\right]\) and \(\left[\frac{c+d}{2}, d\right]\). Since \(A\) is infinite, at least one of \(A \cap\left[c, \frac{c+d}{2}\right]\) or \(A \cap\left[\frac{c+d}{2}, d\right]\) is infinite. Let \(I_{2}=\left[c, \frac{c+d}{2}\right]\) if \(A \cap\left[c, \frac{c+d}{2}\right]\) is infinte and \(I_{2}=\left[\frac{c+d}{2}, d\right]\) otherwise. Continuing in this way, we construct a nested sequence of nonempty cllosed bounded intervals \(\left\{I_{n}\right\}\) such that \(I_{n} \cap A\) is infinite and the length of \(I_{n}\) tends to 0 as \(n \rightarrow \infty\).
We now construct the desired subsequence of \(\left\{a_{n}\right\}\) as follows. Let \(n_{1}=1\). Choose \(n_{2}>n_{1}\) such that \(a_{n_{2}} \in I_{2}\). This is possible since \(I_{2} \cap A\) is infinite. Next choose \(n_{3}>n_{2}\) such that \(a_{n_{3}} \in I_{3}\). In this way, we obtain a subsequence \(\left\{a_{n_{k}}\right\}\) such that \(a_{n_{k}} \in I_{k}\) for all \(k \in \mathbb{N}\).
Set \(I_{n}=\left[c_{n}, d_{n}\right]\). Then \(\lim _{n \rightarrow \infty}\left(d_{n}-c_{n}\right)=0\). We also know from the proof of the Monotone Convergence Theorem (Theorem 2.3.1), that \(\left\{c_{n}\right\}\) converges. Say \(\ell=\lim _{n \rightarrow \infty} c_{n}\). Thus, \(\lim _{n \rightarrow \infty} d_{n}=\lim _{n \rightarrow \infty}\left[\left(d_{n}-c_{n}\right)+c_{n}\right]=\ell\) as well. Since \(c_{k} \leq a_{n_{k}} \leq d_{k}\) for all \(k \in \mathbb{N}\), it follows from Theorem 2.1.5 that \(\lim _{k \rightarrow \infty} a_{n_{k}}=\ell\). This completes the proof. \(\square\)
A sequence \(\left\{a_{n}\right\}\) of real numbers is called a Cauchy sequence if for any \(\varepsilon>0\), there exists a positive integer \(N\) such that for any \(m, n \geq N\), one has
\[\left|a_{m}-a_{n}\right|<\varepsilon.\]
A convergent sequence is a Cauchy sequence.
- Proof
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Let \(\left\{a_{n}\right\}\) be a convergent sequence and let
\[\lim _{n \rightarrow \infty} a_{n}=a.\]
Then for any \(\varepsilon>0\), there exists a positive integer \(N\) such that
\[\left|a_{n}-a\right|<\varepsilon / 2 \text { for all } n \geq N.\]
For any \(m, n \geq N\), one has
\[\left|a_{m}-a_{n}\right| \leq\left|a_{m}-a\right|+\left|a_{n}-a\right|<\varepsilon / 2+\varepsilon / 2=\varepsilon.\]
Thus, \(\left\{a_{n}\right\}\) is a Cauchy sequence. \(\square\)
A Cauchy sequence is bounded.
- Proof
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Let \(\left\{a_{n}\right\}\) be a Cauchy sequence. Then for \(\varepsilon=1\), there exists a positive integer \(N\) such that
\[\left|a_{m}-a_{n}\right|<1 \text { for all } m, n \geq N\]
In particular,
\[\left|a_{n}-a_{N}\right|<1 \text { for all } n \geq N.\]
Let \(M=\max \left\{\left|a_{1}\right|, \ldots,\left|a_{N-1}\right|, 1+\left|a_{N}\right|\right\}\). Then, for \(n=1, \ldots, N-1 \text {, we clearly have } \left|a_{n}\right| \leq M\).Moreover, for \(n \geq N\),
\[\left|a_{n}\right|=\left|a_{n}-a_{N}+a_{N}\right| \leq\left|a_{n}-a_{N}\right|+\left|a_{N}\right| \leq 1+\left|a_{N}\right| \leq M.\]
Therefore, \(\left|a_{n}\right| \leq M\) for all \(n \in \mathbb{N}\) and, thus, \(\left\{a_{n}\right\}\) is bounded. \(\square\)
A Cauchy sequence that has a convergent subsequence is convergent.
- Proof
-
Let \(\left\{a_{n}\right\}\) be a Cauchy sequence that has a convergent subsequence. For any \(\varepsilon>0\), there exists a positive integer \(N\) such that
\[\left|a_{m}-a_{n}\right| \leq \varepsilon / 2 \text { for all } m, n \geq N.\]
Thus, we can find a positive integer \(n_{\ell}>N\) such that
\[\left|a_{n_{\ell}}-a\right|<\varepsilon / 2\).
Then for any \(n \geq N\), we have
\[\left|a_{n}-a\right| \leq\left|a_{n}-a_{n_{\ell}}\right|+\left|a_{n_{\ell}}-a\right|<\varepsilon.\]
Therefore, \(\left\{a_{n}\right\}\) converges to \(a\). \(\square\)
\(\left\{a_{n}\right\}\)Any Cauchy sequence of real numbers is convergent.
- Proof
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Let \(\left\{a_{n}\right\}\) be a Cauchy sequence. Then it is bounded by Theorem 2.4.3. By the Bolzano-Weierstrass theorem, \(\left\{a_{n}\right\}\) has a convergent subsequence. Therefore, it is convergent by Lemma 2.4.4. \(\square\)
It follows from Definition 2.4.1 that \(\left\{a_{n}\right\}\) is a Cauchy sequence if and only if for every \(\varepsilon>0\), there exists \(N \in \mathbb{N}\) such that
\[\left|a_{n+p}-a_{n}\right|<\varepsilon \text { for all } n \geq N \text { and for all } p \in \mathbb{N}.\]
A sequence \(\left\{a_{n}\right\}\) is called contractive if there exists \(k \in[0,1)\) such that
\[\left|a_{n+2}-a_{n+1}\right| \leq k\left|a_{n+1}-a_{n}\right| \text { for all } n \in \mathbb{N}.\]
Every contractive sequence is convergent.
- Proof
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By induction, one has
\[\left|a_{n+1}-a_{n}\right| \leq k^{n-1}\left|a_{2}-a_{1}\right| \text { for all } n \in \mathbb{N}\]
Thus,
\[\begin{aligned}
\left|a_{n+p}-a_{n}\right| & \leq\left|a_{n+1}-a_{n}\right|+\left|a_{n+2}-a_{n+1}\right|+\cdots+\left|a_{n+p}-a_{n+p-1}\right| \\
& \leq\left(k^{n-1}+k^{n}+\cdots+k^{n+p-2}\right)\left|a_{2}-a_{1}\right| \\
& \leq k^{n-1}\left(1+k+k^{2}+\cdots+k^{p-1}\right)\left|a_{2}-a_{1}\right| \\
& \leq \frac{k^{n-1}}{1-k}\left|a_{2}-a_{1}\right|
\end{aligned}.\]for all \(n,p \in \mathbb{N}\). Since \(k^{n-1} \rightarrow 0\) as \(n \rightarrow \infty\) (independently of \(p\)), this implies \(\left\{a_{n}\right\}\) is a Cauchy sequence and, hence, it is convergent. \(\square\)
The condition \(k<1\) in the previous theorem is crucial. Consider the following example. Let \(a_{n}=\ln n\) for all \(n \in \mathbb{N}\).
Solution
Since \(1<\frac{n+2}{n+1}<\frac{n+1}{n}\) for all \(n \in \mathbb{N}\) and the natural logarithm is an increasing function, we have
\[\begin{array}{c} \left|a_{n+2}-a_{n+1}\right|=|\ln (n+2)-\ln (n+1)|=\left|\ln \left(\frac{n+2}{n+1}\right)\right|=\ln \left(\frac{n+2}{n+1}\right) \\
<\ln \left(\frac{n+1}{n}\right)=|\ln (n+1)-\ln n|=\left|a_{n+1}-a_{n}\right|
\end{array}. \nonumber\]
Therefore, the inequality in Definition 2.4.2 is satisfied with \(k=1\), yet the sequence \(\{\ln n\}\) does not converge.
Exercise \(\PageIndex{1}\)
Determine which of the following are Cauchy sequences.
- \(a_{n}=(-1)^{n}\).
- \(a_{n}=(-1)^{n} / n\).
- \(a_{n}=n /(n+1)\).
- \(a_{n}=(\cos n) / n\).
Exercise \(\PageIndex{2}\)
Prove that the sequence
\[a_{n}=\frac{n \cos \left(3 n^{2}+2 n+1\right)}{n+1}. \nonumber\]
has a convergent subsequence
Exercise \(\PageIndex{3}\)
Let \(f:[0, \infty) \rightarrow \mathbb{R}\) be such that \(f(x)>0\) for all \(x\). Define
\[a_{n}=\frac{f(n)}{f(n)+1}. \nonumber\]
Prove that the sequence \(a_{n}\) has a convergent subsequence
Exercise \(\PageIndex{4}\)
Define
\[a_{n}=\frac{1+2^{n}}{2^{n}} \text { for } n \in \mathbb{N}. \nonumber\]
Prove that the sequence \(a_{n}\) is contractive
Exercise \(\PageIndex{5}\)
Let \(r \in \mathbb{R}\) be such that \(|r|<1\). Define \(a_{n}=r^{n}\) for \(n \in \mathbb{N}\). Prove that the sequence \(\left\{a_{n}\right\}\) is contractive
Exercise \(\PageIndex{6}\)
Prove that the sequence \(\left\{\frac{1}{n}\right\}_{n=1}^{\infty}\) is not contractive