Skip to main content

Section 2.3 Bolzano-Weierstrass Theorem

Consider a sequence \((x_n)\text{.}\) For each \(k \in \N\text{,}\) let \(S_k := \{x_n : n \ge k\}\) be the set of all terms of \((x_n)\) with index at least \(k\text{.}\) Clearly \(S_k \supseteq S_{k+1}\text{,}\) and so
\begin{equation*} \inf S_{k} \le \inf S_{k+1} \le \sup S_{k+1} \le \sup S_{k} \end{equation*}
as extended real numbers. Since any two of the sets \(S_k\) differ by only finitely many terms, either all of them are bounded or none of them is. In the bounded case, the real sequence \((\sup S_k)_k\) is decreasing and bounded below by \(\inf S_1\text{.}\) By the Monotone Convergence Theorem, this sequence has a limit. That limit is called the limit superior of \((x_n)\) and is denoted by \(\limsup x_n\text{.}\) Likewise, the limit inferior of \((x_n)\) is defined to be the limit of the real sequence \((\inf S_k)_k\text{.}\)
If the sets \(S_k\) are not bounded above, then each \(\sup S_k = +\infty\text{,}\) and we define \(\limsup x_n = +\infty\text{.}\) Likewise, if the sets \(S_k\) are not bounded below, then each \(\inf S_k = -\infty\text{,}\) and we define \(\liminf x_n = -\infty\text{.}\)

Proof.
Suppose first that \(x_n \to L\text{.}\) Let \(\varepsilon \gt 0\text{.}\) Choose \(N\in\N\) such that for all \(n\ge N\text{,}\)
\begin{equation*} L -\varepsilon \lt x_n \lt L + \varepsilon. \end{equation*}
Then for any \(k \ge N\text{,}\) \(S_k \subseteq (L-\varepsilon, L+\varepsilon)\) and so
\begin{equation*} L-\varepsilon \le \inf S_k \le \sup S_k \le L+\varepsilon. \end{equation*}
Therefore, \(\limsup x_n=\lim_k \sup S_k\) is arbitrary closed to \(L\) and hence must be \(L\text{.}\) Likewise, \(\liminf x_n = L\) as well.
Conversely, suppose \(\liminf x_n = \limsup x_n = L\text{.}\) Since \(x_k\in S_k\text{,}\) we have
\begin{equation*} \inf S_k \le x_k \le \sup S_k \end{equation*}
for every \(k\text{.}\) By the squeeze lemma, \(x_k \to L\text{.}\) Therefore \((x_n)\) is convergent.
One of the main ways in which the completeness of \(\R\) enters the study of sequences is through the following theorem.

Proof.
For a bounded sequence \((x_n)\text{,}\) we construct a subsequence as follows: set \(n_1:=1\text{.}\) Suppose natural numbers
\begin{equation*} n_1 \lt n_2 \ldots \lt n_k \end{equation*}
have been chosen for some \(k \ge 1\text{.}\) By the definition of supremum, there is some natural number \(n_{k+1} \ge n_k + 1 \gt n_k\) such that
\begin{equation} \sup S_{n_k +1} - \frac{1}{k} \le x_{n_{k+1}} \le \sup S_{n_k + 1}.\tag{2.3.1} \end{equation}
This inductively defines a strictly increasing sequence \((n_k)\) of natural numbers and hence a subsequence \((x_{n_k})\) of \((x_n)\text{.}\)
As a subsequence of the convergent sequence \((\sup S_n)\text{,}\) \((\sup S_{n_k+1})\) also converges to \(\limsup x_n\) by Proposition 2.1.10. Since the sequence \((\sup S_{n_k+1} - 1/k)\) has the same limit, the squeeze lemma applied to (2.3.1) shows that \(x_{n_{k+1}} \to \limsup x_n\text{.}\)
In fact, we proved a stronger statement: every bounded sequence contains a subsequence that converges to its limit superior, and likewise a subsequence that converges to its limit inferior. With a slight modification of the proof, one can also show that if a sequence is not bounded above (respectively, below), then it has a subsequence that diverges to \(+\infty\) (respectively, \(-\infty\)).

Proof.
For each \(n\in\N\text{,}\) let \(S_n:=\{x_m:m\ge n\}\text{,}\) and for each \(k\in\N\text{,}\) let
\begin{equation*} T_k:=\{x_{n_j}:j\ge k\}. \end{equation*}
Then \(T_k\subseteq S_{n_k}\text{,}\) so
\begin{equation*} \inf S_{n_k} \le \inf T_k \le \sup T_k \le \sup S_{n_k} \end{equation*}
for every \(k\text{.}\)
If \((x_n)\) is not bounded above, then \(\sup S_n=+\infty\) for every \(n\text{,}\) and so \(\limsup_n x_n=+\infty\text{.}\) In that case the last inequality is automatic. Otherwise \((x_n)\) is bounded above, so \((\sup S_n)\) is a real decreasing sequence converging to \(\limsup_n x_n\text{.}\) Since \((\sup S_{n_k})\) is a subsequence of \((\sup S_n)\text{,}\) it has the same limit by Proposition 2.1.10. Also, \(\sup T_k \le \sup S_{n_k}\) for every \(k\text{,}\) so \(\limsup_k x_{n_k} \le \limsup_n x_n\text{.}\)
Similarly, if \((x_n)\) is not bounded below, then \(\inf S_n=-\infty\) for every \(n\text{,}\) and so \(\liminf_n x_n=-\infty\text{.}\) In that case the first inequality is automatic. Otherwise \((x_n)\) is bounded below, so \((\inf S_n)\) is a real increasing sequence converging to \(\liminf_n x_n\text{.}\) Since \((\inf S_{n_k})\) is a subsequence of \((\inf S_n)\text{,}\) it has the same limit. Also, \(\inf S_{n_k} \le \inf T_k\) for every \(k\text{,}\) so \(\liminf_n x_n \le \liminf_k x_{n_k}\text{.}\)
Finally, \(\inf T_k \le \sup T_k\) for every \(k\text{,}\) so \(\liminf_k x_{n_k} \le \limsup_k x_{n_k}\text{.}\) Combining the three inequalities gives the result.
The follow application of the Bolzano-Weierstrass theorem will be used in our discussion of convergence tests for series later.

Proof.
The middle inequality is trivial. We will proof the right inequality. The left inequality has a similar proof, left as exercise. Let
\begin{equation*} L := \limsup_n \left|\frac{x_{n+1}}{x_n}\right| \qquad\text{and}\qquad s := \limsup_n |x_n|^{1/n}. \end{equation*}
If \(L = +\infty\text{,}\) then the inequality is automatic, so assume that \(L\) is finite.
Let \(L' \gt L\text{.}\) By the definition of limit superior, there exists \(N\in\N\) such that
\begin{equation*} \left|\frac{x_{n+1}}{x_n}\right| \lt L' \end{equation*}
for every \(n\ge N\text{.}\) So by dropping finitely many terms, we can assume the above inequality holds for every \(n\text{.}\) Thus, for every \(n \gt 1\text{,}\)
\begin{equation*} \left| \frac{x_{n}}{x_1}\right| = \left| \frac{x_{n}}{x_n-1}\right| \cdots \left| \frac{x_{2}}{x_1}\right| \lt (L')^{n-1}. \end{equation*}
and so
\begin{equation*} |x_n|^{1/n} \lt c_n:=(L')^{1-1/n}|x_1|^{1/n}. \end{equation*}
Thus, since \(c_n = L' \left( \frac{|x_1|}{L'} \right)^{1/n}\text{,}\) Proposition 2.2.3 implies that \(c_n \to L'\text{.}\) Hence \((|x_n|^{1/n})\) is a bounded sequence and so, by the Bolzano-Weierstrass theorem, contains a subsequence \((|x_{n_k}|^{1/n_k})\) the converges to \(s\text{.}\) Finally, since
\begin{equation*} |x_{n_k}|^{1/n_k} \le c_{n_k} \end{equation*}
for each \(k\text{,}\) letting \(k \to \infty\text{,}\) we conclude that \(s \lt L'\text{.}\)
We conclude this section with a proof of the Bolzano-Weierstrass theorem that uses only convergence, not limit superior or limit inferior. Its main advantage is that it is more visual.

Proof.
Let \((x_n)\) be a real sequence. Call \(n\in\N\) a peak index if \(x_n\ge x_m\) for every \(m\ge n\text{.}\)
First suppose there are infinitely many peak indices. Choose them in increasing order:
\begin{equation*} n_1 \lt n_2 \lt n_3 \lt \cdots. \end{equation*}
Since \(n_k\) is a peak index and \(n_{k+1}\ge n_k\text{,}\) we have \(x_{n_k}\ge x_{n_{k+1}}\) for every \(k\text{.}\) Hence \((x_{n_k})\) is a decreasing subsequence.
Now suppose there are only finitely many peak indices. Then there is some \(n_1\in\N\) such that no \(n\ge n_1\) is a peak index. Since \(n_1\) is not a peak index, there exists \(n_2\gt n_1\) such that \(x_{n_2}\gt x_{n_1}\text{.}\) Since \(n_2\) is not a peak index, there exists \(n_3\gt n_2\) such that \(x_{n_3}\gt x_{n_2}\text{.}\) Continuing inductively, we obtain a subsequence \((x_{n_k})\) with
\begin{equation*} x_{n_1} \lt x_{n_2} \lt x_{n_3} \lt \cdots. \end{equation*}
Therefore \((x_{n_k})\) is increasing.
In either case, \((x_n)\) has a monotone subsequence.
By the lemma above, every sequence has a monotone subsequence. If the original sequence is bounded, then this subsequence is also bounded.
A bounded monotone sequence converges by Theorem 2.1.5. This proves the Bolzano-Weierstrass theorem.
PrevTopNext