The Bolzano-Weierstrass theorem shows that, from every bounded real sequence, it is always possible to extract at least one convergent subsequence.
In the following practice problems, we will apply this result to bounded and unbounded sequences, oscillating examples, accumulation points, and properties of subsequences.
Throughout, we will rely on the fundamental result that every bounded real sequence has at least one convergent subsequence.
Exercise 1 — level ★☆☆☆☆
Determine whether the sequence
\[ x_n=(-1)^n \]
satisfies the hypotheses of the Bolzano–Weierstrass theorem and, if so, exhibit two convergent subsequences.
Answer
The sequence is bounded. Two convergent subsequences are
\[ x_{2k}=1 \longrightarrow 1 \]
and
\[ x_{2k-1}=-1 \longrightarrow -1. \]
Solution
The sequence alternates between the values \(1\) and \(-1\). Indeed, if \(n\) is even, then
\[ x_n=(-1)^n=1, \]
whereas, if \(n\) is odd,
\[ x_n=(-1)^n=-1. \]
Hence every term of the sequence lies in the closed and bounded interval \([-1,1]\). In particular,
\[ -1\leq x_n\leq 1 \]
for every \(n\in\mathbb N\). The sequence is therefore bounded.
By the Bolzano–Weierstrass theorem, it must have at least one convergent subsequence. In this case we can write down two of them explicitly.
Taking the even indices, we obtain
\[ x_{2k}=(-1)^{2k}=1. \]
The subsequence \((x_{2k})\) is constant, so it converges to \(1\).
Taking the odd indices instead, we obtain
\[ x_{2k-1}=(-1)^{2k-1}=-1. \]
This subsequence is constant as well, so it converges to \(-1\).
Exercise 2 — level ★☆☆☆☆
Determine whether the sequence
\[ x_n=\frac{1}{n} \]
has a convergent subsequence.
Answer
Yes. The sequence is bounded and converges to \(0\) on its own. Consequently, every subsequence of it converges to \(0\).
Solution
For every \(n\in\mathbb N\) with \(n\geq 1\) we have
\[ 0<\frac{1}{n}\leq 1. \]
Hence
\[ x_n\in(0,1] \]
for every \(n\geq 1\). The sequence is therefore bounded.
By the Bolzano–Weierstrass theorem, every bounded real sequence has at least one convergent subsequence. In this case we can say more: the whole sequence converges.
Indeed,
\[ \frac{1}{n}\longrightarrow 0. \]
It follows that if \((x_{n_k})\) is any subsequence, then
\[ x_{n_k}=\frac{1}{n_k}. \]
Since \(n_k\to+\infty\), we have
\[ \frac{1}{n_k}\longrightarrow 0. \]
Thus every subsequence converges to the same limit \(0\).
Exercise 3 — level ★☆☆☆☆
Determine whether the sequence
\[ x_n=n \]
has a subsequence that converges in \(\mathbb R\).
Answer
No. The sequence is unbounded, and no subsequence of it converges to a real number.
Solution
The sequence is given by
\[ x_n=n. \]
It is not bounded above, because, given any real number \(M\), there is always an index \(n\) such that
\[ n>M. \]
Hence there is no closed and bounded interval \([a,b]\) containing all the terms of the sequence.
Now consider any subsequence \((x_{n_k})\). It has the form
\[ x_{n_k}=n_k, \]
where
\[ n_1<n_2<n_3<\cdots. \]
Since the indices \(n_k\) form a strictly increasing sequence of natural numbers, we necessarily have
\[ n_k\longrightarrow+\infty. \]
Therefore
\[ x_{n_k}=n_k\longrightarrow+\infty. \]
No subsequence can converge to a real number. This shows that the boundedness assumption in the Bolzano–Weierstrass theorem cannot be dropped.
Exercise 4 — level ★☆☆☆☆
Consider the sequence
\[ x_n=\frac{(-1)^n}{n}. \]
Determine whether it is bounded, and find the limit of a convergent subsequence of it.
Answer
The sequence is bounded and converges to \(0\). Hence every subsequence of it converges to \(0\).
Solution
For every \(n\geq 1\) we have
\[ |x_n|=\left|\frac{(-1)^n}{n}\right|=\frac{1}{n}\leq 1. \]
Hence
\[ -1\leq x_n\leq 1 \]
for every \(n\geq 1\). The sequence is bounded.
By Bolzano–Weierstrass it must have at least one convergent subsequence. In fact, here too the whole sequence converges.
Indeed,
\[ |x_n-0|=|x_n|=\frac{1}{n}\longrightarrow 0. \]
Hence
\[ x_n\longrightarrow 0. \]
Consequently, for every increasing sequence of indices \((n_k)\) we have
\[ x_{n_k}=\frac{(-1)^{n_k}}{n_k}\longrightarrow 0. \]
Exercise 5 — level ★☆☆☆☆
Let \((x_n)\) be a sequence such that
\[ 2\leq x_n\leq 5 \]
for every \(n\in\mathbb N\). Prove that \((x_n)\) has a convergent subsequence.
Answer
The sequence is bounded. By the Bolzano–Weierstrass theorem it has at least one convergent subsequence.
Solution
The hypothesis
\[ 2\leq x_n\leq 5 \]
for every \(n\in\mathbb N\) means that all the terms of the sequence lie in the same closed and bounded interval:
\[ x_n\in[2,5]. \]
Hence the sequence is bounded below by \(2\) and above by \(5\).
Since \((x_n)\) is a bounded real sequence, we may apply the Bolzano–Weierstrass theorem directly.
There therefore exist a strictly increasing sequence of indices
\[ n_1<n_2<n_3<\cdots \]
and a real number \(x_0\) such that
\[ x_{n_k}\longrightarrow x_0. \]
Moreover, since all the terms of the subsequence lie in \([2,5]\), the limit must lie in the same interval:
\[ x_0\in[2,5]. \]
Exercise 6 — level ★★☆☆☆
Consider the sequence
\[ x_n=\frac{(-1)^n n}{n+1}. \]
Find two convergent subsequences together with their limits.
Answer
We have
\[ x_{2k}=\frac{2k}{2k+1}\longrightarrow 1 \]
and
\[ x_{2k-1}=-\frac{2k-1}{2k}\longrightarrow -1. \]
Solution
The sequence is
\[ x_n=\frac{(-1)^n n}{n+1}. \]
Since
\[ \left|x_n\right|=\left|\frac{(-1)^n n}{n+1}\right|=\frac{n}{n+1}<1, \]
we have
\[ -1<x_n<1 \]
for every \(n\in\mathbb N\). The sequence is therefore bounded.
We now examine the even and the odd indices separately.
If \(n=2k\), then
\[ x_{2k}=\frac{(-1)^{2k}\,2k}{2k+1}=\frac{2k}{2k+1}. \]
Since
\[ \frac{2k}{2k+1}=\frac{1}{1+\frac{1}{2k}}, \]
it follows that
\[ x_{2k}\longrightarrow 1. \]
If instead \(n=2k-1\), then
\[ x_{2k-1} = \frac{(-1)^{2k-1}(2k-1)}{2k} = -\frac{2k-1}{2k}. \]
Since
\[ \frac{2k-1}{2k}=1-\frac{1}{2k}\longrightarrow 1, \]
we obtain
\[ x_{2k-1}\longrightarrow -1. \]
Thus the sequence does not converge, but it does have at least two convergent subsequences, one with limit \(1\) and one with limit \(-1\).
Exercise 7 — level ★★☆☆☆
Consider the sequence
\[ x_n=\sin\left(\frac{n\pi}{2}\right). \]
Show that it has convergent subsequences, and find all the limit values arising from the index classes modulo \(4\).
Answer
The values of the sequence repeat periodically:
\[ 1,\ 0,\ -1,\ 0,\ 1,\ 0,\ -1,\ 0,\ldots \]
The constant subsequences converge to \(1\), \(0\) and \(-1\), respectively.
Solution
We compute the values of the sequence by sorting the indices according to their remainder upon division by \(4\).
If \(n=4k+1\), then
\[ x_{4k+1}=\sin\left(\frac{(4k+1)\pi}{2}\right) = \sin\left(2k\pi+\frac{\pi}{2}\right)=1. \]
If \(n=4k+2\), then
\[ x_{4k+2}=\sin\left(2k\pi+\pi\right)=0. \]
If \(n=4k+3\), then
\[ x_{4k+3}=\sin\left(2k\pi+\frac{3\pi}{2}\right)=-1. \]
Finally, if \(n=4k\), then
\[ x_{4k}=\sin(2k\pi)=0. \]
Hence the sequence takes only the values \(-1\), \(0\) and \(1\), and it is certainly bounded.
By Bolzano–Weierstrass it has at least one convergent subsequence. In fact, we can point to several at once:
\[ x_{4k+1}=1\longrightarrow 1, \]
\[ x_{4k+3}=-1\longrightarrow -1, \]
and
\[ x_{4k}=0\longrightarrow 0. \]
Each constant subsequence converges to the constant value it takes.
Exercise 8 — level ★★☆☆☆
Let \((x_n)\) be a bounded real sequence. Prove that every subsequence \((x_{n_k})\) is bounded as well.
Answer
Every subsequence of a bounded sequence is bounded.
Solution
Since \((x_n)\) is bounded, there exist two real numbers \(a\) and \(b\), with \(a\leq b\), such that
\[ a\leq x_n\leq b \]
for every \(n\in\mathbb N\).
Now let \((x_{n_k})\) be a subsequence of \((x_n)\). By definition, every term of the subsequence is also a term of the original sequence.
Indeed, \(n_k\in\mathbb N\) for every \(k\), so the boundedness of \((x_n)\) gives
\[ a\leq x_{n_k}\leq b \]
for every \(k\in\mathbb N\).
Therefore all the terms of the subsequence lie in the same closed and bounded interval \([a,b]\).
Hence \((x_{n_k})\) is bounded.
Exercise 9 — level ★★☆☆☆
Determine whether the sequence
\[ x_n=(-1)^n+\frac{1}{n} \]
has convergent subsequences, and find two of them.
Answer
We have
\[ x_{2k}=1+\frac{1}{2k}\longrightarrow 1 \]
and
\[ x_{2k-1}=-1+\frac{1}{2k-1}\longrightarrow -1. \]
Solution
The sequence is given by
\[ x_n=(-1)^n+\frac{1}{n}. \]
Since
\[ -1\leq (-1)^n\leq 1 \]
and
\[ 0<\frac{1}{n}\leq 1, \]
we obtain
\[ -1<x_n\leq 2. \]
The sequence is therefore bounded. By the Bolzano–Weierstrass theorem it has at least one convergent subsequence.
Let us find two of them explicitly. If \(n=2k\), then
\[ x_{2k}=(-1)^{2k}+\frac{1}{2k} = 1+\frac{1}{2k}. \]
Since
\[ \frac{1}{2k}\longrightarrow 0, \]
it follows that
\[ x_{2k}\longrightarrow 1. \]
If instead \(n=2k-1\), then
\[ x_{2k-1}=(-1)^{2k-1}+\frac{1}{2k-1} = -1+\frac{1}{2k-1}. \]
Since
\[ \frac{1}{2k-1}\longrightarrow 0, \]
we obtain
\[ x_{2k-1}\longrightarrow -1. \]
Exercise 10 — level ★★☆☆☆
Let \((x_n)\) be a real sequence such that \(x_n\in[0,1]\) for every \(n\in\mathbb N\). Prove that there is a convergent subsequence whose limit lies in \([0,1]\).
Answer
There is a subsequence \((x_{n_k})\) such that
\[ x_{n_k}\longrightarrow x_0 \]
with \(x_0\in[0,1]\).
Solution
The hypothesis \(x_n\in[0,1]\) for every \(n\in\mathbb N\) means that
\[ 0\leq x_n\leq 1 \]
for every \(n\in\mathbb N\). Hence the sequence is bounded.
By the Bolzano–Weierstrass theorem, there exist an increasing sequence of indices
\[ n_1<n_2<n_3<\cdots \]
and a real number \(x_0\) such that
\[ x_{n_k}\longrightarrow x_0. \]
It remains to check that the limit \(x_0\) lies in \([0,1]\).
Indeed, every term of the subsequence satisfies
\[ 0\leq x_{n_k}\leq 1. \]
Letting \(k\to+\infty\) in these inequalities, we get
\[ 0\leq x_0\leq 1. \]
Hence
\[ x_0\in[0,1]. \]
Exercise 11 — level ★★★☆☆
Prove that if a bounded real sequence takes only finitely many values, then it has a constant subsequence.
Answer
At least one of the values taken by the sequence occurs infinitely often. Choosing the corresponding indices yields a constant subsequence, which is therefore convergent.
Solution
Suppose the sequence \((x_n)\) takes only the values
\[ a_1,a_2,\ldots,a_m. \]
This means that, for every \(n\in\mathbb N\),
\[ x_n\in\{a_1,a_2,\ldots,a_m\}. \]
Since the sequence has infinitely many terms but can take only finitely many values, at least one of these values must be attained infinitely often.
Indeed, if each of the values \(a_1,\ldots,a_m\) were attained only finitely many times, then the sequence would have only finitely many terms in all — a contradiction.
There is therefore a value \(a_j\) such that
\[ x_n=a_j \]
for infinitely many indices \(n\).
We can then choose a strictly increasing sequence of indices
\[ n_1<n_2<n_3<\cdots \]
such that
\[ x_{n_k}=a_j \]
for every \(k\in\mathbb N\).
The subsequence \((x_{n_k})\) is therefore constant:
\[ x_{n_k}=a_j. \]
Consequently
\[ x_{n_k}\longrightarrow a_j. \]
Exercise 12 — level ★★★☆☆
Let \((x_n)\) be a sequence converging to \(L\). Prove that every subsequence of \((x_n)\) also converges to \(L\).
Answer
Every subsequence of a convergent sequence converges to the same limit as the original sequence.
Solution
Suppose that
\[ x_n\longrightarrow L. \]
By the definition of limit, for every \(\varepsilon>0\) there is an \(N\in\mathbb N\) such that, for every \(n\geq N\),
\[ |x_n-L|<\varepsilon. \]
Now let \((x_{n_k})\) be a subsequence of \((x_n)\). The indices \(n_k\) are strictly increasing, so
\[ n_k\longrightarrow+\infty. \]
In particular, there is a \(K\in\mathbb N\) such that, for every \(k\geq K\),
\[ n_k\geq N. \]
Consequently, for every \(k\geq K\),
\[ |x_{n_k}-L|<\varepsilon. \]
This shows that
\[ x_{n_k}\longrightarrow L. \]
Hence every subsequence of a convergent sequence converges to the same limit.
Exercise 13 — level ★★★☆☆
Prove that the sequence
\[ x_n=\cos(n\pi) \]
does not converge, yet has convergent subsequences.
Answer
Since
\[ \cos(n\pi)=(-1)^n, \]
the sequence does not converge. Nevertheless,
\[ x_{2k}=1\longrightarrow 1 \]
and
\[ x_{2k-1}=-1\longrightarrow -1. \]
Solution
Recall that, for every \(n\in\mathbb N\),
\[ \cos(n\pi)=(-1)^n. \]
Hence the sequence is
\[ x_n=(-1)^n. \]
It alternates between the values \(1\) and \(-1\). In particular it is bounded, since
\[ -1\leq x_n\leq 1. \]
By the Bolzano–Weierstrass theorem it must have at least one convergent subsequence.
We show, however, that the whole sequence does not converge. If it converged to a limit \(L\), then every subsequence of it would have to converge to the same limit \(L\).
But, taking the even indices, we have
\[ x_{2k}=\cos(2k\pi)=1, \]
so
\[ x_{2k}\longrightarrow 1. \]
Taking the odd indices instead, we have
\[ x_{2k-1}=\cos((2k-1)\pi)=-1, \]
so
\[ x_{2k-1}\longrightarrow -1. \]
Since a convergent sequence cannot have two subsequences converging to different limits, the sequence \((x_n)\) does not converge.
Exercise 14 — level ★★★☆☆
Let \((x_n)\) be a bounded real sequence. Prove that there exist a real number \(L\) and a subsequence \((x_{n_k})\) such that
\[ |x_{n_k}-L|<\frac{1}{k} \]
for all \(k\).
Answer
By Bolzano–Weierstrass there is a convergent subsequence \(x_{n_k}\to L\). From the convergence it follows that, by passing to a further subsequence if necessary, one can arrange that
\[ |x_{n_k}-L|<\frac{1}{k}. \]
Solution
Since \((x_n)\) is bounded, by the Bolzano–Weierstrass theorem there exist a subsequence \((x_{m_k})\) and a real number \(L\) such that
\[ x_{m_k}\longrightarrow L. \]
By the definition of convergence, for every \(\varepsilon>0\) there is an index beyond which all the terms of the subsequence lie within \(\varepsilon\) of \(L\).
We now want to build an even sharper subsequence, by imposing the estimate
\[ |x_{n_k}-L|<\frac{1}{k}. \]
Since \(x_{m_j}\to L\), for \(\varepsilon=1\) there is an index \(j_1\) such that
\[ |x_{m_{j_1}}-L|<1. \]
For \(\varepsilon=\displaystyle\frac12\), there is an index \(j_2>j_1\) such that
\[ |x_{m_{j_2}}-L|<\frac12. \]
Proceeding by induction, having chosen \(j_k\), we can choose \(j_{k+1}>j_k\) such that
\[ |x_{m_{j_{k+1}}}-L|<\frac{1}{k+1}. \]
Now set
\[ n_k=m_{j_k}. \]
We obtain a subsequence \((x_{n_k})\) such that
\[ |x_{n_k}-L|<\frac{1}{k} \]
for every \(k\in\mathbb N\). This completes the proof.
Exercise 15 — level ★★★☆☆
Let \((x_n)\) be a bounded real sequence, and suppose that every convergent subsequence of it has limit \(L\). Prove that \(x_n\to L\).
Answer
If a bounded sequence failed to converge to \(L\), one could extract a subsequence that stays away from \(L\). Applying Bolzano–Weierstrass to that subsequence would produce a convergent subsequence with limit different from \(L\), contrary to the hypothesis.
Solution
We want to prove that
\[ x_n\longrightarrow L. \]
We argue by contradiction. Suppose that \((x_n)\) does not converge to \(L\).
Then there is a number \(\varepsilon_0>0\) such that, for every \(N\in\mathbb N\), there is some \(n\geq N\) with
\[ |x_n-L|\geq \varepsilon_0. \]
This means that one can find infinitely many terms of the sequence that stay at distance at least \(\varepsilon_0\) from \(L\).
We therefore construct a subsequence \((x_{n_k})\) such that
\[ |x_{n_k}-L|\geq \varepsilon_0 \]
for every \(k\in\mathbb N\).
Since \((x_n)\) is bounded, the subsequence \((x_{n_k})\) is bounded too.
By the Bolzano–Weierstrass theorem, \((x_{n_k})\) has a convergent subsequence. Denote it by
\[ x_{n_{k_j}}\longrightarrow M. \]
By hypothesis, every convergent subsequence of \((x_n)\) must have limit \(L\). Hence we would have
\[ M=L. \]
Yet, since for every \(j\) we have
\[ |x_{n_{k_j}}-L|\geq \varepsilon_0, \]
letting \(j\to+\infty\) gives
\[ |M-L|\geq \varepsilon_0. \]
In particular \(M\neq L\), a contradiction.
Hence the assumption was false, and therefore
\[ x_n\longrightarrow L. \]
Exercise 16 — level ★★★★☆
Prove that every infinite, bounded set \(A\subseteq\mathbb R\) has at least one accumulation point.
Answer
Choose a sequence of distinct elements of \(A\). It is bounded, so by Bolzano–Weierstrass it has a convergent subsequence. The limit of that subsequence is an accumulation point of \(A\).
Solution
Let \(A\subseteq\mathbb R\) be an infinite, bounded set.
Since \(A\) is infinite, we can choose a sequence of distinct elements of \(A\):
\[ x_1,x_2,x_3,\ldots \]
with
\[ x_n\in A \]
for every \(n\), and
\[ x_n\neq x_m \]
whenever \(n\neq m\).
Since \(A\) is bounded, there exist \(a,b\in\mathbb R\), with \(a\leq b\), such that
\[ A\subseteq [a,b]. \]
Consequently
\[ x_n\in[a,b] \]
for every \(n\). The sequence \((x_n)\) is therefore bounded.
By the Bolzano–Weierstrass theorem, there exist a subsequence \((x_{n_k})\) and a real number \(x_0\) such that
\[ x_{n_k}\longrightarrow x_0. \]
We show that \(x_0\) is an accumulation point of \(A\).
Let \(r>0\). Since \(x_{n_k}\to x_0\), there is a \(K\in\mathbb N\) such that, for every \(k\geq K\),
\[ |x_{n_k}-x_0|<r. \]
Hence
\[ x_{n_k}\in(x_0-r,x_0+r) \]
for every \(k\geq K\).
Since the terms \(x_{n_k}\) are distinct elements of \(A\), every neighborhood of \(x_0\) contains infinitely many points of \(A\). In particular it contains points of \(A\) other than \(x_0\).
Hence \(x_0\) is an accumulation point of \(A\).
Exercise 17 — level ★★★★☆
Using the Bolzano–Weierstrass theorem, prove that a bounded real sequence cannot behave purely by “running off to infinity.”
Answer
A bounded sequence always has a subsequence converging to a real number. Hence not all of its subsequences can diverge to \(+\infty\) or to \(-\infty\).
Solution
To say that a sequence “runs off to infinity” means, precisely, that its terms become arbitrarily large in absolute value.
But if \((x_n)\) is bounded, there is an \(M>0\) such that
\[ |x_n|\leq M \]
for every \(n\in\mathbb N\).
Hence all the terms of the sequence lie in the closed and bounded interval
\[ [-M,M]. \]
By the Bolzano–Weierstrass theorem, there exist a subsequence \((x_{n_k})\) and a real number \(L\) such that
\[ x_{n_k}\longrightarrow L. \]
This subsequence does not run off to infinity, since it converges to a real number.
Thus a bounded real sequence may oscillate, may fail to converge as a whole, and may have several limit values, but it cannot be devoid of any convergent behavior within itself.
The Bolzano–Weierstrass theorem captures exactly this fact: inside every bounded real sequence one can always extract a convergent subsequence.
Exercise 18 — level ★★★★☆
Using nested intervals, construct explicitly a convergent subsequence of a sequence \((x_n)\) contained in \([0,1]\).
Answer
By repeatedly halving \([0,1]\) and, at each step, choosing a subinterval that contains infinitely many terms of the sequence, one obtains a family of nested intervals whose lengths tend to \(0\). Their single common point is the limit of the subsequence so constructed.
Solution
Suppose that
\[ x_n\in[0,1] \]
for every \(n\in\mathbb N\).
Set
\[ I_1=[0,1]. \]
The interval \(I_1\) contains all the terms of the sequence, and hence contains infinitely many of them.
Split \(I_1\) into two closed intervals:
\[ \left[0,\frac12\right], \qquad \left[\frac12,1\right]. \]
At least one of the two contains infinitely many terms of the sequence. Choose such an interval and call it \(I_2\).
We repeat the procedure. Having constructed \(I_k\), split it into two closed intervals of equal length. At least one of the two contains infinitely many terms of the sequence. Choose it and call it \(I_{k+1}\).
In this way we obtain a sequence of closed and bounded intervals
\[ I_1\supseteq I_2\supseteq I_3\supseteq\cdots \]
such that each \(I_k\) contains infinitely many terms of the sequence.
Moreover, the length of \(I_k\) is
\[ \frac{1}{2^{k-1}}, \]
and therefore tends to \(0\).
By the nested interval theorem, there is a unique point \(x_0\in\mathbb R\) such that
\[ \bigcap_{k=1}^{+\infty}I_k=\{x_0\}. \]
We now build the subsequence. Since \(I_1\) contains infinitely many terms, choose \(n_1\) such that
\[ x_{n_1}\in I_1. \]
Since \(I_2\) contains infinitely many terms, we can choose \(n_2>n_1\) such that
\[ x_{n_2}\in I_2. \]
Continuing by induction, we choose \(n_k\) so that
\[ n_1<n_2<\cdots<n_k<\cdots \]
and
\[ x_{n_k}\in I_k. \]
Since \(x_0\in I_k\) as well, the distance between \(x_{n_k}\) and \(x_0\) is at most the length of \(I_k\):
\[ |x_{n_k}-x_0|\leq \frac{1}{2^{k-1}}. \]
Since
\[ \frac{1}{2^{k-1}}\longrightarrow 0, \]
it follows, by comparison, that
\[ |x_{n_k}-x_0|\longrightarrow 0. \]
Hence
\[ x_{n_k}\longrightarrow x_0. \]
Exercise 19 — level ★★★★★
Let \((x_n)\) be a bounded real sequence. Prove that the set of limits of its convergent subsequences is nonempty and bounded.
Answer
The set is nonempty by Bolzano–Weierstrass. It is bounded because any subsequential limit must lie in every closed interval that contains all the terms of the sequence.
Solution
Let \((x_n)\) be a bounded real sequence. Then there exist \(a,b\in\mathbb R\), with \(a\leq b\), such that
\[ a\leq x_n\leq b \]
for every \(n\in\mathbb N\).
Consider the set
\[ E=\{L\in\mathbb R:\text{ there is a subsequence }(x_{n_k})\text{ such that }x_{n_k}\to L\}. \]
We must show that \(E\) is nonempty and bounded.
Since \((x_n)\) is bounded, by the Bolzano–Weierstrass theorem there is at least one convergent subsequence. Hence there is at least one real number \(L\) such that
\[ x_{n_k}\longrightarrow L. \]
Therefore
\[ E\neq\varnothing. \]
Now we show that \(E\) is bounded. Let \(L\in E\). By the definition of \(E\), there is a subsequence \((x_{n_k})\) such that
\[ x_{n_k}\longrightarrow L. \]
Since for every \(k\) we have
\[ a\leq x_{n_k}\leq b, \]
letting \(k\to+\infty\) gives
\[ a\leq L\leq b. \]
Hence every element of \(E\) lies in the interval \([a,b]\). Therefore
\[ E\subseteq[a,b]. \]
It follows that \(E\) is bounded.
Exercise 20 — level ★★★★★
Let \((x_n)\) be a bounded real sequence, and let
\[ \alpha=\liminf_{n\to+\infty}x_n, \qquad \beta=\limsup_{n\to+\infty}x_n. \]
Prove that there exist two subsequences \((x_{n_k})\) and \((x_{m_k})\) such that
\[ x_{n_k}\longrightarrow \alpha \]
and
\[ x_{m_k}\longrightarrow \beta. \]
Answer
The limit inferior and the limit superior of a bounded real sequence are always limits of suitable subsequences.
Solution
Since the sequence \((x_n)\) is bounded, the quantities
\[ \alpha=\liminf_{n\to+\infty}x_n \]
and
\[ \beta=\limsup_{n\to+\infty}x_n \]
are real numbers.
We first prove the existence of a subsequence converging to \(\beta\).
By the definition of limit superior, \(\beta\) is the largest possible limiting value of the tails of the sequence. In particular, for every \(k\in\mathbb N\) there are arbitrarily large indices \(n\) such that
\[ x_n>\beta-\frac{1}{k}. \]
Moreover, again by the definition of \(\limsup\), beyond a certain point the terms of the sequence no longer stay above \(\beta+\displaystyle\frac{1}{k}\). We can therefore choose increasing indices \(m_k\) such that
\[ \beta-\frac{1}{k}<x_{m_k}<\beta+\frac{1}{k}. \]
In other words,
\[ |x_{m_k}-\beta|<\frac{1}{k}. \]
Since
\[ \frac{1}{k}\longrightarrow 0, \]
it follows that
\[ x_{m_k}\longrightarrow \beta. \]
The argument for \(\alpha\) is analogous. By the definition of limit inferior, for every \(k\in\mathbb N\) there are arbitrarily large indices \(n\) such that
\[ x_n<\alpha+\frac{1}{k}. \]
Moreover, these indices can be chosen so that the corresponding terms are not below \(\alpha-\displaystyle\frac{1}{k}\). We thus obtain an increasing sequence of indices \(n_k\) such that
\[ \alpha-\frac{1}{k}<x_{n_k}<\alpha+\frac{1}{k}. \]
Equivalently,
\[ |x_{n_k}-\alpha|<\frac{1}{k}. \]
Letting \(k\to+\infty\), we obtain
\[ x_{n_k}\longrightarrow \alpha. \]
Hence both the limit inferior and the limit superior are limits of suitable subsequences of the original sequence.