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Bolzano-Weierstrass by bisection, the set of subsequential limits, and why the upper and lower limits always exist.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to run the bisection proof of the Bolzano-Weierstrass theorem, extract a convergent subsequence from a bounded sequence by hand, and find the upper and lower limits of a sequence that does not converge. You will also be able to say exactly what the theorem promises — some subsequence, not the sequence and not every subsequence — and why the two limits agree precisely when the sequence converges.
Monotone convergence, which needs the sequence to be monotone, and the nested interval theorem from the completeness lesson. Most sequences are not monotone. This lesson is what remains true when they are not, and the nested intervals are how it is proved.
A subsequence $(a_{n_k})$ is what is left after deleting terms, with $n_1 < n_2 < \cdots$ — the order is kept and infinitely many terms remain. A subsequential limit is a number some subsequence converges to. The upper limit $\limsup a_n$ is the largest subsequential limit and the lower limit $\liminf a_n$ the smallest.
Bolzano-Weierstrass. Every bounded sequence of reals has a convergent subsequence.
Proof by bisection. All the terms lie in some $[-M, M]$. At least one half of it contains $a_n$ for infinitely many $n$; keep that half and bisect again. This gives nested closed intervals $I_1 \supseteq I_2 \supseteq \cdots$, each containing infinitely many terms, with $|I_k| = 2M/2^{k-1} \to 0$. The nested interval theorem gives a single point $c$ in all of them. Now choose $n_1 < n_2 < \cdots$ with $a_{n_k} \in I_k$ — possible precisely because each $I_k$ holds infinitely many indices. Then $|a_{n_k} - c| \le |I_k| \to 0$.
The promise is deliberately weak, and that weakness is its usefulness: boundedness is easy to check, and a convergent piece is enough for most existence arguments.
Upper and lower limits. For a bounded sequence, set $s_N = \sup\{a_n : n \ge N\}$. The $s_N$ decrease and are bounded below, so by monotone convergence they converge; their limit is $\limsup a_n$. Symmetrically for $\liminf$. Both always exist for a bounded sequence, whether or not the sequence converges — and
> $a_n$ converges $\iff$ $\limsup a_n = \liminf a_n$,
in which case the common value is the limit. That equivalence is what makes them the right tool where a limit might not exist, such as the radius of a power series.
Another way: picture
Scatter the terms as dots and draw the narrowest horizontal band that catches all the dots from index $N$ onwards. As $N$ grows the band can only narrow. Its top edge settles at the upper limit and its bottom edge at the lower limit, and the sequence converges exactly when the band closes to a line.
Another way: steps
To find the upper and lower limits:
Every sequence — bounded or not — has a monotone subsequence.
Call $n$ a peak if $a_n \ge a_m$ for every $m > n$. If there are infinitely many peaks, the terms at the peaks form a decreasing subsequence. If there are finitely many, then past the last peak every index is beaten later, and following those later indices gives an increasing subsequence.
Combine this with monotone convergence and Bolzano-Weierstrass falls out in two lines: a bounded sequence has a monotone subsequence, which is bounded, and so converges. Two proofs of the same theorem is not extravagance — the bisection generalises to higher dimensions, and the peak argument does not, so it is worth knowing which is which.
| Sequence | Upper limit | Lower limit | Converges |
|---|---|---|---|
| $1/n$ | $0$ | $0$ | yes, to $0$ |
| $(-1)^n$ | $1$ | $-1$ | no |
| $(-1)^n(1 + 1/n)$ | $1$ | $-1$ | no |
| $n$ | $+\infty$ | $+\infty$ | no, unbounded |
| $\sin n$ | $1$ | $-1$ | no |
The gap between the two columns is a measurement of failure to converge: zero exactly when the sequence converges, and wider the more the sequence oscillates.
The first three rows also show that the upper limit is not the supremum of the terms. For $(-1)^n(1 + 1/n)$ the supremum of the terms is $3/2$ and the upper limit is $1$: the supremum can be achieved once, near the start, and then abandoned.
Reading the theorem as convergence. Bounded gives a convergent subsequence. The sequence itself may converge to nothing.
Reading 'some' as 'every'. $(-1)^n$ has convergent subsequences and divergent ones, and both facts are compatible with the theorem.
Confusing the upper limit with the supremum. The supremum is about all the terms; the upper limit ignores any finite number of them. Changing the first term changes one and never the other.
Reordering instead of deleting. A subsequence keeps the original order. Rearranging a sequence is a different operation with different consequences, as the series lessons will show.
Let $a_n = (-1)^n + 1/n$, which is bounded but has no limit.
The hypothesis holds; the conclusion of convergence does not.
Take the even indices: $a_{2k} = 1 + 1/(2k) \to 1$.
One convergent subsequence, exhibited.
The odd indices give $-1 + 1/(2k+1) \to -1$. So the subsequential limits are $1$ and $-1$, the upper limit is $1$, the lower is $-1$, and they disagree — which is the sequence's divergence, measured.
Two pieces, two limits, one gap.
Suppose $(x_n)$ lies in $[0, 1]$ and $f$ is continuous. Claim: $f$ attains a value arbitrarily close to $\sup\{f(x_n)\}$.
No limit of $x_n$ is assumed.
Choose $x_n$ with $f(x_n)$ approaching the supremum. The $x_n$ are bounded, so some subsequence converges, to $c \in [0,1]$.
Boundedness is all that was needed.
Continuity then transfers the limit to $f(c)$. This is the skeleton of the extreme value theorem, and Bolzano-Weierstrass is the step that produces the point $c$ from nothing but boundedness.
A point produced where none was given.
The terms cycle through $0, -1, 0, 1$ as $n$ runs through $1, 2, 3, 4$, and then repeat.
Identify the repeating pattern.
The indices congruent to $2$ modulo $4$ give the constant subsequence $-1$; those congruent to $0$ give $1$; the odd ones give $0$.
One subsequence per residue class.
So the subsequential limits are $-1$, $0$ and $1$; the upper limit is $1$ and the lower is $-1$. Three subsequential limits is perfectly ordinary — there can be infinitely many, and for a sequence enumerating the rationals in $[0,1]$ every point of the interval is one.
Put the steps of the bisection proof of the Bolzano-Weierstrass theorem into order, for a sequence all of whose terms lie between $-3$ and $3$.
Number the steps in order (write the number in the box):
$a_n = (-1)^n\left(4 + \dfrac{1}{n}\right)$. What is its upper limit?
Answer:
Give the upper and lower limits of each sequence.
| upper limit | lower limit | |
|---|---|---|
| $(-1)^n \cdot 4$ | ||
| $4 + 1/n$ | ||
| $(-1)^n/n$ |
$a_n = (-1)^n 5$ is bounded. What does the Bolzano-Weierstrass theorem entitle you to say about it?
Here is an argument about the sequence $a_n = 2n$. Mark the one step that is not sound.
This task has no paper form; do it on a device.
Give the set of all subsequential limits of $a_n = (-1)^n\left(7 + \dfrac{1}{n}\right)$.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the steps of the bisection proof of the Bolzano-Weierstrass theorem into order, for a sequence all of whose terms lie between $-2$ and $2$.
Number the steps in order (write the number in the box):
You can extract a convergent subsequence from a bounded sequence, state what Bolzano-Weierstrass does and does not promise, and read off upper and lower limits. Say in your own words why the kept half must contain infinitely many terms. Next: proving convergence without ever naming the limit.
10. Your turn: the subsequential limits of $a_n = \cos(n\pi/2)$, step 3