What is a Subsequence?
A subsequence is formed by deleting zero or more terms from the original sequence, keeping the remaining terms in their original relative order.
Formal Definition
Let $(s_n)$ be a sequence. Let $(n_k)$ be a strictly increasing sequence of natural numbers ($n_1 \lt n_2 \lt n_3 \lt \dots$). Then the sequence $(t_k)$ defined by $t_k = s_{n_k}$ is a subsequence of $(s_n)$.
Key Rule: You can skip terms, but you cannot go backwards and you cannot repeat terms (unless they appeared multiple times in the original).
Subsequence Builder
Bolzano-Weierstrass Theorem
Theorem 11.5: Every bounded sequence has a convergent subsequence.
Even if the original sequence is chaotic and doesn't converge, if it's trapped in a box (bounded), there must be some pattern hidden inside it that converges.
Note: The original sequence doesn't have to converge. The theorem guarantees we can find a part of it that does.
Visualizing Bolzano-Weierstrass
Important Theorems
Inheritance of Limits
Theorem 11.3
If a sequence $(s_n)$ converges to $s$, then every subsequence of $(s_n)$ also converges to $s$.
Monotone Subsequence
Theorem 11.4
Every sequence has a monotone subsequence (either increasing or decreasing).
Subsequential Limits
Concept
A number $t$ is a subsequential limit if there exists some subsequence converging to $t$. The set of all such limits is denoted $S$.
Practice Problems
True/False
If a sequence converges to \( s \), then every subsequence converges to \( s \).
True.
This is Theorem 11.3. If the original sequence gets close to \( s \), any selection of terms from it must also get close to \( s \).
If a sequence has a convergent subsequence, then the sequence itself converges.
False.
Counterexample: \( s_n = (-1)^n \). Subsequence of evens converges to 1, but \( s_n \) diverges.
\( \limsup s_n \) is always greater than or equal to \( \liminf s_n \).
True.
The supremum of a set is always greater than or equal to its infimum (provided the set is non-empty).
If \( \limsup s_n = \liminf s_n \), then the sequence converges (possibly to \( \pm \infty \)).
True.
This is a key property. If the "ceiling" and "floor" meet, the sequence is squeezed to that limit.
Fill in the Blank
A subsequence \( (s_{n_k}) \) is formed by selecting terms where indices \( n_k \) are ________.
Answer: strictly increasing
The Bolzano-Weierstrass Theorem states that every ________ sequence has a convergent subsequence.
Answer: bounded
The largest subsequential limit of a sequence is called the ________.
Answer: limit superior (or lim sup)
If a sequence is unbounded above, its \( \limsup \) is ________.
Answer: \( +\infty \)
Full Problems
Find the set of subsequential limits for \( s_n = \sin(n\pi/2) \).
The sequence values repeat: \( 1, 0, -1, 0, 1, 0, -1, 0, \dots \)
Subsequences can converge to 1 (indices \( 4k+1 \)), -1 (indices \( 4k+3 \)), or 0 (indices \( 2k \)).
Set of subsequential limits \( S = \{-1, 0, 1\} \).
Calculate \( \limsup s_n \) and \( \liminf s_n \) for \( s_n = (-1)^n (1 + \frac{1}{n}) \).
For even \( n \), \( s_n = 1 + 1/n \to 1 \).
For odd \( n \), \( s_n = -1 - 1/n \to -1 \).
\( \limsup s_n = 1 \)
\( \liminf s_n = -1 \)
Prove that if \( (s_n) \) converges to \( s \), then every subsequence \( (s_{n_k}) \) converges to \( s \).
Let \( \epsilon \gt 0 \). Since \( s_n \to s \), \( \exists N \) such that \( n \gt N \implies |s_n - s| \lt \epsilon \).
Since \( n_k \) is strictly increasing integers, \( n_k \ge k \).
So if \( k \gt N \), then \( n_k \ge k \gt N \), which implies \( |s_{n_k} - s| \lt \epsilon \).
Thus \( s_{n_k} \to s \).
Let \( (s_n) \) be a bounded sequence. Prove that there exists a subsequence that converges to \( \limsup s_n \).
Let \( v = \limsup s_n \). By definition, \( v = \lim_{N \to \infty} \sup \{s_n : n \gt N\} \).
This implies that for any \( \epsilon \gt 0 \), there are infinitely many terms of the sequence in \( (v - \epsilon, v + \epsilon) \) (or close to it).
We can construct a subsequence inductively. Pick \( n_1 \) such that \( |s_{n_1} - v| \lt 1 \).
Pick \( n_k \gt n_{k-1} \) such that \( |s_{n_k} - v| \lt 1/k \). Such \( n_k \) always exists because \( v \) is a subsequential limit.
Then \( s_{n_k} \to v \).