Subsequences and Bolzano-Weierstrass

Examples

Example 1:

Let $(a_n) = (1, 2, 3, 4, 5, 6, \ldots)$. The subsequence of even-indexed terms is

$$(a_{2k}) = (a_2, a_4, a_6, \ldots) = (2, 4, 6, \ldots)$$

Example 2:

Let $(b_n) = (-1, 1, -1, 1, -1, 1, \ldots)$. The subsequence of odd-indexed terms is

$$(b_{2k-1}) = (b_1, b_3, b_5, \ldots) = (-1, -1, -1, \ldots) \to -1$$

The original sequence $(b_n)$ does not converge, but this subsequence does!

Proof

Suppose $(a_n) \to a$ and let $(a_{n_k})$ be a subsequence. We want to show $(a_{n_k}) \to a$.

Let $\varepsilon > 0$. Since $(a_n) \to a$, there exists $N \in \mathbb{N}$ such that whenever $n \geq N$, we have $|a_n - a| < \varepsilon$.

Now observe: $n_1 < n_2 < n_3 < \cdots$ are strictly increasing. In particular, there exists a term $n_K$ in this sequence such that .

Then for $k \geq K$, we have $n_k \geq n_K \geq N$, so

$$|a_{n_k} - a| < \varepsilon$$

Therefore, $(a_{n_k}) \to a$. $\square$

Proof Setup

Let $(a_n)$ be a bounded sequence. Then there exist $m, M \in \mathbb{R}$ with $m \leq a_n \leq M$ for all $n$.

The Bisection Idea:

We'll recursively divide the interval $[m, M]$ in half, always keeping the half that contains infinitely many terms of $(a_n)$.

Bisection Algorithm

Step 1: Initialize

Set $I_0 = [m, M]$. This interval contains infinitely many terms of $(a_n)$ (all of them).

Step 2: Recursive bisection

For $k = 1, 2, 3, \ldots$:

1. Let $I_{k-1} = [a_k, b_k]$ and set the midpoint $c_k = \dfrac{a_k + b_k}{2}$.
2. The interval $I_{k-1}$ contains infinitely many terms. At least one of $[a_k, c_k]$ or $[c_k, b_k]$ must contain infinitely many terms.
3. Define $I_k$ to be the half-interval containing infinitely many terms. (If both do, pick either one.)

Step 3: Extract the subsequence

For each $k$, choose an index $n_k$ from $I_k$ such that:

• $n_k \in I_k$ (so $a_{n_k}$ is in the interval $I_k$)
• $n_1 < n_2 < n_3 < \cdots$ (indices are strictly increasing)

This is possible because each $I_k$ contains infinitely many terms.

Step 4: Show convergence

The intervals satisfy: $I_1 \supseteq I_2 \supseteq I_3 \supseteq \cdots$, and each $I_k$ has length $\dfrac{M-m}{2^k}$.

As $k \to \infty$, the lengths shrink to zero. The nested intervals shrink to a single point, say $L = \bigcap_{k=1}^{\infty} I_k$.

Since $a_{n_k} \in I_k$ for all $k$, and $I_k \to \{L\}$, we have $(a_{n_k}) \to $ . $\square$

Example 1: Extracting from $(-1)^n$

The sequence $(a_n) = (-1)^n$ does not converge. But it is bounded: $|a_n| \leq 1$ for all $n$.

By Bolzano-Weierstrass, it must have a convergent subsequence. Indeed:

• Even indices: $(a_{2k}) = (1, 1, 1, \ldots) \to $
• Odd indices: $(a_{2k-1}) = (-1, -1, -1, \ldots) \to $

Example 2: A More Complex Bounded Sequence

Let $(b_n) = \dfrac{(-1)^n n}{n+1}$. Then $|b_n| \leq 1$ (bounded). Since $(b_n)$ is bounded:

By Bolzano-Weierstrass, it has a convergent subsequence. Which one?

Even indices:

$$(b_{2k}) = \frac{2k}{2k+1} = \frac{1}{1 + \frac{1}{2k}} \to $

Odd indices:

$$(b_{2k-1}) = \frac{-(2k-1)}{2k} = \frac{-1}{1 + \frac{1}{2k}} \to $

Exercise 1

Let $(a_n) = \sin(n)$ for $n = 1, 2, 3, \ldots$. Explain why this sequence has a convergent subsequence, even though $(a_n)$ itself does not converge.

Exercise 2

For the sequence $c_n = (-1)^n + \dfrac{1}{n}$, find a convergent subsequence and determine its limit.