# Note 6 - Cauchy Sequences and Convergence

## Table of Contents

## Cauchy Sequences and Convergence

Recall: For a bounded sequence \((a_n)\):

- \(I_N = \inf\{a_n, n>N\}\)
- \(S_N = \sup\{a_n, n>N\}\)
- \(I_1 \leq I_2 \leq ... \leq \liminf a_n \leq \limsup a_n \leq ... \leq S_2 \leq S_1\)

**Def**: \((a_n)\) is Cauchy if for all \(\epsilon > 0\) we can find an such that \(\vert a_n - a_m \vert < \epsilon\), for all \(n,m > N\)

**Thm**: A convergent sequence \((a_n)\) implies that \((a_n)\) is Cauchy

**Proof**: \(\forall \epsilon > 0\) s.t. \(\vert a_n - a_m \vert < \epsilon \forall n,m > N\)
\(\implies a_n < a_m + \epsilon \forall n,m > N\)

For any \(n>N\), we have

- \(a_n < a_m + \epsilon\ \forall m > N\)
- \(\implies a_n \leq \inf\{a_m: m > N\} + \epsilon\)
So we have, for any \(n > N\),

- \(a_n \leq I_N + \epsilon\)
- \(\implies \sup\{a_n: n > N\} \leq I_N + \epsilon\)
- \(\limsup_{n \to \infty} a_n \leq S_N \leq I_N + \epsilon \leq \liminf_{n \to \infty} a_n+ \epsilon\)
- \(\implies \limsup a_n \leq \liminf a_n + \epsilon \forall \epsilon > 0\)
- \(\implies \limsup a_n \leq \liminf a_n\), since we always have \(\liminf a_n \leq \limsup a_n < n\)

We proved that bounded and \(\liminf = \limsup\) implies convergence, and convergence implies Cauchy, so now we will prove Cauchy implies \(\liminf = \limsup\).

## Cauchy Implies Infimum = Supremum

**Def**: A subsequence of a sequence \((a_n)\) is a sequence \((a_{k_n}) = (a_{k_1}, a_{k_2}, ...)\) where \(1 \leq k_1 < k_2 < ... < a_m\) are positive integers, e.g.

- \((a_{2n}) = (a_2, a_4, a_6, ...)\)
- \((a_{2n-1}) = (a_1, a_3, a_5, ...)\)

In homework, we proved that \(\lim_{n \to \infty} a_n = 0 \implies \lim_{n \to \infty} a_{2n} = a = \lim_{n \to \infty} a_{2n-1}\)

### Subsequences Approach Sequence Limit

**Question**: Suppose \(\lim_{n \to \infty} a_n = a\), is it true that \(\lim_{n \to \infty} a_{k_n} = a\) for any subsequence \((a_{k_n})\)?

Yes: \(\forall \epsilon > 0, \exists N > 0\) s.t. \(\vert a_n - a \vert < \epsilon \forall n > N\)

The intuition is that there exists a \(k_n\) that is greater than \(N\), at which point the bounds will hold true.

And our sequence is \(a_{k_1}, ...\), which contains strictly increasing \(k_n\), so \(\exists M > 0\) such that \(k_M > N\). Then, for any \(n > M\), we know \(\vert a_{k_n} - a \vert < \epsilon\) since \(k_n > k_M > N\).

The converse is not true. Consider \((0, 1, 0, 1, ...)\). The odd terms converge to \(0\), but the even terms to \(1\). Note that \(\liminf a_n = 0\) and \(\limsup a_n = 1\). We will prove that for any bounded sequence, there exists a subsequence that converges to \(\liminf\) and one that converges to \(\limsup\).

**Another question**: If \(\lim a_{2n} = a = \lim a_{2n - 1}\), does this imply that \(\lim a_n = a\)? It turns out the answer is *yes*, and the proof is left as an exercise.

## Bolzano Weierstrass

**Definition**: Any bounded sequence has a convergent subsequence

It suffices to prove any bounded sequence has a monotone subsequence. If we can bound this monotone subsequence, it will be

monotone and bounded, hence convergent.

**Lemma**: Any bounded subsequence has a monotone subsequence.

**Proof**:

Definition: Given a bounded sequence \((a_n)\), we say \(a_n\) is

bigif it is bigger than all the terms after it, i.e. \(a_n > a_m \forall n < m\).For any bounded sequence \((a_n)\),

- It has infinitely many big terms, or
- It has finitely many big terms.
In the first case, the sequence has infinitely many big terms, so we can just take all the big terms: \(\{ a_{k_1}, a_{k_2}, ...\}\), the sequence of big terms from \((a_n)\). This gives a subsequence of \((a_n)\), a strictly decreasing subsequence (since each term is strictly greater than the term after).

In the second case, the sequence has finitely many big terms, so we take \(N = \max\{k_1, ... k_n\}\). Consider only \(a_n\) for \(n>N\). These \(a_n\) are not big for all of these \(n\)s.

Now we define \(a_{k_1} = a_{N+1}\). Since \(k_1 = N+1 > N\), \(a_k\) is not big. This implies that there exists \(k_2 > k_1\) such that \(a_{k_2} \geq a_{k_1}\). Since \(k_2 > k_1 > N\), \(a_{k_2}\) is not big. Using the same argument, we can construct \(k_3\), and construct the remaining inductively. This creates a monotonically increasing subsequence.

## Bounded to Limsup and Liminf

**Theorem**: For a bounded sequence, there exists two subsequences that converges to \(\liminf\) and \(\limsup\), respectively.

To prove this theorem, first we prove a lemma:

### Lemma: Subsequence iff Infinite Set Within Neighborhood

**Lemma**: Having a subsequence of a sequence converging to \(a\) is equivalent to \(\forall \epsilon > 0\), the set \(\{n \in N: \vert a_n - a \vert < \epsilon\}\) is infinite.

**Proof**:

(→): \((a_{k_n})\) converges to \(a\):

\(\forall \epsilon > 0, \exists N > 0\) s.t. \(\vert a_{k_n} - a \vert < \epsilon \forall n > N\)

which implies that \(\{k_{N+1}, k_{N+2}, ...\} \subset \{n \in N: \vert a_n - a \vert < \epsilon\}\)

(←):

Idea: We’ll find a subsequence (\(a_{k_1}, a_{k_2}, ...\)) such that the differences are less than 1, \(1/2\), … \(1/n\).

- We know that \(\{n \in N: \vert a_n - a \vert < 1\}\) is infinite.
- We know that \(\{n \in N: \vert a_n - a \vert < 1/2\}\) is infinite, so \(\exists k_2 > k_1\) s.t. \(\vert a_{k_2} - a \vert < 1/2\).
- We know that \(\{n \in N: \vert a_n - a \vert < 1/3\}\) is infinite, so \(\exists k_3 > k_1\) s.t. \(\vert a_{k_3} - a \vert < 1/3\).
- etc.

Now back to the original proof:

**Proof**:

Let \(Z := \limsup_{n\to\infty} a_n = \lim_{N\to\infty} S_N = \limsup_{N\to\infty} \{a_n : n > N\}\)

Want to show: \(\forall \epsilon > 0\), the set \(\{n \in N: \vert a_n - z \vert < \epsilon\}\) is infinite.

Proof by contradiction.Suppose that \(\{n \in N: \vert a_n - z \vert < \epsilon\}\) is finite.

- \(\exists N_1 > 0\) such that \(z - \epsilon < \sup \{a_n: n>N\} = S_{N_1} < z-\epsilon\). This implies that \(a_n < z + \epsilon\) for all \(n > N_1\).
- Having a finite set means that \(\exists N_2 > 0\) such that \(z - \epsilon < a_n < z + \epsilon\) is false \(\forall n > N_2\).
Now let \(N := \max\{N_1, N_2\}\). For any \(n > N\), we have

- \(a_n < z + \epsilon\)
- \(z - \epsilon < a_n < z + \epsilon\) is false.
which implies that \(\forall n > N, a_n \leq z - \epsilon\), which means \(\sup\{a_n: n > N\} \leq z - \epsilon\). But recall that \(\sup\{a_n: n > N\} = S_N = \limsup_{n\to\infty}a_n = z\), which is a contradiction.