# Squeeze Theorem/Sequences

## Theorem

There are two versions of this result:

- one for sequences in the set of complex numbers $\C$, and more generally for sequences in a metric space.
- one for sequences in the set of real numbers $\R$, and more generally in linearly ordered spaces (which is stronger).
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### Sequences of Real Numbers

Let $\sequence {x_n}$, $\sequence {y_n}$ and $\sequence {z_n}$ be sequences in $\R$.

Let $\sequence {y_n}$ and $\sequence {z_n}$ both be convergent to the following limit:

- $\ds \lim_{n \mathop \to \infty} y_n = l, \lim_{n \mathop \to \infty} z_n = l$

Suppose that:

- $\forall n \in \N: y_n \le x_n \le z_n$

Then:- $x_n \to l$ as $n \to \infty$

that is:

- $\ds \lim_{n \mathop \to \infty} x_n = l$

Thus, if $\sequence {x_n}$ is always between two other sequences that both converge to the same limit, $\sequence {x_n} $ is said to be**sandwiched**or**squeezed**between those two sequences and itself must therefore converge to that same limit.

### Sequences of Complex Numbers

Let $\sequence {a_n}$ be a null sequence in $\R$, that is:

- $a_n \to 0$ as $n \to \infty$

Let $\sequence {z_n}$ be a sequence in $\C$.

Suppose $\sequence {a_n}$ dominates $\sequence {z_n}$.That is:

- $\forall n \in \N: \cmod {z_n} \le a_n$

Then $\sequence {z_n}$ is a null sequence.

### Sequences in a Linearly Ordered Space

Let $\struct {S, \le, \tau}$ be a linearly ordered space.

Let $\sequence {x_n}$, $\sequence {y_n}$, and $\sequence {z_n}$ be sequences in $S$.

Let $p \in S$.

Let $\sequence {x_n}$ and $\sequence {z_n}$ both converge to $p$.

Let $\forall n \in \N: x_n \le y_n \le z_n$.

Then $\sequence {y_n}$ converges to $p$.

### Sequences in a Metric Space

Let $M = \struct {S, d}$ be a metric space or pseudometric space.

Let $p \in S$.

Let $\sequence {r_n}$ be a null sequence in $\R$.

Let $\sequence {x_n}$ be a sequence in $S$ such that:

- $\forall n \in \N: \map d {p, x_n} \le r_n$.

Then $\sequence {x_n}$ converges to $p$.

## Also known as

This result is also known, in the UK in particular, as the

**sandwich theorem**.