Cauchy Sequence is Bounded/Metric Space

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Theorem

Let $M = \struct {A, d}$ be a metric space.


Then every Cauchy sequence in $M$ is bounded.


Proof

Let $\sequence {x_n}$ be a Cauchy sequence in $M$.

By definition:

$\forall \epsilon > 0: \exists N \in \N: \forall m, n > N: \map d {x_n, x_m} < \epsilon$

Particularly, setting $\epsilon = 1$:

$\exists N_1: \forall m, n > N_1: \map d {x_n, x_m} < 1$

Note that since $N_1 \ge N_1$, this means that:

$\forall n \ge N_1: \map d {x_n, x_{N_1} } < 1$


To show $\sequence {x_n}$ is bounded, we need to show that there exists $a \in A$ and $K \in \R$ such that $\map d {x_n, a} \le K$ for all $x_n \in \sequence {x_n}$.


Let $K'$ be the maximum distance from $x_{N_1}$ to any of the earlier terms in the sequence.

That is, $K' = \max \set {\map d {x_{N_1}, x_1}, \map d {x_{N_1}, x_2}, \ldots, \map d {x_{N_1}, x_{N_1 - 1} } }$


Then:

Each $x_n$ for $n \ge N_1$ satisfies $\map d {x_{N_1}, x_n} \le 1$ by choice of $N_1$ as mentioned above
Each $x_n$ for $n < N_1$ satisfies $\map d {x_{N_1}, x_n} \le K'$ by choice of $K'$.


Thus, taking $a = x_{N_1}$ and $K = \max \set {K', 1}$, we have shown that each $x_n$ satisfies $\map d {a, x_n} \le K$.


So, $\sequence {x_n}$ is bounded.

$\blacksquare$


Sources

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