Squeeze Theorem/Sequences/Metric Spaces

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Theorem

Let $M = \left({S, d}\right)$ be a metric space or pseudometric space.

Let $\left\langle{x_n}\right\rangle$ be a sequence in $S$.

Let $p \in S$.

Let $\left\langle{r_n}\right\rangle$ be a sequence in $\R_{\ge 0}$.

Let $\left\langle{r_n}\right\rangle$ converge to $0$.

For each $n$, let $d \left({p, x_n}\right) \le r_n$.


Then $\left\langle{x_n}\right\rangle$ converges to $p$.


Proof

Let $\epsilon > 0$.

Since $r$ converges to $0$, there exists an $N$ such that $n > N \implies r_n < \epsilon$.

But then by Extended Transitivity, $n > N \implies d \left({p, x_n}\right) < \epsilon$.

Thus $\left\langle{x_n}\right\rangle$ converges to $p$.

$\blacksquare$