Limit of Sequence to Zero Distance Point

Theorem

Let $S$ be a non-empty subset of $\R$.

Let the distance $\map d {\xi, S} = 0$ for some $\xi \in \R$.

Then there exists a sequence $\sequence {x_n}$ in $S$ such that $\ds \lim_{n \mathop \to \infty} x_n = \xi$.

If $S$ is bounded above, then there exists a sequence $\sequence {x_n}$ in $S$ such that:

$\ds \lim_{n \mathop \to \infty} x_n = \sup S$

If $S$ is unbounded above, then there exists a sequence $\sequence {x_n}$ in $S$ such that $x_n \to +\infty$ as $n \to \infty$.

First it is shown that:

$\forall n \in \N_{>0}: \exists x_n \in S: \size {\xi - x_n} < \dfrac 1 n$

Aiming for a contradiction, suppose that:

$\exists n \in \N_{>0}: \not \exists x \in S: \size {\xi - x} < \dfrac 1 n$

Then $\dfrac 1 n$ is a lower bound of the set $T = \set {\size {\xi - x}: x \in S}$.

This contradicts the assertion that $\map d {\xi, S} = 0$.

$\ds \lim_{n \mathop \to \infty} \dfrac 1 n = 0$

So as $\size {\xi - x_n} < \dfrac 1 n$ it follows from the Squeeze Theorem for Real Sequences that:

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

Hence the result.

$\blacksquare$