Limit Point is Limit of Convergent Sequence

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $E \subseteq X$ be a subset of $X$.

Let $p$ be a limit point of $E$.


Then there exists a sequence $\left\langle{x_n}\right\rangle \subseteq E$ which converges to a limit:

$\displaystyle \lim_{n \mathop \to \infty} x_n = p$

where $\displaystyle \lim_{n \mathop \to \infty} x_n$ is the limit of the sequence $\left\langle{x_n}\right\rangle$.


Proof 1

We may construct a sequence $\left\langle{p_n}\right\rangle$ using finite recursion.

Let $p_0$ be some arbitrary point in $E$ not equal to $p$.

$p_{n+1}$ is some arbitrary point in $E \cap B_{\frac{d \left({p, p_n}\right)} 2} \left({p}\right)$ where $B_{\frac{d \left({p, p_n}\right)} 2} \left({p}\right)$ denotes the open $\dfrac{d \left({p, p_n}\right)} 2$-ball of $p$.

We can be assured that such a point exists by the definition of a limit point.

Then:

$d \left({p_n, p}\right) < 2^{-n} d \left({p_0, p}\right)$

Since $d \left({p_0, p}\right)$ is fixed:

$d \left({p_n, p}\right) < 2^{-n} d \left({p_0, p}\right) < r$

for some $n$ because the natural numbers are Archimedean.

Therefore $p$ is the limit of the sequence $\left\langle{p_n}\right\rangle$ by definition.

$\blacksquare$


Proof 2

From Metric Induces Topology, a metric space induces a topological space.

Then Equivalence of Definitions of Limit Point can be applied.

$\blacksquare$


Sources