📂MetricSpace

Convergence of Sequences in Metric Spaces

Definitions¹

If there exists a point $p \in X$ such that the sequence $\left\{ p_{n} \right\}$ of points in a metric space $(X,d)$ satisfies the following condition, the sequence $\left\{ p_{n} \right\}$ is said to converge to $p$, and it is denoted by $p_{n} \rightarrow p$ or $\lim \limits_{n\to \infty}p_{n}=p$.

$$\forall \varepsilon >0,\ \exists N\in \mathbb{N}\ \mathrm{s.t}\ n\ge N \implies d(p_{n},p)<\varepsilon$$

If $\left\{ p_{n} \right\}$ does not converge, it diverges. Furthermore, the set of all $p_{n}$s is called the range of $\left\{ p_{n} \right\}$. If the range of $\left\{ p_{n} \right\}$ is bounded, the sequence $\left\{ p_{n} \right\}$ is said to be bounded.

Theorems

Let $\left\{ p_{n} \right\}$ be a sequence in a metric space $(X,d)$.

(a) A necessary and sufficient condition for $p_{n}\to p$ is that every neighborhood of $p$ contains all terms of $\left\{ p_{n} \right\}$ except for a finite number.

(b) If $p_{n} \to p$ and $p_{n} \to p^{\prime}$, then $p=p^{\prime}$.

(c) If $\left\{ p_{n} \right\}$ converges, it is bounded.

(d) Given $E\subset X$. If $p$ is a limit point of $E$, there exists a sequence $\left\{ p_{n} \right\}$ from $E$ that satisfies $p=\lim \limits_{n \to \infty}p_{n}$. Furthermore, if $\left\{ p_{n} \right\}$ is a set of distinct points, the converse also holds.

Proofs

(a)

• $(\implies)$

  Suppose $p_{n} \to p$. Given any positive number $\varepsilon >0$, let $V$ be a neighborhood of $p$ with radius $\varepsilon$. According to the definition of a neighborhood, the following holds:

  $$d(p,q)<\varepsilon\quad \implies q\in V$$

  However, by assumption, there exists a $N$ that satisfies the following condition for the given $\varepsilon$:

  $$\forall n \ge N,\ d(p_{n},p) <\varepsilon$$

  Therefore, all points $p_{n}$, except for a finite number, are included in $V$.

• $(\impliedby)$

  Assume every neighborhood of $p$ contains all but a finite number of points $\left\{ p_{n} \right\}$. Given any positive number $\varepsilon>0$, let $V$ be a neighborhood of $p$ with radius $\varepsilon$. Then, by assumption, a $N$ exists that satisfies the following condition:

  $$n \ge N \implies p_{n}\in V$$

  Since $V$ is a neighborhood of $p$, the following holds:

  $$\forall n \ge N,\quad d(p_{n},p)<\varepsilon$$

  Therefore, $p_{n}\to p$.

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(b)

Given any positive number $\varepsilon >0$. By assumption, there exist two positive numbers $N$, $N^{\prime}$ that satisfy the following condition:

$$\begin{align*} n\ge N & \implies d(p_{n},p) <\frac{\varepsilon}{2} \\ n\ge N^{\prime} & \implies d(p_{n},p) <\frac{\varepsilon}{2} \end{align*}$$

Then, for $n \ge \max(N,N^{\prime})$, the following equation holds:

$$d(p,p^{\prime}) \le d(p,p_{n}) + d(p_{n},p^{\prime})<\varepsilon$$

Since $\varepsilon$ is any positive number,

$$d(p,p^{\prime})=0$$

and by the definition of distance, $p=p^{\prime}$.

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(c)

Assume that $\left\{ p_{n} \right\}$ converges to $p$. By assumption, there exists a positive number $N$ that satisfies the following equation:

$$n \ge N \implies d(p_{n},p)<1$$

Now, let

$$r=\max \left\{ 1,\ d(p_{1},p),\ \cdots,\ d(p_{N},p) \right\}$$

Then, for all $n$,

$$d(p_{n},p)\le r$$

so $\left\{ p_{n} \right\}$ is bounded.

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(d)

• $(\implies)$

  Assume $E\subset X$ and that $p$ is a limit point of $E$. By definition of a limit point, for each $n$,

  $$d(p_{n},p) < \frac{1}{n}$$

  exists a $p_{n}\in E$. Now, given any positive number $\varepsilon >0$ and $N\varepsilon>1$ satisfying $N$, the following holds for $n >N$:

  $$d(p_{n},p)< \frac{1}{n}<\frac{N}{n}\varepsilon<\varepsilon$$

  Therefore, $\left\{ p_{n} \right\}$ converges to $p$.

• $(\impliedby)$

  Assume there exists a sequence $\left\{ p_{n} \right\}$ of distinct points from $E$ that satisfies $p=\lim \limits_{n\to\infty}p_{n}$. Then, for all positive numbers $\varepsilon >0$,

  $$n \ge N \implies d(p_{n},p)< \varepsilon$$

  exists a $N$. Let $V_{\varepsilon}$ be a neighborhood of $p$ with radius $\varepsilon$. Thus, $V_{\varepsilon}$ includes $p_{n} \in E (n\ge N)$ which is not $p$, proving $p$ is a limit point of $E$.

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