# 數學求救信

A set $S$ is dense in $\mathbb{R}$ if and only if every number $x$ is the limit of a sequence in $S$.

## Definition

Let $X$ be a metric space.

• A neighborhood of a point $p$ is a set $N_r(p)$ consisting of all points q such that $d(p, q) < r$. The number $r$ is called radius of $N_r(p)$.
• A point $p$ is a limit point of the set $E$ if every neighborhood of $p$ contains $q \neq p$ such that $q \in E$.
• A set $E$ is dense in $X$ if every point of $X$ is a limit point of $E$, or a point of $E$(or both).

## Proof

If $S$ is dense in $\mathbb{R}$ then every point $x \in \mathbb{R}$ is a limit point of $S$, or a point of $S$(or both).

If $x \in S$, then it is trivial that $x$ is the limit of sequence ${ x }$.

Otherwise choose sequence ${ q_i \mid q_i \in S, d(x, q_i) < 1/i, i = 1, 2, \dots }$. For every $\epsilon > 0$ choose an integer $N = \lceil\frac{1}{\epsilon}\rceil \geq \frac{1}{\epsilon}$ such that $n \geq N$ implies

$$d(x, q_n) < \frac{1}{n} \leq \frac{1}{N} \leq \epsilon.$$

Then $x$ is the limit of a sequence in $S$.

Conversely, consider $x$ is the limit of a sequence ${q_i}$ in $S$ and $x \notin S$, then given arbitrary radius $r$, there exists some integer $N$ such that $d(x, q_N) < r$, i.e. $x$ is a limit point of $S$. Since every point is either in $S$ or a limit point of $S$(or both), then $S$ is dense in $\mathbb{R}$. $\blacksquare$