數學求救信

之前(2017 年)在刷 Email 看有沒有人寄面試邀請函時居然看到了數學求救信 😑

題目是這樣的:

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

老實說我連 dense set 的定義是什麼都忘了啊~ 🤣

Definition

言歸正傳,根據 Walter Rudin 寫的 Principle of Mathematical Analysis 3rd edition 裡提到的定義:

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).

另外 limit of a sequence 的定義這個就不需贅述了。

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 $\mathbb{R}$. $\blacksquare$

結論

因為這題根據定義就能做出來!所以沒有回信給這位求救的朋友,但還是手癢把它寫出來。如果有緣的話在茫茫 Google 海中自會再相見。 😂

updatedupdated2021-10-222021-10-22