证明集合$S = \{<x,y>\,:\,xy=1\}$在$\mathbb{R}^2$中为闭集（开集定义法）

To show $S$ is closed, we'll prove its complement $U = \mathbb{R}^2 \setminus S$ is open. By definition, a set is open if every point in it has an open rectangle (or open disk) neighborhood entirely contained in the set. Our goal is to show for any $(a,b) \in U$, there exists an open rectangle around $(a,b)$ where every $(x,y)$ in the rectangle satisfies $xy \neq 1$.

Key Observations #

• For $(a,b) \in U$, $ab \neq 1$, so let $c = |ab - 1| > 0$. We need to ensure that for all $(x,y)$ in our rectangle, $xy$ stays far enough from 1 that it never equals 1. Using the triangle inequality:

  $|xy - 1| = |(xy - ab) + (ab - 1)| \geq \left| |ab - 1| - |xy - ab| \right| = c - |xy - ab|$

• If we can make $|xy - ab| < c$, then $|xy -1| > 0$, so $xy \neq 1$. That's our target.

Case Analysis for $(a,b) \in U$ #

Case 1: $a \neq 0$ #

Let's pick $\delta$ and $\epsilon$ to bound $|xy - ab|$ below $c$:

1. Choose $\delta = \min\left(1, \frac{c}{2(|b| + 1)}\right)$. This ensures:
   • $|x - a| < \delta \implies |x| \leq |a| + \delta \leq |a| + 1$
   • $|b||x - a| < |b| \cdot \delta \leq \frac{c}{2}$ (since $\delta \leq \frac{c}{2(|b|+1)}$)
2. Choose $\epsilon = \frac{c}{2(|a| + 1)}$. Then:
   • $|y - b| < \epsilon \implies |x||y - b| \leq (|a| +1) \cdot \epsilon = \frac{c}{2}$
3. Combining these, $|xy - ab| = |x(y - b) + b(x - a)| \leq |x||y - b| + |b||x - a| < \frac{c}{2} + \frac{c}{2} = c$.

The open rectangle $(a-\delta, a+\delta) \times (b-\epsilon, b+\epsilon)$ is entirely contained in $U$.

Case 2: $a = 0$ #

Since $a=0$, $ab=0 \neq 1$, so $c=1$. We split into two subcases:

• Subcase 2a: $b \neq 0$:
  Choose $\delta = \frac{1}{2(|b| +1)}$ and $\epsilon = \frac{1}{2}$. For any $(x,y)$ in $(-\delta, \delta) \times (b-\epsilon, b+\epsilon)$:
  • $|x| < \delta < \frac{1}{|b| +1}$
  • $|y| \geq |b| - \epsilon = |b| - \frac{1}{2}$. If $|b| > \frac{1}{2}$, $|y| >0$, so $|xy| < \frac{1}{|b|+1} \cdot (|b| + \frac{1}{2}) <1$. If $|b| \leq \frac{1}{2}$, $|y| \leq |b| + \frac{1}{2} \leq1$, so $|xy| < \delta \cdot1 <1$. Either way, $xy \neq1$.
• Subcase 2b: $b =0$:
  Choose $\delta=1$ and $\epsilon=1$. The rectangle $(-1,1) \times (-1,1)$ contains only points where $|xy| <1$, so $xy \neq1$.

Conclusion #

For every point in $U$, we've constructed an open rectangle neighborhood entirely within $U$. This means $U$ is open, so its complement $S$ is closed in $\mathbb{R}^2$.

内容的提问来源于stack exchange，提问作者tobiasbriones