Closed subset of the $\mathbb{R}^n$

Question

I want to show that $U = \{(x, y) \in \mathbb{R}^2|xy ≤ 1\}$ is a closed subset of $\mathbb{R}^2$.

Yes there are (easy) ways to do this using functions, but what's the (easiest) way to prove this without using continuous functions? I thought that when $U$ is closed, $U^c = \{(x, y) \in \mathbb{R}^2|xy > 1\}$ must be open, and we could maybe choose a random point $(x, y) \in C^c$ and cleverly choose an $\epsilon > 0$, and then show that $B_\epsilon(x, y) \subseteq U$. But how to choose $\epsilon$, and how to show that the open ball for a fitting $\epsilon$ really is a subset of U? Or are there other, maybe easier ways that don't involve functions? Thanks in advance!

Answer 1

Like you did, consider $U^c$ and we'll show it's an open set.

take $(x,y) \in U^c$ such that $xy= 1+\delta$

Now, as $y\neq 0$, $x= \frac{1+\delta}{y}$

Consider the norm $\|(x,y)\| = |x|+|y|$

Take $(x',y')$ such that $\|(x',y')\|< \epsilon$

$$c = (x+x')(y+y') = xy + xy'+x'y+x'y' $$ $$1+\delta - \epsilon |x| - \epsilon |y| + \epsilon^2 \leq c $$

$$1+\delta - (\epsilon\|(x,y)\|) + \epsilon^2 \leq c $$

Now take $\epsilon$ such that $$\epsilon\|(x,y)\|- \epsilon^2 \leq \delta$$

And the claim is proved

Answer 2

Choose a point $(x,y)\in \mathbb{R^2}$ such that $xy \gt 1$. Then Take $\epsilon=\frac{1}{2}min\{|x-\frac{1}{y}|,|y-\frac{1}{x}|\}$. Then you will be done.

Answer 3

Suppose $(x_n,y_n)$ is a sequence in $U$ and $(x_n,y_n)\to (a,b).$ We want to show $(a,b)\in U.$ As we know, $(x_n,y_n)\to (a,b)$ implies $x_n\to a, y_n \to b.$ By the product rule for limits, $x_ny_n \to ab.$ But $x_ny_n \in (-\infty, 1]$ for all $n.$ Because $(-\infty, 1]$ is closed in $\mathbb {R},$ the limit of $x_ny_n$ is in $(-\infty, 1].$ I.e., $ab\le 1$ and $(a,b)\in U$ as desired.