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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!

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  • $\begingroup$ You can't do this without basically proving that the function in question is continuous. You might not use continuity directly, but anything you do will essentially prove continuity to some extent. So you will be reinventing the wheel. Why wouldn't you want to use continuity? Using $\epsilon$-$\delta$ arguments amounts to that and is excruciating besides. $\endgroup$ – MPW May 1 '15 at 13:36
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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

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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.

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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.

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