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What is the closure in $\mathbb{R}^2$ of the set $$ \left\{ \ x \times y \ \in \mathbb{R} \times \mathbb{R} \ \colon \ x > 0, \ y = \frac{1}{x} \ \right\}? $$

I know that each point of the set is also a limit point. Does this set have any limit points in addition to its elements?

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No, there are no other limit points: this set is closed. Suppose $(x, y)$ is a limit point. Then there is a sequence $x_n$ of real numbers such that $x_n \to x$ and $\frac{1}{x_n} \to y$. Then $xy = \lim x_n \frac{1}{x_n} = 1$, because the multiplication map $\mathbb{R}^2 \to \mathbb{R}$ is continuous. So $x \neq 0$ and $y = \frac{1}{x}$. (It's clear that $x$ isn't negative since it's a limit of positive numbers).

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