Show this curve is a closed set in $R^2$ by using the definition

Question

Let $S = \{<x,y>\,:\,xy=1\}$, prove S is closed in $R^2$. A similar exercise is proving a line is closed. This is requested to prove by using the definition that says that the open sets are the open rectangles (it could also be open discs) so I'm requested to prove it that way, I made I proof for any smooth curve using calculus but I don't have a good idea about demonstrating this exercise as requested, the point is to show the complement of S is open, that is, the union of open rectangles.

Answer 1

I'm with you: I'd do it using the topological definition of continuity. We can always use continuity to help inform our method, even if we don't use it in the end. Since we want to work with rectangles forming our topological basis, I suggest we start proving $f(x, y) = xy$ is continuous, where $\mathbb{R}^2$ is endowed with the $\infty$-norm: $$\|(x, y)\| := \max \lbrace |x|, |y| \rbrace.$$ I chose this because the balls of $\| \cdot \|$ are square (and the sides are parallel to the axes).

Fix $\varepsilon > 0$ and $(x_0, y_0) \in \mathbb{R}^2$. Note first that, $$\|(x, y) - (x_0, y_0)\| = \max \lbrace |x - x_0|, |y - y_0| \rbrace \ge |x - x_0|, |y - y_0|.$$ Consider, \begin{align*} |f(x, y) - f(x_0, y_0)| &= |xy - x_0 y_0| \\ &= |xy - x_0 y + x_0 y - x_0 y_0| \\ &\le |x - x_0| |y| + |y - y_0||x_0| \\ &= |x - x_0| |y - y_0 + y_0| + |y - y_0||x_0| \\ &\le |x - x_0| |y - y_0| + |x - x_0||y_0| + |y - y_0||x_0| \\ &\le \|(x, y) - (x_0, y_0)\|^2 + (|y_0| + |x_0|)\|(x, y) - (x_0, y_0)\|. \end{align*} If, in particular, we additionally assume $\|(x, y) - (x_0, y_0)\| \le 1$, then $$\|(x, y) - (x_0, y_0)\|^2 \le \|(x, y) - (x_0, y_0)\|,$$ hence $$|f(x, y) - f(x_0, y_0)| \le (1 + |y_0| + |x_0|)\|(x, y) - (x_0, y_0)\|.$$ So, setting, $$\delta = \min \left\lbrace \frac{\varepsilon}{1 + |y_0| + |x_0|}, 1 \right\rbrace,$$ yields $$\|(x, y) - (x_0, y_0)\| < \delta \implies |f(x, y) - f(x_0, y_0)| < \varepsilon.$$

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How does this help us? The process of coming up with these estimates gives us the estimates we want for the question. Fix a point $(x_0, y_0) \notin S$. Denote $\varepsilon = |f(x_0, y_0) - 1| = |x_0 y_0 - 1|$. From the above argument, we know to set $$\delta = \min \left\lbrace \frac{|x_0 y_0 - 1|}{1 + |y_0| + |x_0|}, 1 \right\rbrace.$$ Then, following the above logic, we therefore have, \begin{align*} (x, y) \in B((x_0, y_0); \delta) &\implies \|(x, y) - (x_0, y_0)\| < \delta \\ &\implies |f(x, y) - f(x_0, y_0)| < |x_0 y_0 - 1| \\ &\implies |(x_0 y_0 - 1) - (xy - 1)| < |x_0 y_0 - 1| \\ &\implies |x_0 y_0 - 1| - |xy - 1| < |x_0 y_0 - 1| \\ &\implies |xy - 1| > 0 \\ &\implies xy \neq 1 \\ &\implies (x,y) \notin S. \end{align*} Therefore, $B((x, y); \delta)$ is contained in the complement of $S$, for some $\delta > 0$ and every $(x, y) \notin S$. This implies $\mathbb{R}^2 \setminus S$ is open, hence $S$ is closed.