Prove that $S=\{ ( x,y) :| xy| < 1\}$ is open in standard Euclidean metric?

Question

The problem I always face as a fresher is how to think about a defined $\epsilon$ for defined $\left(x,\,y\right)$ so that I can show the open ball with radius $\epsilon$ lies within the set.

Answer 1

Let $|x_0y_0| <1$. Suppose $\|(x,y)-(x_0,y_0)\| <\epsilon$ where $\epsilon =\min \left(\frac {1-|x_0y_0|}{1+|x_0|+|y_0|},1\right)$. Then $$|xy|\leq\left(|x_0|+\epsilon\right)\left(|y_0|+\epsilon\right)=|x_0||y_0| +\epsilon^{2}+\epsilon\left(|x_0|+|y_0|\right)<|x_0||y_0|+\epsilon +\epsilon\left(|x_0|+|y_0|\right)<1.$$

Answer 2

Consider the continuous map \begin{align} (x,y)\mapsto |xy| \end{align} and the preimage of the open set $(-\infty,1)$.

Answer 3

Since the product topology on $\mathbb{R}^2$ is generated by the basis $$B = \{U \times V: U, V \ \text{are open in} \ \mathbb{R} \}$$

and since any open interval is open in $\mathbb{R}$, the cartesian product of open intervals is open in $\mathbb{R}^2$. It suffices to verify that $$S = \bigcup_{m \in \mathbb{R}_{>0}}\left( (-m, m) \times (-\frac{1}{m}, \frac{1}{m}) \right) $$

Edit: Just to be thorough, $\mathbb{R}^2$ equipped with the product topology is metrizable with the Euclidean metric. Thus, a set will be open in $\mathbb{R}^2$ with respect to the product topology iff it is open with respect to the metric topology induced by the Euclidean metric.