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If we consider $M \subseteq \mathbb{R}^2$, where

$$M = \{(x, y) \in (0, \infty) \times (0, \infty)| \space x y < 1\}$$

I want to prove that M is an open subset of $\mathbb{R}^2$.

Now it may seem kind of obvious that this is indeed the case, but how can it be properly proven? In order to be an open subset, for any $(x, y)$, there needs to exist an open ball with center (x, y) that we might call $ B_\epsilon(x, y)$ that is still a subset of $M$. Now I thought that maybe we can choose $\epsilon < |xy - 1|$, but I don't know how to proceed from there.

Another solution that came to my mind is: if $U \subseteq X$ and $V \subseteq Y$, and both U and V are open, then the Cartesian product $U \times V$ is an open subset of $X \times Y$. If we choose a fixed $x \in (0, \infty)$, then y must satisfy $y < \frac{1}{x}$, and the set of choosable $y$ is therefore open for a fixed x, and vice versa for a fixed y. Would this be a valid solution? It doesn't seem so elegant, and I still prefer my first attempt above, so a continuation of my first try would be appreciated.

Thanks in advance.

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    $\begingroup$ Consider the function $f:(x,y)\mapsto xy$, it is continuous and $f^{-1}((-\infty,1))$ must therefore be open. $\endgroup$ – Gregory Grant Apr 29 '15 at 14:17
  • $\begingroup$ @GregoryGrant Of course, one must also establish the (easy) fact that $(0, \infty) \times (0, \infty)$ is open in $\Bbb R^2$. $\endgroup$ – Travis Willse Apr 29 '15 at 14:23
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The function $f$ defined by $f(x,y) = xy$. $f^{-1}(-\infty ,1)=M$. $f$ is continous therefore $M$ is open as the preimage of an open set.

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    $\begingroup$ Or even youmay take $f^{-1}(0,1)$ as $x>0$ and $y>0$. $\endgroup$ – Nizar Apr 29 '15 at 14:19
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The boundaries of $M$ are $x=0$, $y=0$ and $xy=1$. As $M$ doesn't contain any of these points, and there are no discrepancies in $M$'s interior, it is open.

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