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Let $A=\left\{(x,y)\in\mathbb{R}^2 : x\neq 0, \ y=\frac{1}{x}\right\}$. Prove that $A$ is closed.

So far in the most difficult examples of this type I had to use a trick with inverse-image under a continuous function. Can I also use that in this case? How?

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up vote 4 down vote accepted

Yes you can, with $f(x,y)=xy$.

[Added] A little more detail: $f$ is continuous (which should be well-known) and $\{1\}$ is closed in $\mathbb{R}$, so you only need to show that $A = f^{-1}[\{1\}]$ to see $A$ is closed: if $(x,y) \in A$ then in particular $y = \frac{1}{x}$, so $xy = 1$ and so $(x,y)$ is in the inverse image, while if $(x,y)$ is in the inverse image, we know $f(x,y) = xy = 1$ which implies that $x$ (and $y$) are $\neq 0$, and so we can divide both sides by $x$ to see $y =\frac{1}{x}$, so $(x,y) \in A$.

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That seems brilliant. So there is absolutely no problem here? Thomas wrote about some nuance, that I don't understand and I'm wondering if there is no some catch here. – xan May 31 '12 at 20:42
@xan: You could write out the details to make sure that there is no problem. – Jonas Meyer May 31 '12 at 20:49
@JonasMeyer I see no details to write down :D – AD. May 31 '12 at 21:35
@AD.: Imagine if a student's complete turned in solution to this problem were "$f(x,y)=xy$" (taking into account the level). How much mercy would you give for the complete lack of explanation? :) – Jonas Meyer May 31 '12 at 21:46
@JonasMeyer Sure I guess you are right. I do however remember a student who solved something like "Find a solution to $dy/dx+y=0$" with $y(t)= e^t +t^5$, he got credits (no extra though) and a little talk... :) – AD. May 31 '12 at 22:22

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