Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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 $(x,y) \in A$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.