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For each of following sets decide whether it is open, closed, compact or bounded:

(a) $\left\{ (x,y)\in\mathbb{R}^2 : \sin\left( \sin\left( \cos(xy)\right)\right)=\sin\left(\sin(x+y) \right)\cdot y , \ x\ge -1 , \ y\le x \right\}$

(b) $\left\{ (x,y)\in\mathbb{R}^2 : e^{x+y^2}=\ln \frac{1}{1+x^2+y^2} \right\}$

The above formulas are so complicated that I guess there is some smart way to test everything without solving equations. Can anybody help?

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2 Answers 2

up vote 3 down vote accepted

To continue answering the other questions, it's clear that the set in (b) is empty since $e^{x+y^2}>0$ but $\ln(\frac{1}{1+x^2+y^2})\leq 0$. That would mean it is open, closed, compact and bounded.

The only tricks I saw to (a) so far are: Since $\mathbb{R}^2$ only has the two trivial clopen sets, it would not be open.

(Withdrawing second observation, I don't think it's correct.) If you can show there is an upper bound to the $x$'s, and bounds on the $y$'s, then you will have shown it is bounded, and with closedness, this implies compactness. If it is unbounded it is not compact.

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a) closed

b) closed

In both cases you have continuous functions $f(x,y),g(x,y)$ on both sides of the "=", and so $f(x,y)-g(x,y)$ is also continuous. You can rewrite your sets as $\{(x,y)\mid f(x,y)-g(x,y)=0\}$. Since the inverse image of $0$ under a continuous map is closed, you have that your sets are closed.

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thank you rschwieb :) –  El Angel Exterminador May 31 '12 at 19:15
1  
In (a) you have $x\ge -1$ and $y\le x$, instead of $\text{something}=0$. Still, these are closed sets, and you're looking for their inverse-images under a continuous function, so the conclusion is still that you're getting closed sets. And if they're also bounded, then they're compact. –  Michael Hardy May 31 '12 at 19:32

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