Check for compactness of subsets $A =\{(x, y) :xy = 1\}$ $B =\{(x, y) :x^2y^2 = 1\}$ of $\mathbb{R}^2$.

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I am working on following mathematical problem:

Check for compactness of following subets of $\mathbb{R}^2$.

$A =\{(x, y) :xy = 1\}$

$B =\{(x, y) :x^2y^2 = 1\}$

My attempt: I know that all I have to check whether subsets $A$ and $B$ are closed and bounded in $\mathbb{R}^2$ or not?

Checking for boundedness: Can we say that subsets A and B are not bounded in $\mathbb{R}^2$ since they contains points $(n, 1/n)$, where $x = n \in \mathbb{N}$ tends to infinity. I am not entirely confident about the correctness of my approach. Could you please review my approach and let me know if I am on the right track or not? Additionally, if you have any alternative approaches to suggest, I would greatly appreciate your insights.

Checking for closedness

I feel both sets are closed but how can I proved that?

PS: I have asked the same question before, but this time I have some additional doubts regarding my approach. Thank you very much

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Yes, both sets are closed and both sets are not bounded, hence they are not compact.

For the boundedness, your idea works. $(n,1/n)$ belongs to $A$ for all $n$, hence the norm of the elements of $A$ is not bounded. Similarly for $B$ with the same counterexample.

For closedness, just notice that $A=f^{-1}(\{1\})$ where $f:\mathbb R^2\to \mathbb R$ is the map $f(x,y):=xy$, which is continuous. Again, similarly for $B$ with the map $g(x,y):=x^2y^2$.