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Which of the following subsets of $ \mathbb{R^2}$ are compact?

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

  2. $\displaystyle \{(x, y) : x^{\large\frac{2}{3}} + y^{\large\frac{2}{3}} = 1\}$

  3. $\displaystyle \{(x, y) : x^2 + y^2 < 1\}$

    I am stuck on this problem. Can anyone help me please?

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marked as duplicate by Dietrich Burde, Fabian, Aloizio Macedo, ᴡᴏʀᴅs, RecklessReckoner Jan 6 at 20:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What is the definition of compact you are using? Do you have any theorems about compact sets which you think may be able to help you? If you include your thoughts on the problem, it is easier for people to write a response which will help you understand the answer. – Michael Albanese Dec 27 '12 at 14:13
Can you show that $[0, 1] \subset \mathbb{R}$ is compact using this definition? – Michael Albanese Dec 27 '12 at 14:24
If you know that a set in $\mathbb{R}^k$ is compact $\iff$ it is closed and bounded, then you should add this to your question. We have been assuming you need to use the definition you give in your comments. We have no way of knowing what you know unless you expand on your question a bit, so we can HELP you. As Michael originally asked: what theorems about compact sets can you use? – amWhy Dec 27 '12 at 14:31
My thought process: "It is not a circle, but the equation has some superficial similarities. Circles are bounded. Maybe this is bounded." Is it? – Eric Stucky Dec 27 '12 at 14:37

For $1)$ use the fact that $|(x,y)| \ge \max \{|x|, |y| \}$.

For $2)$, recall that a continuous mapping of a compact set is compact. This set looks awfully similar to a very well-known compact set. Can you identify it, and can you think of a continuous surjection from that set onto yours?

For $3)$ recall that $(0,1) \subset \mathbb{R}$ is not closed because we may find a convergent sequence (say $\{\frac{1}{n}\}_{n=1}^{\infty}$) which converges but not to an element of $(0,1)$. See if this can be adapted.

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As mentioned in the comments, it helps us help you if you tell us what you've tried and what tools are at your disposal so we can best direct our help.

For example, do you know that in $\mathbb{R}^2$, that a set $S$ is compact if and only if it is closed and bounded? If so, use this characterization of compactness (in $\mathbb{R}^2$) to reason as follows (each time $S$ represents the set you gave):

(1) This is the graph of a hyperbola. $S$ is certainly unbounded so by what I cited above, $S$ cannot be compact.

(2) Here's a plot of $S$:

Mathematica graphics

$S$ here is closed and bounded (be sure you understand why), so $S$ is compact.

(3) This is the interior of the unit disk, but with its boundary $x^2+y^2=1$ excluded.

Mathematica graphics

$S$ here is bounded but not closed, so $S$ is not compact.

Hope that helps.

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