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.

I have to check for compactness of given subets of $\mathbb{R}^2$.

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

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

$C =\{(x, y) :e^x = \cos y\}$

$D =\{(x, y) :\mid x\mid +\mid y \mid \leq 10^{100}\}$

The purpose of asking above question is not just to get answers. I need concepts to deal with these kind of problems. Let me explain where I face difficulties.

Take set $A$; Intuitively this is clear to me that $A$ is not a compact subset of $\mathbb{R}^2$ as it is closed but unbounded. My problem is I am having trouble with checking boundedness or unboundedness of given subsets. Here I know that set $A$ consists of points which lies on rectangular hyperbola. So I have no difficulty in judging that set $A$ is unbounded. But I am not sure about others. Since I am not able to figure out set them.

Edit: I don't want graphical approach to solve these problems. Because quite often i face problems where I find myself unable to visualize graph of given functions. I think there must be available some mathematical tool to deal with this.

I need help to understand this. I would be very much thankful to all of you.

share|improve this question
2  
Draw pictures! For example, in C) $x$ can be arbitrarily large negative. Just choose $y$ close to $\pi/2$, but a tiny bit below. –  André Nicolas Jun 6 '12 at 12:38
    
@AndréNicolas I am not good with graphs. That's why I am looking for alternate approach. –  srijan Jun 6 '12 at 12:41
    
As for boundedness, notice that in B and C, $x$ can be a very large negative number, while such a thing cannot happen in D, neither for $x$ nor for $y.$ –  Ehsan M. Kermani Jun 6 '12 at 12:44
add comment

2 Answers 2

up vote 1 down vote accepted

Draw pictures, or at least visualize. In set $B$, $x$ can be arbitrarily big. If you pick any $x\ne 0$, there is a $y$ such that $x^2y^2=1$. So there is no disk with centre the origin that contains all of $B$.

Also in set $C$, $x$ can be arbitrarily large negative. Just choose $y$ close to $\pi/2$, but a tiny bit below. Or else note that $y$ can be anything that makes $\cos y$ positive, and there are arbitrarily large $y$ with this property, such as $y=2n\pi$ where $n$ is any positive integer.

The set $D$ is clearly bounded, it is inside the disk with centre the origin and radius $10^{100}$. So you need to check whether or not $D$ is closed.

share|improve this answer
    
In A , also x can be arbitrarily large? Can we say that if any of variable can be made arbitrarily large implies that set is unbounded? Thanks for giving answer. That's too helpful to me. –  srijan Jun 6 '12 at 12:50
    
Certainly. I didn't do it because your post seemed to indicate you knew what was happening, you get a rectangular hyperbola. Yes, $x$ can be arbitrarily large positive or negative. And answer to your second question is yes. A set $S$ in the plane is bounded if there is a circle with centre the origin that contains $S$. –  André Nicolas Jun 6 '12 at 12:57
    
Set Set $D$ is closed as its complement must be open(union of two open sets). Am I right? Thanks again. –  srijan Jun 6 '12 at 13:04
1  
Yes, it is the union of two opens. But I think you have wrong picture. It says $|x|+|y|\le 1$, so $D$ is the stuff on or inside a certain square. The complement is the stuff outside the square, and it is open. I don't know how detailed you are supposed to be about showing that it is open. Take a point $(a,b)$ such that $|a|+|b|>10^{100}$, say equal to $10^{100}+t$ where $t$ is positive. Show there is a little circle about $(a,b)$ which is all outside $D$. –  André Nicolas Jun 6 '12 at 13:13
    
Nicolas I understand your point sir. Can we show that $D$ is closed beacuse it is inverse image of closed set $f^-1\{(10^{100})\}$. where f is continuos map sending $(x,y)\rightarrow \mid x\mid + \mid y \mid$? Sorry if I am being curiostic enough? –  srijan Jun 6 '12 at 13:21
show 2 more comments

For this you need the Heine-Borel Theorem which says that a subset of $\mathbb{R}^n$ is compact iff it is closed and bounded.

For $B$, note that the positive branch of $xy = 1$ is a subset of $B$. Since it is not bounded, $B$ fails to be compact.

For $D$ the constant $10^{100}$ is a red herring; the result is the same for any positive constant. Draw what $|x| + |y| \le 1$ and you will get the idea quickly.

For $C$, look at the graph of $x=\log(\cos(y))$

share|improve this answer
    
For C , can't we do without visualizing graph? That's what I want. Thanks for answering. –  srijan Jun 6 '12 at 12:53
add comment

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.