# Homework Help: Is it compact?

1. Jan 16, 2006

### twoflower

Hi,

I want to find global extremes of function

$$f(x,y,z) = xy^{2}y^3$$

on the set

$$M = \left\{[x,y,z] \in \mathbb{R}^{3}, x+2y+4z = a, x,y,z > 0\right\}$$

I need to show that this is compact. Because I'm in $\mathbb{R}^{n}$ it is sufficient to show it is closed and bounded. Closeness (is this the right noun?) is obvious. How to show it is bounded as well? Is it trivial, ie. may I just say that it is some bounded plane or how?

Thank you.

Last edited: Jan 16, 2006
2. Jan 16, 2006

### Muzza

If (x, y, z) is in M, then x, y, z > 0, so we must have that 0 < x < a (and similarly for y and z). This implies that M is bounded.

(And if you have to ask, it's clearly not trivial. ;))

Last edited: Jan 16, 2006
3. Jan 16, 2006

### matt grime

I'm not sure that closedness is obvious since you have a strictly greater then zero condition, and strictly greater than something tends to make things open if you aren't careful. Since boundedness is genuinely trivial, and you didn't spot it, I think you shuold check closure.

Consider the easier to typeset example of x+y=1, x,y>0

the points (0,1) (and (1,0)) satisfy the condition, and are not in the set, yet the sequence (1/n,1-1/n) is in the set and tends to (0,1)....

4. Jan 16, 2006

### twoflower

Well, I thought that closedness is obvious because of this:

$M$ is continuous image of closed set $\left\{0\right\}$ and thus is also closed.

5. Jan 16, 2006

### twoflower

Thank you Muzza, that's...very nice :)

6. Jan 16, 2006

### twoflower

And one more question to this matter: when I'm given such $M$ which is not bounded, how to do that? I recall unclearly our professor telling us something about this case...could it be something like "Close the set in something larger (eg. some cube), find extremes on it and then...?" Hm, I remember less than I thought :(

7. Jan 16, 2006

### matt grime

It isn't obvious, and what you've written is indeed false. Nothing requires the image of a closed set to be closed. The inverse image of a closed set is closed, under a continuous function, not that that matters, since I cannot see anything that is the image or inverse image of a closed set. Did you not understand the counter example which showed your 'obvious' reasoning to be false?

8. Jan 16, 2006

### twoflower

More precisely, I'm using this:

$$f: P \rightarrow Q\mbox{ .Then these statements are equivalent:}$$

$$\mbox{(i) f is continuous on P}$$

$$\mbox{(ii) }f^{-1}(F)\mbox{ is closed for each closed F \subset Q}$$

In my case, let

$$G := x + 2y + 4z - a$$

Then

$$\left\{0\right\}\mbox{ is closed and }G = f^{-1}\left\{0\right\}\mbox{ and thus G is also closed}$$

Last edited: Jan 16, 2006
9. Jan 16, 2006

### matt grime

But your set is not the image of a closed set under the inverse of a continuous function. It is most definitely not the inverse image of the zero set under that function, eg the element (a,-2,1) is in the inverse image set but not in the set you want since last time I checked -2 is a negative number. So why have you for the second time ignored the counter example I gave you? And just reiterated the same incorrect argument?

Last edited: Jan 16, 2006
10. Jan 16, 2006

### twoflower

Sorry matt. Of course I wrote it wrongly...Let me try it again, then your because your counterexample referred to my misinterpreting of the theorem.

$$g(x,y,z) := x + 2y + 4z - a$$

Then

$$M = \left\{[x,y,z] \in \mathbb{R}^3, g(x,y,z) = 0\right\}$$

And because $g$ is continuous, $\left\{0\right\}$ is closed and

$$M = g^{-1}\left\{0\right\}$$

Then $M$ is also closed.

Hope this is finally correct, I also found it in the examples our professor did for us.

Last edited: Jan 16, 2006
11. Jan 16, 2006

### matt grime

So? You are restricting to the subset of the inverse image such that all entries are strictly positive. This is not a closed subset as I've said, ooh, what 4 or 5 times now, the theorem is completely irrelevant. The set you said was obviously closed but not obviously bounded is in fact obviously bounded but not actually closed.

12. Jan 16, 2006

### twoflower

Please, can you tell me, in which sense I don't meet the requirements of that theorem?

$\left\{0\right\}$ is always closed and I can't see any other troubles using that theorem.

13. Jan 17, 2006

### shmoe

This set M' (note I've renamed it) is closed. However this is not the same as the M from your first post. M = M' intersect {(x,y,z)|x,y,z>0}

Do you have a picture in your head of what this looks like? It's not complicated. What if you change the >0 condition to >=0?

14. Jan 17, 2006

### twoflower

I suspect it's because of the strict inequality and thus it is not closed because I can always make a ball around some point around zero?

15. Jan 17, 2006

### shmoe

What does "a ball around some point around zero" mean?

Do you understand now why your theorem doesn't apply?

Do you know what this set M looks like?

16. Jan 17, 2006

### matt grime

It's not closed because it is not closed, and as demonstration for the example x+y=1, x,y>0 I showed a sequence in the set that converged to something not in the set. And that was in post 3.

17. Jan 18, 2006

### twoflower

I don't know whether I'm expressing it correctly in English, but we had defined open set as the set in which around every point there exists ball (or sphere is the right word?) which whole belongs to the set.

And because M satisfies it, it won't be closed (I know OPEN and NOT CLOSED is not equivalent, but I think here it applies).

18. Jan 18, 2006

### twoflower

I'm stupid. Of course I've read your counterexample, but I didn't consider it to be the PROOF of the set not being closed, but rather some hint to me to figure it out. Anyway, now I found the theorem about characterization of closed sets which poses right this condition. Sorry for bothering you, you helped me much.