The worst part is that I only stepped in for goalie because it was a small-sided game and no one else wanted to play. Anyway...

I was talking about the complement of the closed subset as a subset of the compact set. Whenever we talk about open covers of a compact set, we really mean a collection of sets that are "open in the compact set." That's a basic topological notion you may not have encountered yet, but the point is that compactness is a property that is independent of where the set lies, whether it can be embedded in a metric space M or not. Of course, if you are given a compact set without a metric space in which it's embedded, someone also has to tell you what the open sets of that set are; that's called specifying the topology on the set. In this case, the open sets of your compact set (i.e. its topology) are induced because it is a subset of M. By definition (this is called the subspace topology), U is open in K (compact) if and only if U can be written as the intersection of V with K for some open set V of M. You will find this confusing at first because U may not look like an open set of M. For example, if M = R and K = [0,1] (closed and bounded subset of R, so compact), the open sets of K are all open intervals inside (0,1), and also all intervals (a,1] and [0,b) for a,b in [0,1]. That's right, [0,1] = K is also an open set of K. Of course, you can also take unions and intersections of these to get even more open sets.

The point is that what exactly is meant by an open cover of a compact subset is often glossed over in the initial stages. You should convince yourself that it's not even possible to find an open cover of [0,1] if you are only allowed to use sets that are open in R (how will you get the endpoints?). In practice, you will be fine if you allow the sets in your open covers to extend beyond the set K you are trying to cover, since in the subspace topology you will intersect each of them with K anyway.

Ah, that makes much more sense now. Although Im still slightly confused it makes 10x more sense than before. Thank you.

The next two I was having problems with were:

Let R+ denote the nonnegative real numbers and let f : (R+ )^3 → R+ be given by
f (x, y, z) =
xyz/x^p +y^p +z^p , (x, y, z) = (0, 0, 0)
0, (x, y, z) = (0, 0, 0)

where p ≥ 0. For what values of p is f continuous on all of (R+ )^3 ? Prove your claim.

Now, my thinking on this was that I need to find what values of p give us that:

Lim x,y,z -> 0 f(x,y,z) = 0

Here is where I get stuck. How do I go about finding such values of p? Taking the natural log wouldnt help either. So, Im stuck. Is this even a correct starting point?

Also, Im supposed to find all discontinuities of sin(x) · χQ (x), where Chi_Q is defined as:

1, x in Q
0, x in R\Q

Wouldnt it be discontinuous at every rational? The only point where sin(x) would be defined as 0 is an irrational so it is continuous at that point, but at all the rationals the limits dont match up.

~BD

For the first question, I'm thinking you definitely can't have p > 3 because then the function blows up to infinity as you approach the origin. If p = 3 it's also not continuous because then approaching along the line (z,z,z) gives a limit of z^3 / 3z^3 = 1/3 which is not zero. On the other hand, the function should be continuous for all 0 < p < 3 (and obviously p = 0 too). I believe this because the numerator is a cubic and therefore approaches zero a lot faster than any sub-cubic (try graphing y = x^3 vs. y = x^2 around x=0 and you will see this). Heuristically speaking, this means that the function should get crunched down to zero near the origin as long as the denominator is sub-cubic.

For the second question, you are correct in saying that the function is definitely not continuous at any rational point, since sin(x) would be nonzero there and the rational is surrounded by irrationals. On the other hand, the function is not continuous at every irrational. Let x = pi/2 so sin(x) = 1. While f(x) = 0, x is surrounded by rationals at which f is not only nonzero, but approaching 1. Therefore, f is not continuous at x. In fact, f is only continuous at y such that sin(y) = 0, which is a subset of the irrationals.

For part 1, as I worked on it I got that clearly p>3 doesnt work, as you said, but how would I go about proving that it works for 0<p<3?

And for the second part: That makes more sense. So for any y = n*pi for n in Z would yield sin(y) -> 0 from both the left and the right. Gotcha.

Thanks tons. I should pay you for the help you're giving me.

You open to a few more questions? I think I have like 2-3 more if you're up for it

~BD

Let me see if I can put you on the right track. For 0 ≤ p < 3, we want to show that f approaches zero from all directions as we approach the origin. This is always more difficult than producing a counterexample to show that it is not continuous, and there is no universal method. Here's my idea. Fix epsilon > 0 and let x^2 + y^2 + z^2 = epsilon^2. Then xyz ≤ epsilon^3 / 3^(3/2) (the max of xyz occurs when x = y = z so then 3x^2 = epsilon^2 => x = epsilon / 3^(1/2) ). The minimum of the denominator depends on whether p<2, p=2, or p>2. If p<2 then the minimum is achieved at y = 0 = z (really any two out of x,y,z), giving a minimum value of epsilon^p; if p=2 the denominator is constant at epsilon^2; if p>2 the minimum is achieved at x=y=z, with minimum value equal to 3 epsilon^p. You can prove these claims using calculus. I've basically chosen the max value for numerator and the min value for denominator, subject to the constraint x^2 + y^2 + z^2 = epsilon^2, in order to get a maximum value of f subject to that constraint.

Anyway, in each of these cases you have f ~ epsilon^(3-p) which goes to zero precisely when 0 ≤ p < 3.

Ask away. I don't have any problem sets out at the moment and I can't really play SS with one hand.

That's an interesting way to go about that.

Well, I have like 5 problems left that Im struggling with (granted, our professor told us that its not necessary to do every problem before the exam but I figure I need all the prep I can get haha) so, here they are:

1.) Let f : K ⊂ Rn → R be continuous on a compact set K, and let M = {x ∈ K|f (x) is the maximum of f on K}.
Prove that M is a compact set.

2.) Let A and B be subsets of R. Suppose that the Cartesian product A × B ⊂ R^22 is connected.

(a) Prove that A must be connected.

(b) Generalize this result to the case of A, B ⊂ M , a general metric space with metric d.

3.) Suppose we have two sequences ak → a ∈ R and bk → b ∈ R. Let ck = ak * bk be the sequence of products.

Give a proof using continuity of the function f (x, y) = xy.

Im really lost on this one. Not quite sure how continuity comes into play. I understand quite well that I can prove it using the fact that epsilon > 0 there is some N etc etc but using continuity confuses me.

4.) Let f : R2 → R be given by f (x, y) = x.

(a) Prove that f is uniformly continuous on R.

(b) Show that there exists a closed set A ⊂ R2 such that f (A) is not closed in R.

This one just confused me since it was in R^2. Its clear for a Im supposed to use an epsilon-delta proof but my book doesnt really go into detail for any spaces beyond R.

5.) A set A ⊂ Rn is called relatively compact when cl(A) is compact. Prove that A is relatively compact
if and only if every sequence in A has a subsequence that converges to a point in Rn .

Thats all of em. Granted, on most a small hint will get me going in the right direction and again, thanks for all the help. It will definitely show on my test.

~BD

1. Prove that K \ M is open, so M is closed. Every closed subset of a compact set is compact. You can do this by showing there's an open ball around every point in K \ M.

2.
a) The proof is easy by contrapositive. If A is disconnected, then there are nonempty, disjoint open subsets A1, A2 covering A. It should be easy to show that (A1 x B) and (A2 x B) are nonempty, disjoint open subsets covering A x B.
b) The proof doesn't change here.

3. Since f is continuous, the image of a convergent sequence under f is a convergent sequence. Therefore, f(ak, bk) converges to f(a,b) = ab.

4. Look back at one of the older problems we discussed where we needed a notion of continuity for a function on an arbitrary metric space M. I made a bit of a deal out of how the elements here are perhaps vectors, perhaps something more exotic, and therefore we usually use || - || to denote the norm instead of absolute value signs.
a) For all epsilon > 0, we can take delta = epsilon and have || (x0, y0) - (x1, y1) || < delta implies | x_0 - x_1 | < epsilon. In particular, there exists delta > 0 for which this is true.
b) I can't think of any way to motivate this without actually giving you an example. You will have to think critically about the following in order to get anything out of this. The example is the graph of y = 1/x, whose image under f is the real line minus the origin (not closed). You can convince yourself the graph is closed in R^2 a number of ways, including that it contains all its limit points, or that its complement is open.

5. If A is relatively compact, then A has bounded closure. Having bounded closure implies that we can enclose A in a large closed ball (which is bounded hence compact), and so any sequence in A has a subsequence that converges in the ball (property of compactness). On the other hand, if every sequence in A has a subsequence converging to a point in R^n (in particular, to a point in cl(A) by definition), you should be able to prove that cl(A) is compact. I haven't worked it out in full detail, but my idea for this is the following. Take any sequence a_k in cl(A). We can exclude sequence with an infinite number of terms in A since these have convergent subsequences in cl(A) by hypothesis. Then throw away the remaining finite number of terms in A. Now define a collection of sequences {a_k_j} in A converging to each of the remaining elements of our sequence a_k. We now have a new sequence in A {a_j_j} which has a convergent subsequence {a_j'_j'} in cl(A). Prove that {a_j'} converges to the same thing as {a_j'_j'}. (I'm pretty proud of this little proof!)

Feel free to ask more questions if you need elaboration on any of these. And... while I had never considered this myself, I guess if you wanted to pay me I'd accept ingame creds, however much you deem appropriate. I'd say this service is at least as valuable as the avatar-creation that Copernicus was charging for. But it's really up to you, and I'll honestly continue to answer your questions either way.

Heh, I feel as though I should do something to compensate you for all your work. Im sure answering all my questions isnt "enjoyable" for you haha.

I havent had a chance just yet to look over your elaborations due to other aggravating homework (damn quantum and thermodynamics..) but Im sure they're as helpful as always.

Anywho, I havent been subbed since the semester started, I find that SS and school dont mix well, one usually gets forgotten about and I'd rather not risk that being school. I cant afford that, both metaphorically and physically.

But maybe over Christmas break I'll pop and and see if I can repay you in some form or other.

You cant imagine how much you've helped, truly.

~BD

Technically I should be able to answer your quantum physics questions as well, but... oh god I don't want to. It's been too many years.

Good luck on that exam! You know where to find me if you have more questions later. And yes, I'm well acquainted with the SS-school dilemma. I fondly recall skipping lecture to get on a weapons 20 run several years ago... it was totally worth it but also totally unacceptable.

Yeah, quantum just doenst make any physical sense, in my opinion. It seems like its all made up.. Oh well. My roommate and I can usually figure it out.

Anyway, I was still a bit confused about:

1.) Let f : K ⊂ Rn → R be continuous on a compact set K, and let M = {x ∈ K|f (x) is the maximum of f on K}.
Prove that M is a compact set.

I dont really even understand the set (I feel like this happens a lot haha). Is M just the single maximum point of f(x)?

~BD

Also, I've been thinking about your example of f(x) = 1/x and it still confuses me. I've been attempting to work it out but I find myself confused everytime haha.

So, we are looking for a closed subset A in R^2, this is defined as y = 1/x from say x = [-1, 1] and this is closed? Whereas the function of this .. well, theres where I get confused. Isnt f(A) in R^2 and A in R?

~BD

There's nothing keeping f from achieving its maximum value more than once. Consider f(x) = cos x, which achieves its maximum for x = 2πk for all integers k. Note that the set of integers is indeed closed in R.

More generally, you could have a whole hyperplane on which f is maximal (a hyperplane of R^n is a (n-1)-dimensional plane), or any lower dimensional space as well. For example, the function f(x,y) = 1 - x^2 has maximal set {(x,y) | x = 0}, which is a line in R^2. You'll have to think harder if you want to visualize the phenomenon in any higher dimension than that, although the next step up should be doable at least.

In your proof, you might want to remember that if f is constant then M = R^n, the whole space. So if you talk about the complement of M, don't forget that it might be empty.

As to your next question, I'm talking about the set {(x,y) | y = 1/x} in R^2. I originally meant the entire graph, but you can truncate it for x in [-1, 1] if you want. The image of this subset of R^2 under f is then [-1, 0) union (0, 1], which is not closed. You are getting confused because you're assuming I meant f(x) = 1/x when I wrote y = 1/x. Really I was just using shorthand to denote the graph of this relation in the plane. I expect you will now see that the image of the graph under f is not closed. That just leaves the proof that graph itself is closed. Let me know if you have trouble with that.

Okay, as you suggested, Im having a little trouble seeing how the graph is closed with its asymptotes floating around here..

This is in reference to y = 1/x.

~BD

Here are two ways to prove this. Let C denote the set satisfying y = 1/x.

1. Show that its complement is open in R^2. Any (x,y) not satisfying y = 1/x lies some positive minimum distance d from C. Therefore, there's an open ball of radius d/2 around that point that lies entirely in the complement of C. Therefore, C is open.

2. (trickier but uses problem number 1) Define a new function g: R^2 -> R by g(x,y) = - (y - 1/x)^2. This is obviously maximized when the quantity inside the square is zero, which happens precisely on C. Therefore, C is the maximal set of g so it must be closed.

I know this is much easier than what you guys are doing, but i just cant figure it out.

A solid cylinder with circular ends of radius 30mm has a total surface area of 52,800mm^2, to the nearest 100mm^2. Find the lenght of the cylinder.

The method i was trying to use was something like this.

Area of the 2 circles = 2 * pi * 30(r)^2 = 5654mm^2
Then the curved surface(flatterned out rectanlge) = Lenght * Height

Height = Perimeter of circle ? which i assumed was 2 * pi * r
= (2 * pi * 30)
= 188mm^2

2 x pi x r^2 + 2 x pi x r x (x)

2 x pi x 30^2 + 2 x pi x 30 x (x)

5654 + 188 * x = 52 800 (Surface Area)

Now i am stumped what to do from here.

The method used in the book is

Let the lenght of the cylinder be x mm
Area of the two circles : 2*pi*30^2
Area of curved surface : 2*pi*30*x mm^2
Hence 2 * pi * 30^2 + 2*pi*30* x = 52800
Take 1800*pi from each side : 60*pi*x = 52800 - 1800*pi

52800 - 1800*pi/60*pi

=250mm

I understand it up to the part where they subtract 1800*pi from SA, which is the bit that makes this confusing.

Hopefully this is understandable and im not talking jibberish.

How to solve this, or some info on where the 60*pi x = 52800 - 1800*pi comes from or an easier method would be very helpfull.

Thanks

Z