Measuring Functions in L2: Closed, Bounded, and Not Compact?

[Mystic998]

Forgive the lack of TeX. I'm too lazy to type it out right now. For reference, these are problems 13 and 14 in chapter 11 of Rudin's Principles of Mathematical Analysis, slightly reworded for lack of pretty mathematical symbols.

Homework Statement

Problem one: I need to show that the set of functions sin(nx), where n is a natural number (excluding 0, just to be clear) and x is in the closed interval -pi to pi, is closed and bounded in L2, but not compact.

Problem two: A complex function f is measurable iff for every open set V in the plane, the pre-image of V under f is measurable.

Homework Equations

Problem one: Probably too many to list.

Problem two: f is measurable iff Re(f) is measurable and Im(f) is measurable. Also, we can assume that all open sets are equivalent to a countable union of open rectangles.

The Attempt at a Solution

Problem one: Boundedness is relatively easy. As for closed, my only idea is to prove that any sequence (apart from sequences that are constant for infinitely many of the terms obviously) of these functions fails to converge, and therefore, the set has no limit points and is closed. But I'm not sure that's even right.

Compactness: I was thinking maybe take the open cover as the set of all functions in L2 representable by a Fourier sine series. Then each function would be its own series, and you couldn't have a finite subcover. But I'm not sure that the given cover is open, nor would I know how to prove it.

Problem two: I'm basically completely lost on this one. I think it would follow relatively easily if I knew that the Cartesian product of two open sets is open, but unfortunately that remains unproved in this particular class, so I can't really use it except as a last resort.

 

[Dick]

For problem one the functions are orthogonal and have constant norm. That should really be enough to make the rest of the problem easy.

 

[Mystic998]

It does feel like this should be simple, but I'm really not sure why the set's being closed follows from that.

Am I at least on the right track with what I started with and just need to use the mentioned properties to show what I said is true, or am I completely off base?

 

[Dick]

Call f_n=sin(nx). L2 is a metric space. What's the distance from f_n to f_m? That's really a big hint.

 

[Mystic998]

Thanks. I finally got it after I thought about it a little longer. Any of those functions where n and m differ have to have distance sqrt(2pi) in L2. So nothing other than a constant sequence can be Cauchy. Hence, no convergence.

Of course, I'm sure the rest of the answers I finally came up with are imperfect, but it'll do for a 3 problem homework assignment that has little effect on the class grade. Anyway, thanks again.