# Real analysis, subsequences

1. Nov 9, 2008

### kbfrob

1. The problem statement, all variables and given/known data
Show that for any L$$\in$$[-1,1] there exists a subsequence of cos(n) such that that subsequence converges to L

3. The attempt at a solution
I have no idea.
I suppose that the ultimate goal would be to find a subsequence nk so that nk converges to x, where x = cos-1(L) + 2$$\pi$$k
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 9, 2008

### Staff: Mentor

Think about the graph of y = cos(x) and the parts of this graph that lie between the horizontal lines
$$y = L + \epsilon$$
and
$$y = L - \epsilon$$.
Some of the values of cos(n) will lie in this band.

3. Nov 9, 2008

### kbfrob

how can you guarantee that there is a value of n in that band? n is a natural number so i'm not sure how you can say that it is in there.

4. Nov 9, 2008

### Dick

The easiest way is to show that n mod 2*pi is dense in [0,2pi]. Then use that cos is continuous. To show the density use that pi is irrational. If r is irrational then n*r mod 1 is dense in [0,1]. I KNOW this is true. For some reason the proof doesn't stick in my head. I've looked it up more than once. Do I have to do it again, or can you figure it out? I know it involves the pigeonhole principle.