Advanced calculus

http://uploadpic.org/v.php?img=XpbEW3EXp
I think it should be part of real analysis , right?

2011-12-20 18:18:10 補充：
some mistake:
the first line of 1(b) :
we know 1/2 <= 2x , for all x in [n-1/n , 1/n] ( a part )

2011-12-20 18:18:31 補充：
we claim sup f([n-1/n,1]) = 2.Now, define S = {2x : x in [0,1) }
consider a real sequence w_n = 1 - 1/n , this sequence is in [0,1)
so , 2 - 2/n is in S , having limit 2 .
also , it is obvious that 2 is an upper bounded of S .
then , by supremum limit theorem , sup S = 2

2011-12-20 18:18:35 補充：
and so , sup f([n-1/n,1]) = sup ( S U { f(1) } ) = 2
this proves the claim.

2011-12-21 08:59:33 補充：
sorry again , the range of x should be [(n-1)/n , 1 ) in set S .
the real sequence should be w_k = 2[ 1 - 1/nk ]
It is easy to check as k = 1,2,3, .... , w_k is in S and have limit 2