Could someone verify that for the following statement, the proof given below it is correct:
Statement: The subspace of l-infinity (space of bounded sequences with the sup norm) consisting of all scalar sequences converging to zero [let's call this subspace c0] is closed.
Proof:
For any point x in cl(c0) (the closure of c0) there exists a sequence (x_n) such that x_n -> x, (x_n's in c0) Let x be represented as (d_1, d_2, ... ) and x_n be represented as (d(n)_1, d(n)_2, ... ) Then x_n -> x implies (taking into account the defn of the sup metric in l-infty) for all e > 0, there exists an N such |d_j - d(n)_j| < e for all j and all n > N
Also, since every convergent sequence is Cauchy for all e > 0, there exists an N such that |d(n)_j - d(m)_j| < e for all j and all m, n > N Now, |d_j| <= |d_j - d(n)_j| + |d(n)_j - d(m)_j| +|d(m)_j| In RHS of the above inequality, 1st term can be made < e/3 for all j and all n > some N1 2nd term can be made < e/3 for all j and all m, n > some N2 3rd term < e/3 for all j > some Nj, since x_d -> 0
Hence |d_j| < e for all n > max(N1, N2) and all j > Nj implying that x -> 0, hence x in c0 Since x was chosen arbitrarily, every x in cl(c0) is in c0 => c0 is closed
> Could someone verify that for the following statement, the proof given > below it is correct:
> Statement: The subspace of l-infinity (space of bounded sequences with > the sup norm) consisting of all scalar sequences converging to zero > [let's call this subspace c0] is closed.
> Proof:
> For any point x in cl(c0) (the closure of c0) there exists a sequence > (x_n) such that x_n -> x, (x_n's in c0) > Let x be represented as (d_1, d_2, ... ) and x_n be represented as > (d(n)_1, d(n)_2, ... ) > Then x_n -> x implies (taking into account the defn of the sup metric > in l-infty) > for all e > 0, there exists an N such |d_j - d(n)_j| < e for all j and > all n > N
> Also, since every convergent sequence is Cauchy > for all e > 0, there exists an N such that |d(n)_j - d(m)_j| < e for > all j and all m, n > N > Now, > |d_j| <= |d_j - d(n)_j| + |d(n)_j - d(m)_j| +|d(m)_j| > In RHS of the above inequality, > 1st term can be made < e/3 for all j and all n > some N1 > 2nd term can be made < e/3 for all j and all m, n > some N2 > 3rd term < e/3 for all j > some Nj, since x_d -> 0
> Hence |d_j| < e for all n > max(N1, N2) and all j > Nj > implying that x -> 0, hence x in c0 > Since x was chosen arbitrarily, every x in cl(c0) is in c0 > => c0 is closed
> End of proof
> Thanks...
I haven't read your proof carefully, but one thing to notice is you are reproving a result you've probably seen before. Leaving out details, the result is: If f_n -> f uniformly, then lim_x->a lim_n->oo f_n(x) = lim_n->oo lim_x->a f_n(x)
