> 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...