closure of a subspace of l-infinity (space of bounded sequences)

[zill...@googlemail.com]

Hello,

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...

[The World Wide Wade]

In article
<4381c95b-5a17-4910...@8g2000hse.googlegroups.com>,
"zill...@googlemail.com" <zill...@googlemail.com> wrote:

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)