consider a sequence in , say which converges to some , we can write
now, is this in ?
Consider the following
and is closed
I'm stuck on this problem:
Let be a normed vector space of complex sequences that converge to zero, with the sup-norm:
I need to show that Z is closed in X, but I'm getting a bit confused about sequences of sequences... here's what I got so far:
We will use the fact that Z is closed (in a metric space, as is a NVS) iff it contains all its limit points, i.e.
we want to show that whenever is a sequence in Z, which converges to some then we have that .
So, we take an , which converges to some . By definition, this means that
But how do I use this to show that ?
Possibly by contradiction - suppose is not in Z, then at least one of its members will not satisfy , say, this is the k-th member; so we set and show that this contradicts convergence? But I'm still getting very confused in passing from the infinity-norm to the Euclidean norm...
Is this right, where do I go from here, and is there any way to prove this without contradiction?
Thanks a lot for that!
One more question on the same problem:
I have to show that the element
is in the closure of Y+Z but not in Y+Z.
Now, I've written it as the limit of:
where the former is in Y (the last '1' occurs at the 2n-th place) and the latter in Z (the last nonzero entry is in the 2n-th place, again). So, I've shown (xj) is in the closure, as I've written it as the limit of a sequence which is in the sum, and closures are closed.
But how do I show it is *not* in the sum Y+Z?