Show that the set of convergent sequences is an infinite-dimensional subspace of the vector space of all sequences of real numbers.
Sorry, but I'm really lost on this one, any hints? Thanks.
Show that the set of convergent sequences is an infinite-dimensional subspace of the vector space of all sequences of real numbers.
To show it is a subspace you need to prove three things:
1. The zero vector of is in
2. is closed under addition
3. for any scalar in the underlying field of and that
In addition to these you will need to show that it is not of finite dimension.
RonL
RonL
proof.
Let S be the set of convergent sequences.
Now, the sequence {0} converges to 0, implies {0} is in S.
Assume the sequences of {x} and {y} be in S, with {x} converges to x and {y} converges to y.
{x} + {y} = {x+y} converges to x + y, so it is in S.
Let r be a scalar of real number, r{x} = {rx} converges to (r)x, so it is in S.
Thus S is a subspace.
Is this right?
Let be the sequence that is zero except for the th term which is . These are obviously convergent and so in
Suppose the subspace were finite dimensional of dimension .
Obviously , .. are linearly independent and therefore constitute a basis for the subspace. But cannot be written as a linear combination of them Contradicting our assumption that the subspace was of finite dimension.
RonL
It depends on what you know about finite dimensional vector spaces.
A space is of dimension then any set of linear independent vectors span the space and no smaller set will span the space. Because is not in either it is not a convergent sequence (but it is) or , which contradics our assumption.
RonL