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 $\displaystyle \bold{C}$ of convergent sequences is an infinite-dimensional subspace of the vector space $\displaystyle \bold{S}$ of all sequences of real numbers.
To show it is a subspace you need to prove three things:
1. The zero vector of $\displaystyle \bold{S}$ is in $\displaystyle \bold{C}$
2. $\displaystyle \bold{C}$ is closed under addition
3. for any scalar $\displaystyle a$ in the underlying field of $\displaystyle \bold{S}$ and $\displaystyle \bold{u} \in \bold{C}$ that $\displaystyle a \bold{u} \in \bold{C}$
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 $\displaystyle e_i$ be the sequence that is zero except for the $\displaystyle i $th term which is $\displaystyle 1$. These are obviously convergent and so in $\displaystyle C$
Suppose the subspace were finite dimensional of dimension $\displaystyle N$.
Obviously $\displaystyle e_1$, .. $\displaystyle e_N$ are linearly independent and therefore constitute a basis for the subspace. But $\displaystyle e_{N+1}$ 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 $\displaystyle N$ then any set of $\displaystyle N$ linear independent vectors span the space and no smaller set will span the space. Because $\displaystyle e_{N+1}$ is not in $\displaystyle span(e_1, .., e_N)$ either it is not a convergent sequence (but it is) or $\displaystyle span(e_1, .., e_N) \ne C$, which contradics our assumption.
RonL