# Thread: Convergent sequence implies subspace

1. ## Convergent sequence implies subspace

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.

2. Originally Posted by tttcomrader
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 $\bold{C}$ of convergent sequences is an infinite-dimensional subspace of the vector space $\bold{S}$ of all sequences of real numbers.

To show it is a subspace you need to prove three things:

1. The zero vector of $\bold{S}$ is in $\bold{C}$
2. $\bold{C}$ is closed under addition
3. for any scalar $a$ in the underlying field of $\bold{S}$ and $\bold{u} \in \bold{C}$ that $a \bold{u} \in \bold{C}$

In addition to these you will need to show that it is not of finite dimension.

RonL

RonL

3. 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?

4. Originally Posted by tttcomrader
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?
Yes, you stillneed to show that it is not finite dimensional

RonL

5. I don't quite understand how to show it. A convergent sequence can of course contain infinite coordinates as long as they all converge, right?

6. Originally Posted by tttcomrader
I don't quite understand how to show it. A convergent sequence can of course contain infinite coordinates as long as they all converge, right?
Let $e_i$ be the sequence that is zero except for the $i$th term which is $1$. These are obviously convergent and so in $C$

Suppose the subspace were finite dimensional of dimension $N$.

Obviously $e_1$, .. $e_N$ are linearly independent and therefore constitute a basis for the subspace. But $e_{N+1}$ cannot be written as a linear combination of them Contradicting our assumption that the subspace was of finite dimension.

RonL

7. I understand the proof until the $e_{N+1}$ part. For exactly which rule does it violated in order to draw a contradiction?

I'm sorry, I know I'm pretty weak in understanding this stuff here.

Thanks!

8. Originally Posted by tttcomrader
I understand the proof until the $e_{N+1}$ part. For exactly which rule does it violated in order to draw a contradiction?

I'm sorry, I know I'm pretty weak in understanding this stuff here.

Thanks!
It depends on what you know about finite dimensional vector spaces.

A space is of dimension $N$ then any set of $N$ linear independent vectors span the space and no smaller set will span the space. Because $e_{N+1}$ is not in $span(e_1, .., e_N)$ either it is not a convergent sequence (but it is) or $span(e_1, .., e_N) \ne C$, which contradics our assumption.

RonL