1. ## Finite Dimensional

I am having a fair amount of trouble with what would appear to be a very straight forward question. The question states:

Show that the set consists of all real sequences converging to zero is a vector space and show that it is not finite dimensional.

If anyone could give me a rough idea how to get started on this question or how to approach it, I would be very appreciative. Thanks.

Edit: I should also point out that the english is very bad lol, it should probably read something like: Show that the set "which" consists ....

2. pointers on how I tried

Part 1: If X,Y are real sequences converging to 0, it is easy to show aX + bY, where a,b are real will also converge to 0. So set of S of all such seqences is a vector space.

Part 2: Attempt by contradiction. Assume the space is finite dimensional, get a basis and then construct a sequence which though converges to 0, cannot be a expressed as a linear combination of the basis we selected earlier. Hence proving the space is infinite dimensional.

As a further hint any n-dimensions vector space is isomorphic to $\displaystyle \Re^n$

Hope it helps

3. I am trying to figure it out and I understand how it is not finite dimensional but how can i prove it is a vector space when one of the properties of a vector space is that it must contain the zero vector. However, the question is stating that it is converging to zero, so wouldn't that mean that there couldn't be a zero vector since it is only converging?

4. No - {0,0,0,0,0,0,0,0,0,0,0,0.......................} by all means converges to 0. It perfectly satisfies the definition of a seq converging to 0

5. The sequence $\displaystyle (0,0,0,...)$ converges to 0 trivially.

To solve the other part of the problem, note that for any $\displaystyle n$, any sequence of the form $\displaystyle (x_1,...,x_n,0,0,0,..)$ is admissible, where the $\displaystyle x_i$ are any numbers you wish. In other words, for any $\displaystyle n$ your vector space contains a subspace isomorphic to $\displaystyle \mathbb{R}^n$, so it can't be finite-dimensional.

6. Thank you both. I greatly appreciate the help.