Vector space of infinite sequences

Homework: Define $V=\{(a_1,\ldots,a_n,\ldots): a_i \in {\bf R}, a_n$ converges$\}$.

1. Is this a vector space in some sense?

Define

• 1. $(a_1,\ldots,a_n,\ldots)+(b_1,\ldots,b_n,\ldots) = (a_1+b_1,\ldots,a_n+b_n,\ldots)$

Since $a_i$ converges, $b_i$ converges, hence $a_i+b_i$ converges. (That comes from calc 2.)

• 2. $r(a_1,\ldots,a_n,\ldots) = (ra_1,\ldots,ra_n,\ldots$)

Since $a_i$ converges, then $ra_i$ converges.

So the operations are well defined.

Observe: $$V \subset U = \{(a_1,\ldots,a_n) \colon a_i \in {\bf R} \} = {\bf R}^{\omega}.$$ So we can borrow the operations from $U$. Still have to show $V$ is closed under them.

2. Define $$S=\{(1,0,\ldots), (0,1,0,\ldots), \ldots \}$$ Is this a basis of $V$?

We can think of $S$ as $S = \{e_1,e_2,\ldots \}$. It's infinite!

First, $S$ is linearly independent (by definition, every finite subset $F$ of $S$ is linearly independent).

Suppose $S$ is linearly dependent, then there is a finite subset $F \subset S$ which is linearly dependent; then $F \subset \{e_1,\ldots,e_n\}$ for some $n$ where $n = {\rm max}\{i \colon e_i \in F\}$.

Then $\{e_1,\ldots,e_n\}$ is linearly dependent, so there is $a_1,\ldots,a_n$ with $a_1e_1+\ldots+a_ne_n=0$. Hence $a_1=a_2+\ldots=a_n=0$. Contradiction.

Note: There are no infinite sums in linear algebra (that would need calculus).

But, $S$ does not span $V$!

Test $e=(1,1,\ldots,1,\ldots) \in {\rm span \hspace{3pt}} S$.

$=$ the set of all linear combinations of finite subsets of $S$.

Suppose there is a finite set $F$ of $S$ such that $e=$ linear combinations of $F$.

Let $n={\rm max}\{i \colon e_i \in F\}$. Then any linear combination of $F$ will have $(n+1)^{\rm th}$ entry $=0$, $(n+2)\ldots$ So it can't be $e=(1,\ldots,1,\ldots)$.