Let $$V:=\bigcup_{n=1}^{\infty}\left\{(x_i)_{i=1}^{\infty}:x_i\in\mathbb{C},\space{}x_i=0\space{}\forall{}\space{}i>n\right\}.$$
I am trying to show that $V$ is infinite dimensional and that its Hamel basis is $(e_i)_{i=1}^{\infty}$.
I tried showing this by contradiction but didn't manage to:
I defined $V_n:=\{(x_1,...,x_n,0,...)|x_i\in\mathbb{C}\}$ which has a basis $E_n=\{e_1,...,e_n\}$ and its clear that $V_n\subset{}V_{n+1}$ hence $E_n$ spans all $V_i$ for all $i=1,...,n$.
But am not sure where to go with this.
Also a side question:
I understand that a Hamel basis can be infinite, but only considers finite sums. What is meant by simply a 'basis' in texts, does this then include infinite sums (is this referred to a schauder basis)?
any hints are appreciated.
Suppose $v_1, ... ,v_n \in V.$. For each $i$ with $1 \le i \le n$ let $i_*$ be such that $v_i(j)=0 \text { if } j \ge i_*$, Let $$M=\max\{i_*, 1 \le i \le n \}.$$ Define $v$ by $$v(M+1)=1, v(i)=0 \text { for } i \ne M+1.$$ Then $$v \notin\text {span}(v_1, ... ,v_n)$$ so no finite set can span $V.$ On the other hand, if $v \in V$ then there exists $K$ such that $v(i)=0 \text { for } i >K$ so $v \in \text {span}(e_1, ... ,e_K) \subseteq \text {span}(e_1,...).$ Since no non-trivial finite linear combination of $e_1,...$ can be 0, the set $\{e_1, ... \}$ is a Hamel basis of $V.$