I have the following problem:
Problem: Suppose that $\pi$ is a nondegenerate representation of $A$ on $\mathcal{H}$. Let $\{e_\lambda\}$ be an approximate identity for $A$. Prove that $\lim_\lambda\pi(a)h=h$ for all $h\in\mathcal{H}$.
Now, the author states that $\pi$ is a nondegenerate representation of $A$ on $\mathcal{H}$ if $$ [\pi(A)\mathcal{H}]:= \overline{\text{span}}\{\pi(a)h:a\in A\text{ and }h\in\mathcal{H}\}= \mathcal{H}. $$
Then in the proof of the problem the author proceeds to say the following: "Let $h\in\mathcal{H}$ and $\epsilon>0$ be given. By assumption there are vectors $h_1,\dots,h_n\in\mathcal{H}$ and elements $x_1,\dots,x_n\in A$ such that $\|h-\sum_{i_1}^n\pi(x_i)h_i\|<\epsilon$." I do not see why this is true. I mean, I understand that for $h\in\mathcal{H}$ there is a sequence in $\text{span}\{\pi(a)h:a\in A\text{ and }h\in\mathcal{H}\}$ such converges to $h$. However, I do not understand why the author is saying that such a sequence is of the form $\sum_{i_1}^n\pi(x_i)h_i$.
Once again, I have difficulty locating the issue, so let me try to be very detailed.
Let $h \in \mathcal{H} = \overline{\operatorname{span}}\{\pi(a)h: a \in A, h \in \mathcal{H}\}$ and $\epsilon > 0$. Hence, there is $y \in \operatorname{span}\{\pi(a)h: a \in A, h \in \mathcal{H}\}$ such that $\|h - y\| < \epsilon$.
Now, by definition of the linear span, $$y = \sum_{i=1}^n \lambda_i\pi(a_i)h_i= \sum_{i=1}^n \pi(\lambda_i a_i)h_i$$ for certain $a_i \in A, h_i \in \mathcal{H}$ and $\lambda_i \in \mathbb{C}$.
Putting $x_i := \lambda_i a_i\in A$ gives the desired form.