I am working through the wiki proof that every finite dimensional subspace of a normed linear space is closed. The proof goes as follows:
Let $(V ,\| \cdot \|)$ be a normed linear space. Let $W$ be a finite dimensional subspace of $V$ with basis $\{e_1, \cdots , e_n\}.$ Assume to the contrary that $W$ is not closed. Then there exists a sequence $(w_n)$ in $W$ such that $w_n \to w$ but $w \in V \setminus W.$ Then, the set $\{ e_1,\cdots , e_n, w\}$ is linearly independent. Consider the subspace $W^* = \operatorname{span}\{e_1, e_2, \ldots, e_n, w\}.$
I am ok up to this point, but I think I am lost in the notation a bit. From here the proof reads:
Using the sequence $(w_n)$ from before, write: $$w_k=(w^{(1)}_k,w^{(2)}_k,…,w^{(n)}_k,0)∈W^*$$ and: $$w=(0,0,…,0,1)∈W^∗.$$ We necessarily have: $$w^{(j)}_k \to 0$$ for each $1\leq j \leq n$.
However, we would also have:
$0 \to 1.$
Clearly this is impossible, so we have derived a contradiction.
So $W$ is necessarily closed.
My questions are: What is $w^{(j)}_k$ and why does $w^{(j)}_k \to 0?$