Let $\mathbf 0$ be the sequence of real numbers with all the components equal to $0$ and, for each $n \in \Bbb N$, let $δ_n$ be the sequence of real numbers whose n-th component equals $1$ and all other components are $0$.
Show that for any $r > 0$ and for any $n \in \Bbb N$, we have that $\frac r2\delta_n$ $\in N(0,r)$.
For this proof, I'm not sure how to begin. It seems like there's not much to prove because it is self explanatory.
The previous parts of this problem establishes that $0 \in l^2$ and that for any $a \in \Bbb R$, $a\delta_n \in l^2$.
So here $r>0$ then $a\delta_n \in l^2$ so let $a = \frac r2$ which is in the $N(0,r)$. I feel that this is just my reasoning and is not rigorous enough.
Can anyone help me formulate this?
Assuming $N(0,r)$ is the set $N(0, r) = \{ x = (x_n) \in \mathcal{l}^2 \; : \; \|x\| = \Big(\sum_{n=1}^\infty |x_n|^2\Big)^{1/2} < r\}$, i.e. the ball centered at $0$ with radius $r$, we need to show that
$\|\frac{r}{2} \delta_n\| < r$,
since we have already shown that $\frac{r}{2} \cdot \delta_n \in \mathcal{l}^2$ for all $n$. Since $\delta_n = (0, 0, \ldots, 1, 0, \ldots)$, it has norm $1$ because $0^2 + 0^2 + \cdots + 1^2 + 0^2 + \cdots = 1$. By the absolute homogeneity of the norm, we have $\|\frac{r}{2} \delta_n\| = \frac{r}{2} \cdot \|\delta_n\| = \frac{r}{2} < r$.
So $r/2 \cdot \delta_n$ belongs to $N(0,r)$ for all $n \in \mathbb{N}$.