Let $(V,\|\cdot\|)$ be a (possibly infinite-dimensional) normed space and $V_1,V_2 \subseteq V$ be subspaces such that $V = V_1 \oplus V_2$. Let $B^i_{r_i} \subseteq V_i$ be open balls in $V_i$, wrt the induced norms $\|\cdot\|_i$, of radius $r_i > 0$. Define the sum of these two sets as $$ B_{r_1,r_2} := B^1_{r_1} + B^2_{r_2} := \{y_1 + y_2 \in V\ |\ y_i \in B^i_{r_i}, i = 1, 2\} $$ My question is whether this set is open or not.
My idea is: If we have $r := \inf \{\|x\|\ |\ x \in \overline{B^i_{r_i}} + \partial B^j_{r_j}\} > 0$, $i \neq j$, then $B_{r} \subseteq B_{r_1,r_2}$. But I don't know how to show exactly, or even if it helps or the set is open at the first place.
Suppose $V$ is Banach and $V_1, V_2$ are closed. Then the map $$V_1\times V_2\mapsto V, \quad (v_1,v_2)\mapsto v_1+v_2$$ is a continuous linear bijection of Banach spaces and so an open map. In particular the image of any open set is open.
If $V$ is not Banach it is not necessarily true, even if $V_1, V_2$ are closed:
Let $V=c_{00}(\Bbb N)$ the space of finitely supported sequences in $\Bbb C$, give it supremum norm. Let $V_1$ the space of sequences so that $2^nx_{2n}=x_{2n-1}$ and $V_2$ the space so that $2^n x_{2n}=-x_{2n-1}$.
Let $N>0$ be even, define: $$x_n=\begin{cases}\epsilon & n\text{ even}, n\leq N\\ 0 & \text{else}\end{cases}$$ Then $x= v_1 + v_2$ where $\|v_1\|\geq \epsilon 2^{N/2}$ while $\|x\|=\epsilon$. This prevents the image of the unit ball of $V_1\times V_2$ from being open in $V$.