I have a question about the proof of Theorem 1.41 in Rudin, Functional Analysis, 2/e. The theorem states
Let $N$ be a closed subspace of a topological vector space (t.v.s.) $X$. Let $\tau$ be the topology of $X$ and define $\tau_N$ as the quotient topology of $X/N$. Then $\tau_N$ is a vector topology on $X/N$; the quotient map $\pi:X\to X\backslash N$ is linear, continuous, and open.
The text first proves that $\tau_N$ is a topology, and that $\pi$ is linear, continuous, and open, which are all fine. Then it attempts to show that the addition map in $X\backslash N$ is continuous:
Proof. ... If now $W$ is a neighborhood of $0$ in $X\backslash N$, there is a neighborhood $V$ of $0$ in $X$ s.t. $$V+V\subset \pi^{-1}(W).$$ Hence $\pi(V)+\pi(V)\subset W.$ Since $\pi$ is open, $\pi(V)$ is a neighborhood of $0$ in $X\backslash N$. Addition is therefore continuous in $X\backslash N$.
I understand every step in the proof, but I think this only shows that $\pi$ is continuous at $0$, not that it is continuous in $X$.
I know that by Theorem 1.17 in the book, if $X$, $Y$ are t.v.s.'s and $\Lambda:X\to Y$ is a linear mapping continuous at $0$, then it is continuous. But this requires that $X$ and $Y$ are t.v.s.'s, and if we know that $X\backslash N$ is a t.v.s., there would be nothing to prove here!
Am I missing something here? My question is: Does continuity at $0$ imply continuity in this problem? If so, how? Any help is appreciated!
Yes, continuity of a group product at 0 implies continuity everywhere for a group equipped with a topology that is translation bi-invariant: Meaning if $U$ is open then $g \cdot U$ and $U \cdot g$ are open for any $g \in G$. In particular, a vector space under addition is an abelian group and so this theorem applies. (Hopefully, by construction it is clear that this topology satisfies the translation-invariant property).
EDIT: Here's an explicit proof of the vector space case, which easily generalizes: Suppose $V$ is a vector space equipped with a topology such that $+$ is continuous at $0$, and whenever $U$ is an open set and $v \in V$, $v + U$ is also open.
Now, we'll show $+:V \times V \rightarrow V$ is continuous at $(v,w)$. Let $U \subset V$ be an open set containing $v + w$. Then, the set $\tilde{U} = -(v + w) + U$ is an open set containing $0$, and so there exist open sets $U_1$ and $U_2$ containing $0$ such that $U_1 + U_2 \subset \tilde{U}$ (by continuity of addition at 0). Then the sets $v + U_1$ and $w + U_2$ are open sets containing $v$ and $w$ respectively, and $(v + U_1) + (w + U_2) \subset (v + w) + \tilde{U} = U$.