Definition of continuity in Rudin's Functional Analysis book

Let $\left(X, \mathcal{T}_{X}\right)$ and $\left(Y, \mathcal{T}_{Y}\right)$ be topological spaces. then a function $f: X \rightarrow Y$ is said to be continuous if the inverse image of every open subset of $Y$ is open in $X$. In other words, if $V \in \mathcal{T}_{Y}$, then its inverse image $f^{-1}(V) \in \mathcal{T}_{X}$,

But in Rudin Functional Analysis he has

To say that addition is continuous means, by definition, that the mapping $$ (x, y) \rightarrow x+y $$ of the cartesian product $X \times X$ into $X$ is continuous: If $x_{i} \in X$ for $i=1,2$, and if $V$ is a neighborhood of $x_{1}+x_{2}$, there should exist neighborhoods $V_{i}$ of $x_{i}$ such that $$ V_{1}+V_{2} \subset V $$

Where $X$ is a vector space and the addition is the map $+:X\times X \rightarrow X$ that satisfies the usual addition properties.

My question is how these two definition of continuity are realated?

Solution 1:

Let $f\colon X\times X\to X$ be the addition on $X$, i.e. $f(x_1,x_2) = x_1+x_2$. Take $V\subseteq X$ open. We want to show that $f^{-1}(V)$ is open in $X\times X$. Since topology on $X\times X$ is generated by sets $V_1\times V_2$ for $V_{1,2}$ open in $X$, it is enough to show that for any point $(x_1,x_2)\in f^{-1}(V)$ we can find $V_1\ni x_1$, $V_2\ni x_2$ open such that $V_1\times V_2\subseteq f^{-1}(V)$, which is equivalent to $V_1 + V_2\subseteq V$.

Edit: If for any point $z = (x_1,x_2)\in f^{-1}(V)$ we can find open $V^z_1\ni x_1$ and $V^z_2\ni x_2$ such that $V^z_1\times V^z_2 \subseteq f^{-1}(V)$, then one can easily show that $$f^{-1}(V) = \bigcup_{z\in f^{-1}(V)} V^z_1\times V^z_2$$ and since all $V^z_1\times V^z_2$ are open in $X\times X$, $f^{-1}(V)$ is open as union of open sets.

This is analogous to how in metric space to show that $U$ is open, it is enough to show that for any point $x\in U$ there is an open ball $B(x,r_x)\subseteq U$. Then you can show that $U = \cup_x B(x,r_x)$.