Continuity of Translation Operator on Topological Vector Spaces I was reading Rudin's functional analysis book on topological vector spaces (tvs). According to it, a topological vector space $X$ is a vector space together with a topology $\tau$ which is $T_{1}$ and in which the maps $s: X \times X \to X$ and $p: \mathbb{K}\times X \to X$ given by $s(x,y) := x+y$ and $p(\alpha, x) := \alpha x$ are continuous. Here $\mathbb{K}$ is either $\mathbb{R}$ or $\mathbb{C}$. 
Then, Rudin introduces the translation operator $T_{a}$, which is a map $T_{a}: X \to X$ given by $T_{a}(x) := a+x$. The proposition that follows these definitions states that $T_{a}$ is an homeomorphism of $X$ onto $X$. Well, it is clear that $T_{a}$ is bijective and has inverse $T_{-a}$ due the axioms of vector space. For the continuity part, Rudin states that it follows immediately from the hypothesis that $s$ is continuous and this is a little tricky to me because $s$ is defined on the cartesian product $X \times X$  while $T_{a}$ is a map defined on $X$. So, I'd like to do it a little more carefully. My attempt to prove it is described in the following.
Let $s|_{\{a\}}$ the restriction of $s$ to $\{a\}\times X$, that is, $s|_{\{a\}}: \{a\}\times X \to X$ is given by $s|_{\{a\}}(a,x) := a+x$. Treating $\{a\}\times X$ as a topological subspace of $X \times X$, it  follows that $s|_{\{a\}}$ is continous (restrictions of continuous maps are continuous). Furthermore, we can define an inclusion $i: X \to \{a\}\times X$ by $x \mapsto i(x) :=(a,x)$. Then, $T_{a} = s|_{\{a\}}\circ i$ and, thus, to prove $T_{a}$ is continuous is enough to prove that $i$ is continuous. But $i$ is indeed continuous since, given $x \in X$ and a neighborhood $V$ (open set containing) of $(a,x) \in \{a\}\times X$, we can find a neighborhood $U_{x} \in \tau$ of $x$ so that $(a,x) \in \{a\}\times U_{x}\subset V$, since the sets $\{a\}\times U$, $U \in \tau$, forms a basis of the topology of $\{a\}\times X$. Thus, we have $i(U_{x}) = \{i(x): x \in U_{x}\} = \{(a,x): x \in U_{x}\} = \{a\}\times U_{x} \subset V$. Thus, because $V$ and $x$ is arbitrary, it follows that $i$ is continuous.
Is my reasoning correct? Could it be simpler? Any thoughts would be appreciated! Thanks!
 A: Your work seems, well, tedious. Break it up into simpler parts.
1) Since $X \times X$ has the product topology (you did not say that), the mapping $X \to {a} \times X$ is continuous for each a in X. 
2) For every continuous function $f\colon Y \to Z$ between topological spaces and subset $S$ of $Y$, the restriction map $S \to Z$ is continuous when $S$ is given the subspace topology since this map is the composition of the inclusion $S \hookrightarrow Y$, which is continuous, and the original function $Y \to Z$, and the composition of continuous maps is continuous. 
I left out some details to emphasize the main points. If you are at the stage of learning about topological vector spaces then I think the omitted details (e.g., continuity of the inclusion of a subset) should be familiar from general topology. That is why Rudin skipped an explanation. If the details are not familiar from general topology, okay, but still it is worth highlighting the main points to grasp. Review the product and subspace topologies so such properties are automatically familiar and you do not reinvent the wheel when you have to apply them in new situations.
