In the book "Funktionalanalysis" by Werner were locally convex spaces defined by seminorms. One statement is that a locally convex space is Hausdorff, if there exist a neighborhoodbasis of zero $\mathcal{U}$ s.t. $\cap_{U\in\mathcal{U}}U=\{0\}$ (Page 429)
It was proven by: Let $x\neq y$. Choose $U\in\mathcal{U}$ s.t. $x-y\notin\mathcal{U}$. Since taking the difference is continuous, there exist neighborhoods of zero $V,W$ with $W-V\subset U$. It follows that $(x+V)\cap(y+W)=\emptyset$.
I don't understand, how the continuity of taking the difference implies $V-W\subset U$ and how that implies $(x+V)\cap(y+W)=\emptyset$
Thanks for your help.
Let's denote the locally convex space by $X$.
Let $S(p, q) = p - q$ be the subtraction map. Then $S$ is continuous and maps $(0, 0)$ to $0$.
If $U \subseteq X$ is a neighbourhood of $0$, then $S^{-1}(U)$ contains a neighbourhood of $(0, 0)$ in the product space $X \times X$. Without loss of generality, this neighbourhood takes the form $W \times V$, for open sets $W$ and $V$ (given these form a basis for the product space). And, since $(0, 0)$ is contained in this neighbourhood by assumption, we have $0 \in W$ and $0 \in V$. We would have, by construction $$W \times V \subseteq S^{-1}(U),$$ or equivalently, $$S(W \times V) \subseteq U.$$ But, the subtraction map, when applied to $V \times W$, is simply the set $$\{w - v : (w, v) \in W \times V\} = \{w - v : w \in W, v \in V\} = W - V.$$ Hence, $W - V \subseteq U$.
Now, it is claimed that if $U$ fails to contain $x - y \neq 0$, then $(x + V) \cap (y + W) = \emptyset$. Let's prove this!
If $z \in (x + V) \cap (y + W)$, then $z - x \in V$ and $z - y \in W$. Hence, $(z - y) - (z - x) \in W - V \subseteq U$. But, $(z - y) - (z - x) = x - y$. Thus, $x - y \in U$, which is a contradiction. Hence, no such $z$ can exist, and the intersection is empty.
Hope that helps.