# Locally convex spaces are topological vector spaces?

I've come across the below definition of a 'locally convex space' and am trying to prove that addition and multiplication are continuous with respect to the locally convex topology generated by the family of seminorms i.e. that it is a topological vector space:

Let $$V$$ be a vector space over $$\mathbb{K}\in\lbrace\mathbb{R},\mathbb{C}\rbrace$$ and $$P=\lbrace p_i:i\in I\rbrace$$ be a family of seminorms on $$V$$. Define $$p_{i,x_0}:V\to[0,\infty)$$ by $$p_{i,x_0}(x):=p_i(x-x_0)$$ and define $$\mathcal{T}_P$$ to be the smallest topology on $$V$$ making $$p_{i,x_0}$$ continuous for each $$x_0\in V$$, $$i\in I$$. A locally convex space is then defined to be a pair $$(V,\mathcal{T}_P)$$, where $$V$$ is a $$\mathbb{K}$$-vector space and $$P$$ is a family of seminorms on $$V$$.

I have managed to show that this works if, for all $$x_0\in V$$, $$i\in I$$ and $$a\in\mathbb{R}$$, we have that $$\lbrace(x,y)\in V\times V:p_i(x+y-x_0) $$\lbrace(x,y)\in V\times V:p_i(x+y-x_0)>a\rbrace$$ are both open in $$V\times V$$ and $$\lbrace(\lambda,x)\in \mathbb{K}\times V:p_i(\lambda x-x_0) $$\lbrace(\lambda,x)\in \mathbb{K}\times V:p_i(\lambda x-x_0)>a\rbrace$$ are both open in $$\mathbb{K}\times V$$. But I am having trouble finding how to express these sets in terms of the the subbases of the product topologies on $$V\times V$$, $$\mathbb{K}\times V$$. Any ideas? Am I missing something?

Suppose $$x,y$$ are such that $$p_i(x+y-x_0). Then for all $$(x',y')\in V\times V$$ you have $$p_i(x'+y'-x_0) ≤ p_i(x+y-x_0)+p(x-x')+p(y-y')$$ by applying the triangle inequality. Now the first term in the sum is less than $$a$$ and let $$\epsilon = \frac{a-p_i(x+y-x_0)}2$$, then take $$U_{(x,y)}= B_\epsilon(x;p_i)\times B_\epsilon(y;p_i)$$, where $$B_\epsilon(x;p_i):=\{\xi\in V \mid p_i(\xi-x)<\epsilon\}$$ Then $$U_{(x,y)}$$ is open in $$V\times V$$, for all $$(x',y')\in U_{(x,y)}$$ the above inequality becomes: $$p_i(x'+y'-x_0) and the relevant inequality is satisfied for every point in $$U_{(x,y)}$$. In particular $$F:=\{(x,y)\in V\times V\mid p_i(x+y-x_i) is open.
For the other inequality on $$V\times V$$ look at:
$$a-p_i(x'+y'-x_0)≤ a-p_i(x+y-x_0) -p(x-x')-p(y-y')$$
and do the same. If $$a-p_i(x+y-x_0)$$ is positive let $$\epsilon = \frac{p_i(x+y-x_0)-a}2$$ and let $$U_{(x,y)}=B_\epsilon(x;p_i)\times B_\epsilon(y;p_i)$$.
For the sets in $$\Bbb K\times V$$ the inequality you should start with is $$p_i(\lambda' x'- x_0) ≤ p_i(\lambda x-x_0) + |\lambda| p_i( x-x') + |\lambda-\lambda'|p_i(x')$$ and you can do something morally equivalent to the first step.