Seminorms in a LCS also open? Assume that we have a bunch of seminorms generating the topology of a LCS $X$, does this mean that they are also open maps $p:X \rightarrow [0,\infty)$?
 A: Every nonzero seminorm on a topological vector space is an open map; it need not even be continuous.  (Obviously the zero seminorm $p(x) \equiv 0$ is not open, assuming you consider it to be a seminorm at all.)
Let $p$ be a nonzero seminorm and let $U$ be open in $X$.  Let $a \in p(U)$.  We will show $a$ is an interior point of $p(U)$.  Consider two cases: $a = 0$ and $a \ne 0$.
Suppose $a = 0$.  Then there exists $x \in U$ with $p(x) = 0$.  Since $p$ is not the zero seminorm, there exists $y \in X$ with $p(y) \ne 0$; by rescaling, let us assume $p(y) = 1$.  Since addition and scalar multiplication are continuous on $X$, there exists $\epsilon$ so small that for all $t \in [0,\epsilon)$ we have $x+ty \in U$. Now by the triangle inequality and homogeneity of $p$, for any such $t$ we have 
$$p(x+ty) \le p(x) + t p(y) = t$$
and
$$t = p(ty) = p(x + ty - x) \le p(x+ty) + p(-x) = p(x+ty)$$
so that in fact $p(x+ty) = t$ and thus $t \in p(U)$.  So $[0,\epsilon) \subset p(U)$.
Suppose $a \ne 0$.  Then there exists $x \in U$ with $p(x) = a$.  Since scalar multiplication is continuous on $X$, there exists $\epsilon$ so small that $tx \in U$ for all $t \in (1-\epsilon, 1+\epsilon)$.  Since $p(tx) = ta$ by homogeneity, we have  $((1-\epsilon) a, (1+\epsilon)a) \subset p(U)$.
