I'm trying to prove that if a separating family of seminorms $D = \{\sigma_{\lambda}\}$ has a finite supremum for each $\lambda$ on $X$, then the map $f: x \mapsto \sup_{\lambda \in \Lambda} \sigma_{\lambda}$ is a norm on $X$.
I've nearly done so by proving all of the norm axioms for $\sigma$, however I'm stuck on proving that $x = 0 \implies \sigma(x) = 0$. The opposite direction is quite simple as it follows directly from the definition of a family of separating norms, but I cannot see why $x = 0$ must imply that each seminorm in my family is also equal to $0$. What is restricting some member of my family of seminorms from having a value $\neq 0$ for $x = 0$? To me, it doesn't look to really be any result from the definition.
For clarity, my given definition of separating is that for each $x \neq 0$ in $X$, there exists $\lambda \in \Lambda$ such that $\sigma_{\lambda}(x) \neq 0$.
If $s$ is a semi norm, then, by definition we have $s(0) = 0$. The difference between norm and semi norm is that a semi norm may be zero for non zero argument.
Since $\sigma_\lambda(0) = 0$ for all $\lambda$ we have $f(0) = \sup_\lambda \sigma_\lambda(0) = 0$.