Let $X$ be a topological vector space,given any seminorm $p$.the semiball is defined as $U_p = \{x\mid p(x)<1\}$.
Prove if it's open at origin(that is exist a neiborbood of origin $B_0\subset U_p$ ) , then in fact the semiball is open everywhere.
My attempt:
Our goal is to show any $y\ne 0$ there exist neiborhood around it contains in $U_p$
since for topological vector space,any $\epsilon B_0$ is also basis at origin for $\epsilon>0$.Moreover if $\epsilon\le1$ we further has $\epsilon B\subset B$,due to the neiborhood can be choosen to be balanced.
Now given any point $y\ne 0$ with $p(y)<1$,we have $0<1-p(y)\le 1$ hence $(1-p(y))B_0\subset B_0\subset U_p$ any it's open by homeomorphism of the non zero scalar
Hence we can find neiborhood at $y$ as $y+(1-p(y))B_0 \subset U_p$
To show to last inclusion note that any point in the neiborhood of $y$ has the form $y+(1-p(y))z$ with $z\in B_0\subset U_p$(hence $p(z)<1$) hence $p(y+(1-p(y))z))\le p(y)+ (1-p(y))p(z)< p(y) + (1-p(y))\le1$ hence complete the proof.
Is my proof correct?