Let $X$ be a topological vector space and $p$ be a semi-norm on $X.$ Show that the following statements are equivalent.
$(1)$ $p$ is continuous.
$(2)$ $p$ is continuous at $0.$
$(3)$ The set $\{x \in X\ |\ p(x) \lt 1 \}$ is open in $X.$
How do I prove $(2) \implies (3)$ and $(3) \implies (1)\ $? If we can show that $(2) \implies (1)$ then we are done with $(2) \implies (3).$ Because the set in $(3)$ is just $p^{-1} (B(0,1)).$
$(2) \implies (1)$ is trivial. Because $p$ is same as a norm except the fact $p(x) = 0$ need not imply $x = 0.$ So it satisfies reverse triangle inequality i.e. $|p(x) - p(y)| \leq p(x - y),$ for all $x,y \in X.$ So we have $(2) \implies (3).$ How do I prove $(3) \implies (1)\ $?