Restricting an Extended Norm to elements of Finite Norm form a Normed Vector Space
# Statement
Let $V$ be a Vector Space over $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$. Suppose $||\cdot||: V \to \overline{\mathbb{R}_{\geq0}}$ is an Extended Norm on $V$. Then $$X := {x \in V : ||x|| < \infty}$$ is a Normed Vector Space and a Vector Subspace of $V$.
# Proof
We can quickly see that $X$ is a Vector Subspace of $V$ because for $c \in \mathbb{K}$ and $u,v \in X$: $$||c u + v || \leq |c| ||u|| + ||v|| < \infty.$$ Thus, $cu + v \in X$ and $X$ is a Vector Subspace because A Subset of a Vector Space is a Subspace iff it is closed under scaling and addition. This also makes $X$ a Vector Space in its own right.
It follows that $X$ is a Normed Vector Space because finiteness is preserved by the 3 defining Axioms of a Norm. $\blacksquare$