In Rudin's Functional Analysis, Thm 4.1, we have Banach spaces $X$ and $Y$, and the set of bounded linear mappings $B(X,Y)$.
The first steps of the theorem in question argue that $||T|| = \sup \{ ||Tx||: x \in X, ||x|| \le 1\}$ satisfies $||T||<\infty$ for each $T \in B(X,Y)$.
His justification is
Subsets of a normed spacre are bounded if and only if they lie in some multiple of the unit ball.
I'm trying to understand this in more detail. I'm stumbling here, and I've gotten stuck on these types of arguments before. Mostly, I think it's just a matter of translating the different notions of a bounded subset.
$T$ is bounded by the definition of $B(X,Y)$, so $T$ maps bounded sets of $X$ to bounded sets of $Y$. In particular, $$A = \{x \in X: ||x|| \le 1\}$$ is a bounded set of $X$, so $$T(A) = \{Tx: x \in A\} = \{Tx: ||x|| \le 1\}$$ is a bounded set in $Y$.
By the definition Rudin uses for a bounded set in a topological vector space, this means that for every open neighbourhood $V$ of $0$ in $Y$, there exists a $t$ such that $T(A) \subset tV$. How does the fact that $||T|| < \infty$ follow from this?
Take V to be open unit ball. Then $||T(x)|| \le t $ for all $x\in A$ (same notation as the question). Therefore $||T|| \le t $