[proofplan] Use the definition of a bounded [linear map](/page/Linear%20Map) to obtain a global constant controlling $\|T(v)\|_W$ by $\|v\|_V$; the linearity of $T$ is part of this hypothesis, although the argument only needs the resulting global norm estimate. Use the definition of a [bounded set](/page/Bounded%20Set) to obtain a uniform bound for $\|a\|_V$ over $a\in A$. Then every element of $T(A)$ has the form $T(a)$ for some $a\in A$, so combining the two bounds gives a uniform bound on $T(A)$. [/proofplan] [step:Choose bounds for the operator and for the set] Since $T:V\to W$ is bounded, there exists a constant $C\geq 0$ such that, for every $v\in V$, \begin{align*} \|T(v)\|_W\leq C\|v\|_V. \end{align*} Since $A\subset V$ is bounded, there exists a constant $M\geq 0$ such that, for every $a\in A$, \begin{align*} \|a\|_V\leq M. \end{align*} [guided] The proof uses the two relevant definitions and then applies them to an arbitrary element of the image. First, by the definition of a bounded linear map, the map $T:V\to W$ is linear and there is a single constant $C\geq 0$ such that, for every input $v\in V$, \begin{align*} \|T(v)\|_W\leq C\|v\|_V. \end{align*} The linearity is part of the phrase "bounded linear map," but this particular preservation-of-boundedness argument only uses the displayed global norm estimate. Second, by the definition of a [bounded set](/page/Bounded%20Set), the subset $A\subset V$ is contained in a ball centered at $0$, so there is a single constant $M\geq 0$ such that, for every $a\in A$, \begin{align*} \|a\|_V\leq M. \end{align*} The word "single" is important in both estimates: $C$ is independent of $v$, and $M$ is independent of $a$. Now let $y\in T(A)$. By the definition of the image of a set under a function, there exists an element $a\in A$ such that \begin{align*} y=T(a). \end{align*} Using this identity, then the boundedness estimate for $T$, and finally the bound for elements of $A$, we get \begin{align*} \|y\|_W=\|T(a)\|_W\leq C\|a\|_V\leq CM. \end{align*} The number $CM$ is non-negative because $C\geq 0$ and $M\geq 0$. Since the same bound $CM$ works for every $y\in T(A)$, the set $T(A)$ is bounded in $W$. [/guided] [/step] [step:Combine the two bounds on each element of the image] Let $y\in T(A)$. By the definition of the image of a set under a function, there exists $a\in A$ such that \begin{align*} y=T(a). \end{align*} Using first this identity, then the boundedness estimate for $T$, and then the bound on $A$, we obtain \begin{align*} \|y\|_W=\|T(a)\|_W\leq C\|a\|_V\leq CM. \end{align*} Thus every $y\in T(A)$ satisfies $\|y\|_W\leq CM$. Since $CM\geq 0$, this is a uniform bound for $T(A)$ in $W$. Therefore $T(A)$ is bounded. [/step]