[proofplan] Use the definition of boundedness in a [normed vector space](/page/Normed%20Vector%20Space) to choose one radius that controls all elements of $A$. For the translate, represent an arbitrary element of $A+v$ as $a+v$ and apply the triangle inequality. For the dilation, represent an arbitrary element of $\lambda A$ as $\lambda a$ and apply absolute homogeneity of the norm; a harmless positive bound covers the case $\lambda=0$ as well. [/proofplan] [step:Choose a uniform norm bound for $A$] Since $A$ is bounded in the normed [vector space](/page/Vector%20Space) $(V,\|\cdot\|_V)$, there exists a real number $R>0$ such that \begin{align*} \|a\|_V\le R \end{align*} for every $a\in A$. [/step] [step:Bound every element of the translate $A+v$] Let $y\in A+v$. By the definition of $A+v$, there exists $a\in A$ such that $y=a+v$. Applying the triangle inequality for the norm $\|\cdot\|_V$ and then the bound on $A$, we get \begin{align*} \|y\|_V=\|a+v\|_V\le \|a\|_V+\|v\|_V\le R+\|v\|_V. \end{align*} Define $C:=R+\|v\|_V$. Since $R>0$ and $\|v\|_V\ge 0$, we have $C>0$. Thus every $y\in A+v$ satisfies $\|y\|_V\le C$, so $A+v$ is bounded. [guided] We want to prove that $A+v$ is bounded, so we must produce one positive number that bounds the norm of every element of $A+v$. Let $y\in A+v$ be arbitrary. The definition of the translated set says precisely that there is some element $a\in A$ with \begin{align*} y=a+v. \end{align*} The element $a$ is controlled because $A$ is bounded: $\|a\|_V\le R$. The fixed vector $v$ contributes the fixed amount $\|v\|_V$. Applying the triangle inequality in $V$ gives \begin{align*} \|y\|_V=\|a+v\|_V\le \|a\|_V+\|v\|_V. \end{align*} Using $\|a\|_V\le R$, this becomes \begin{align*} \|y\|_V\le R+\|v\|_V. \end{align*} Define $C:=R+\|v\|_V$. This number is positive because $R>0$ and norms are nonnegative. Since the same $C$ works for every $y\in A+v$, the set $A+v$ is bounded in $V$. [/guided] [/step] [step:Bound every element of the scaled set $\lambda A$] Let $z\in \lambda A$. By the definition of $\lambda A$, there exists $a\in A$ such that $z=\lambda a$. Applying absolute homogeneity of the norm and then the bound on $A$, we obtain \begin{align*} \|z\|_V=\|\lambda a\|_V=|\lambda|\|a\|_V\le |\lambda|R. \end{align*} Define $D:=\max\{1,|\lambda|R\}$. Then $D>0$ and $\|z\|_V\le D$ for every $z\in\lambda A$. Hence $\lambda A$ is bounded. [/step] [step:Conclude both transformed sets are bounded] The previous two steps give positive constants $C$ and $D$ such that \begin{align*} \|y\|_V\le C \end{align*} for every $y\in A+v$, and \begin{align*} \|z\|_V\le D \end{align*} for every $z\in\lambda A$. Therefore $A+v$ and $\lambda A$ are bounded subsets of $V$. [/step]