Prove that if $A$ is a matrix, and we defined $$\|A\|=\sup\left\lbrace\frac{\|Ax\|}{\|x\|}:x\neq 0\right\rbrace $$ then $$\|A\|=\inf\left\lbrace c>0:\|Ax\|\leq c\|x\| \right\rbrace $$
Attempt: $\|A\|$ is a lower bound, because $\|Ax\|\leq \|A\|\|x\|$, for all $x\in \mathbb{R}^n$ so $$ \|A\|\leq \inf\left\lbrace c>0:\|Ax\|\leq c\|x\| \right\rbrace$$ if I were able to prove that $\|A\|\geq \inf\left\lbrace c>0:\|Ax\|\leq c\|x\| \right\rbrace$ then this will be done! Or another criteria will be show that $$\forall \varepsilon>0, \exists \tilde{c}\in \left\lbrace c>0:\|Ax\|\leq c\|x\| \right\rbrace, \ \mathrm{such\ that}\ \|A\|+\varepsilon>\tilde{c} $$ I tried and the only thing that I achieved $$\forall \varepsilon>0,\exists\tilde{x}\in\mathbb{R}^n,\ \mathrm{such\ that},\ \|A\|-\varepsilon< \frac{\|A\tilde{x}\|}{\|\tilde{x}\|}$$ from the definition of $\sup$ of $\|A\|$, any help or hint?
Let's call this infimum $I$. For all $x$ we have $||Ax||\leq ||A||\cdot ||x||$, and so $I\leq ||A||$.
Conversely, if $||Ax||\leq C||x||$ for all $x$, then for all $x\ne 0$ we have $\frac{||Ax||}{||x||}\leq C$. By taking the supremum on $x\ne 0$ we obtain $||A||\leq C$. Now we take the infimum on all such values of $C$ to get $||A||\leq I$.