Why is the operator norm $||T||_{op} = \text{sup}\{\frac{||Tv||}{||v||} = \text{sup}\{||Tv|| : ||v|| \leq 1\}$

Is this just coming from the fact that $||Tv|| \leq R||v|| \underset{||v|| \leq 1}{\leq} R$? and then just taking the sup? If so (or if not) I can't seem to wrap my head round this. Thank you!

Since $T$ is linear, take any $v\neq 0$, since it holds that $$ \frac{\|T v\|}{\|v\|} = \bigg\| T\bigg(\frac{v}{\|v\|}\bigg)\bigg\|,\quad \text{and}\quad \left\|\frac{v}{\|v\|}\right\| = 1 $$ You can directly see that $$ \sup\left\{\frac{\|Tv\|}{\|v\|}\right\} = \sup\big\{\|Tv\|\mid \|v\| = 1\big\} $$

By the same reasoning you can reduce $$ \sup\big\{ \|Tv\| \mid \|v\|\leq 1\big\} = \sup\big\{\|Tv\| \mid \|v\| = 1\big\} $$ This shows that the three formulations are equivalent $$ \|T\| =\sup\left\{\frac{\|Tv\|}{\|v\|}\right\} = \sup\big\{ \|Tv\| \mid \|v\|\leq 1\big\} = \sup\big\{\|Tv\| \mid \|v\| = 1\big\} $$

Edit: This hold if we assume that the vector space we work on is not trivial, i.e. $V\neq \{0\}$.