Show that a compact operator is bounded

Question

Definition: A linear operator $T: V \to W$ is compact if and only if the image of the unit ball in $V$ is precompact (= every sequence has a cauchy subsequence $\iff $ totally bounded).

Prove: Let $T: V \to W$ be a compact linear operator. Show that $T$ is bounded.

My attempt:

Suppose $T$ is not bounded. Then,

$$\forall M > 0: \exists v_M \in V: \Vert T v_M \Vert > M \Vert v_M \Vert$$

I then tried to construct a sequence without cauchy sequence in $\{Tv \mid \Vert v \Vert \leq 1\}$

So, let $p > q$. Then, $$\left\Vert T \frac{v_p}{\Vert v_p \Vert} - T \frac{v_q}{\Vert v_q \Vert}\right\Vert \geq \left|\frac{\Vert Tv_p \Vert}{\Vert v_p \Vert} - \frac{\Vert Tv_q \Vert}{\Vert v_q \Vert}\right|$$

but was unable to conlude something because off the minus sign.

Any ideas?

EDIT: This is not a duplicate, as other posts use other definitions of compact operators.

Answer 1

If $T$ is unbounded, there is a sequence $(x_n)_{n\in\mathbb N}$ of elements of the unit ball in $V$ such that$$(\forall n\in\mathbb N):\bigl\|T(x_n)\bigr\|>n.$$Therefore, the sequence $\bigl(\|T(x_n)\|\bigr)_{n\in\mathbb N}$ is not a Cauchy sequence.

Answer 2

It follows directly from the definitions. Since $T$ is compact, the image $T(V_1)\subset W$ of the closed unit ball of $V$ is precompact. Consider the cover $W\subset \bigcup_n W_n$, where $W_n$ is the ball of radius $n$. As $\overline{T(V_1)}$ is compact, the cover has a finite subcover, which means that there exists $m$ with $\overline{T(V_1)}\subset W_m$. In particular, $\|Tv\|\leq m$ for all $v\in V_1$, which leads to $$ \|Tv\|\leq m\|v\|,\ \ \ v\in V.$$