Show that the operator norm can be defined in terms of a minimum 
Let $A$ be a bounded linear map. The operator norm is defined as follows:
  \begin{aligned}
\Vert A\Vert&=\inf\{\alpha\in\mathbb R:\Vert Av\Vert\leq\alpha\Vert v\Vert\text{for all }v\in V\}\\
&=\min\{\alpha\in\mathbb R:\Vert Av\Vert\leq\alpha\Vert v\Vert\text{for all }v\in V\}.
\end{aligned}

I've tried to show that $\inf=\min$ by using contradiction, but I did not succeed:
Assume the infimum of this set is $m$, and this it can't be replaced by the minimum. That means that for each $v\in V:\Vert Av\Vert<m\Vert v\Vert$. I can't say that there exists $\epsilon>0$ such that $\Vert Av\Vert\leq m\Vert v\Vert-\epsilon$, because for all I know we have that $A$ gets arbitrarily close to $m$.
So how can I show that this infimum is in fact a minimum?
EDIT
I would like to use that a compact subset of a vector space is closed and bounded, because then I know that the linear map, which is continuous, will attain a minimum on the unit sphere. However, I will ask this specific question in a different post.
 A: Let $(\alpha_n)$ be a minimizing sequence in $X_A:=\{\alpha\in\mathbb R\,:\,\|Av\|\le\alpha\|v\|\text{ for all }v\in V\}$, that is, $\alpha_n\in X_A$ and $\alpha_n\to\|A\|$ as $n\to\infty$. Fix $v\in V$. We have $\|Av\|\le\alpha_n\|v\|$ for each $n$. Letting $n\to\infty$, we find $\|Av\|\le\|A\|\|v\|$. Since $v$ was arbitrary, it follows that $\|A\|\in X_A$, that is the infimum is attained inside the set, so it is a minimum.
A: I'm not sure about the context of your question, but I think it must be something like this : 
The set $A$: 
$\{ \enspace \alpha\in\mathbb R: \enspace \forall \enspace v\in V , \enspace \Vert Av\Vert\leq\alpha\Vert v\Vert  \enspace \}\\
$
Has this as its complement ${\overline{A}}$: 
$\{ \enspace \alpha\in\mathbb R:  \enspace  \exists \enspace v\in V , \enspace \Vert Av\Vert\gt\alpha\Vert v\Vert  \enspace \}\\
$
Set ${\overline{A}}$ cannot contain its own boundary ( if :  $\Vert Av\Vert\gt\alpha\Vert v\Vert$ , then we can find small enough $\epsilon$ with $\Vert Av\Vert\gt(\alpha+\epsilon)\Vert v\Vert$  ). Therefore ${\overline{A}}$ is open and $A$ must be closed . It follows that $\inf{A}=\min{A}$
