In the proof of the following result, there's a little step (marked with bold letters) that I can't understand
Let $V,W$ Hilbert spaces and $A$ a continuous and linear operator from $V$ to $W'$ with the property that $$||Av||_{W'} \geq ||v||_V$$ Then the image of $A$ is closed.
Proof: Take $\{w_n\}_n \in Im(A)$ Cauchy, therefore there exists $w$ s.t. $w_n \rightarrow w \in W'$.
For every $w_n \in Im(A)$ we have of course a $v_n$ such that $Av_n = w_n$ and by hypothesis we have $$||w_n||_{W'} \geq ||v_n||_V$$
Now, since $\{ w_n\}_n$ is Cauchy, also $\{v_n\}_n$ is Cauchy
I can't understand why the bold sentence is true.
EDIT: I think I got why it's true: pick $\varepsilon >0$ and $m,n \in \mathbb{N}$
$$||w_n - w_m|| = ||A(v_n - v_m)|| \geq ||v_n - v_m||$$
and now, for $m,n > N$ for $N$ large enough we have $||w_n-w_m|| < \varepsilon$, which implies
$$\varepsilon >||w_n - w_m|| = ||A(v_n - v_m)|| \geq ||v_n - v_m||$$
and hence $\{v_n\}_n$ is a Cauchy sequence.
$||Av_n - Av_m||=||A(v_n-v_m) || \geq ||v_n - v_m|| $ and the result follows as $(Av_n) _n $ is Cauchy.