Relaxation of the knapsack constraints A set $\mathcal{A}$ is the relaxation of another set $\mathcal{B}$, if $\mathcal{B} \subseteq \mathcal{A}$.

I have a set of points defined as the knapsack constraint

$$
\mathcal{X} = \{x \in \mathcal{Z}^n: w^{\top}x \leq b \}
$$
where $\mathcal{Z}^n$ is the n-dimensional 0-1 vectors and $w \in \Re^n_+$ and $b \in \Re_+$.

I read that one possible relaxation of the above set is 
$$
\mathcal{X} = \{x \in \mathcal{Z}^n: \lfloor w\rfloor ^{\top}x \leq \lfloor b\rfloor \}
$$

where $\lfloor \rfloor$ represents the floor function.

It is very counter-intuitive to me.  $w^\top x \leq b$ represents a half space with left or lower space of the hyperplane defined by $w^\top x = b$. By taking the floor function, we are in fact moving the hyperplane lower and shrinking the size of the set.   

Can anyone explain to me why the floor function terms represent a relaxed set ?
Let me call $
\mathcal{X}_1 = \{x \in \mathcal{Z}^n: w^{\top}x \leq b \}
$
and
$
\mathcal{X}_2 = \{x \in \mathcal{Z}^n: \lfloor w\rfloor ^{\top}x \leq \lfloor b\rfloor \}
$ to avoid confusion.

Let $x \in \mathcal{X}_1$. Then: $$b \ge w^T x = \sum_{i : x_i = 1} w_i$$ which implies:
 $$\lfloor b \rfloor \ge  \left\lfloor  \sum_{i : x_i = 1} w_i \right\rfloor \ge \sum_{i : x_i = 1} \lfloor w_i \rfloor = \lfloor w \rfloor^T x.$$

Therefore $x \in \mathcal{X}_2$, and this shows that $\mathcal{X}_1 \subseteq \mathcal{X}_2$.