Denote a set by $x = (A, W, b)$ with $A\in R^{p \times q}$, $W \in R^{q \times r}$ and $b\in R^ {q}$.
I try to define the norm(a kind of Euclidean norm) of the set $$ ||x|| = \sqrt{||A||_F^2 + ||W||_F^2 + ||b||_2^2}. $$ where $||\cdot||_F$ is the Frobenius norm for a matrix. Is this a proper way norm? My intuition says 'yes' because matrix can be viewed as a collection of vectors. so the above set is a just collection of vectors....
In general, if we have three normed linear spaces $(X_1, \|\cdot\|_1)$, $(X_2, \|\cdot\|_2)$, and $(X_3, \|\cdot\|_3)$ over the same field, then $X:= X_1 \times X_2 \times X_3$ is a vector space over the given field and $$\|(x_1, x_2, x_3)\| := \sqrt{\|x_1\|_1^2 + \|x_2\|_2^2 + \|x_3\|_3^2}$$ is a norm. I won't prove the vector space, but I will prove the norm axioms.
Positive-definiteness
Certainly $\|(x_1, x_2, x_3)\| \ge 0$ for all $(x_1, x_2, x_3) \in X$. If $\|(x_1, x_2, x_3)\| = 0$, then $$0 = \sqrt{\|x_1\|_1^2 + \|x_2\|_2^2 + \|x_3\|_3^2} \implies 0 = \|x_1\|_1^2 + \|x_2\|_2^2 + \|x_3\|_3^2.$$ If $\|x_1\|_1 > 0$, $\|x_2\|_2 > 0$, or $\|x_3\|_3 > 0$, then the right hand side would be strictly positive, hence $\|x_1\|_1 = \|x_2\|_2 = \|x_3\|_3 = 0$, which implies $x_1 = 0$, $x_2 = 0$, and $x_3 = 0$, since the three norms are positive-definite.
Homogeneity
Suppose $\lambda$ is a scalar. Then, \begin{align*} \|\lambda(x_1, x_2, x_3)\| &= \|(\lambda x_1, \lambda x_2, \lambda x_3)\| \\ &= \sqrt{\|\lambda x_1\|_1^2 + \|\lambda x_2\|_2^2 + \|\lambda x_3\|_3^2} \\ &= \sqrt{|\lambda|^2\|x_1\|_1^2 + |\lambda|^2\|x_2\|_2^2 + |\lambda|^2\|x_3\|_3^2} \\ &= \sqrt{|\lambda|^2}\sqrt{\|x_1\|_1^2 + \|x_2\|_2^2 + \|x_3\|_3^2} \\ &= |\lambda| \|(x_1, x_2, x_3)\|. \end{align*}
Subadditivity
Suppose $(x_1, x_2, x_3), (y_1, y_2, y_3) \in X$. Then, \begin{align*} \|(x_1, x_2, x_3) + (y_1, y_2, y_3)\| &= \|(x_1 + y_1, x_2 + y_2, x_3 + y_3)\| \\ &= \sqrt{\|x_1 + y_1\|_1^2 + \|x_2 + y_2\|_2^2 + \|x_3 + y_3\|_3^2} \\ &\stackrel{(1)}{\le} \sqrt{(\|x_1\|_1 + \|y_1\|_1)^2 + (\|x_2\|_2 + \|y_2\|_2)^2 + (\|x_3\|_3 + \|y_3\|_3)^2} \\ &\stackrel{(2)}{\le} \sqrt{\|x_1\|_1^2 + \|x_2\|_2^2 + \|x_3\|_3^2} + \sqrt{\|y_1\|_1^2 + \|y_2\|_2^2 + \|y_3\|_3^2} \\ &= \|(x_1, x_2, x_3)\| + \|(y_1, y_2, y_3)\|, \end{align*} where $(1)$ is due to the subadditivity of the three norms $\| \cdot \|_1, \| \cdot \|_2, \| \cdot \|_3$, and $(2)$ is due to the subadditivity of the Euclidean norm on $\Bbb{R}^3$.