Why is "weight clipping" needed for Wasserstein GANs?

I am reading the original paper on the Wasserstein GAN:

https://arxiv.org/pdf/1701.07875.pdf

and I came across this paragraph:

I don't understand the statement: "$$\mathcal{W}$$ is compact implies that all the functions $$f_w$$ will be $$K$$-Lipschitz for some $$K$$ that only depends on $$\mathcal{W}$$". Here, we are talking about a family of functions $$\{f_w\}_{w \in W}$$. Why does the index coming from a compact space means that the functions will be $$K$$-Lipschitz continuous? If I can understand this, then I can understand why we need to clip the weights to a compact space such as a box.

Certainly this statement is not always true strictly as written: letting $$\sigma$$ be the logistic function $$\sigma(x) = 1 / (1 + \exp(-x))$$, consider a (very simple) network of the form $$f_w(x) = \begin{cases} \sigma\left(\frac{x}{w}\right) & w \ne 0 \\ \mathrm{sgn}(x) & w = 0 \end{cases} .$$ Letting $$\mathcal W = [-1, 1]$$, $$\mathcal W$$ is compact, yet $$f_0$$ is not Lipschitz, and each other $$f_\epsilon$$ is Lipschitz but with some constant that becomes infinite as $$\epsilon \to 0$$.
But: consider a more typical network $$f_w^L$$ given recursively by $$f_w^{(0)}(x) = x \qquad f_w^{(\ell)}(x) = \sigma_\ell(W_\ell f_w^{(\ell-1)}(x) + b_\ell) ,$$ where $$w$$ contains all of the parameters $$W_\ell$$, $$b_\ell$$ for each layer $$\ell$$ (and each $$\sigma_\ell$$ is some fixed Lipschitz activation function). Then we have that $$\lVert f_w^{(L)} \rVert_\mathrm{Lip} \le \lVert \sigma_{L} \rVert_\mathrm{Lip} \; \lVert W_L \rVert_\mathrm{op} \lVert f_w^{(L-1)} \rVert_\mathrm{Lip} \le \prod_{\ell=1}^L \lVert \sigma_{\ell} \rVert_\mathrm{Lip} \; \lVert W_\ell \rVert_\mathrm{op} .$$ Now, if $$\mathcal W$$ is compact, then there is some single constant $$D$$ such that $$\lVert W_\ell \rVert_\mathrm{op} \le D$$ for every $$w \in \mathcal W$$.* We've also assumed that each $$\lVert \sigma_\ell \rVert_\mathrm{Lip}$$ is constant and independent of $$w$$. Thus, for any $$w \in \mathcal W$$, we have that $$\lVert f_w^{(L)} \rVert_\mathrm{Lip} \le \prod_{\ell=1}^L \lVert \sigma_{\ell} \rVert_\mathrm{Lip} \; \lVert W_\ell \rVert_\mathrm{op} \le D^L \prod_{\ell=1}^L \lVert \sigma_{\ell} \rVert_\mathrm{Lip} ,$$ a constant independent of the particular choice of $$w$$.
* $$\mathcal W$$ being compact, assuming we're making reasonable decisions about what topology we mean "compact" in, implies that the set of valid $$W_i$$ in $$\mathcal W$$ is also compact, which implies that the operator norm is bounded.