Consider $\{0,1\}$ with the discrete topology and $\Sigma=\{0,1\}^{\mathbb{N}}$ with the product topology. We know that this product topology is generated by the metric $$ d(x,y)=\sum_{n=1}^{\infty}\dfrac{|x_i-y_i|}{2^i}. $$
Denote by $C(\Sigma)$ the space of all continuous functions $f:\Sigma \to \mathbb{R}$ and $C^{\gamma}(\Sigma)$ the subset of $C(\Sigma)$ of all $f$ such that $$ Hol(f)=\sup_{x\neq y}\dfrac{|f(x)-f(y)|}{d(x,y)^{\gamma}}<\infty $$
We consider on $C(\Sigma)$ the sup norm, i.e, $\|f\|=\sup_{x\in \Sigma}|f(x)|$ and on $C^{\gamma}(\Sigma)$ the "Hölder norm" defined by $\|f\|_{\gamma}=\|f\|+Hol(f)$.
Fact: $C^{\gamma}(\Sigma)$, $0\leq\gamma <1$ is dense in $C(\Sigma)$ in the uniform norm.
New setting: Let $(M,d)$ be a compact metric space, and $\Sigma=M^{\mathbb{N}}$ with the product topology. $\Sigma$ is compact by Tychonov's Theorem. The product topology is generated by the metric
$$
d_{\Sigma}(x,y)=\sum_{n=1}^{\infty}\dfrac{d(x_i,y_i)}{2^i}
$$
In an analogously way we the spaces $C(\Sigma)$ and $C^{\gamma}(\Sigma)$
and the norms $\|\cdot\|$ and $\|\cdot\|_{\gamma}$ over $C(\Sigma)$ and $C^{\gamma}(\Sigma)$ respectively, they are both well defined because $\Sigma$ is compact.
My Question: Is, for $0<\gamma<1$, $C^{\gamma}(\Sigma)$ dense in $C(\Sigma)$ in the uniform norm?
