homeomorphism between zero-dimensional Hausdorff and two-point space Given $\left\{g_{\alpha}: \alpha\in T\right\}$ consists of all continuous functions from $A$ to $\{0,1\}$ ($A$ is a zero-dimensional Hausdorff space). Let $G =\prod_{\alpha\in T} g_{\alpha}: A\rightarrow \left\{0,1\right\}^{T}$, so $G(x) = (g_{\alpha} (x))_{\alpha\in T}$. 
Show that $G$ gives a homeomorphism between $A$ and $\{0,1\}^{T}$.
My progress. By the definition of $G$, it's easy to see that $G$ is $1$-$1$ and onto as well, so $G$ is a bijection. In addition, since all the components of $G$ are continuous functions, by the universal mapping property, $G$ is also continuous. It remains to show that $G^{-1}$ is continuous as well. Consider the neighborhood $(a,b)$ around the point $0$ (similar for $1$). Now, it's clear that $(G^{-1})^{-1}(a,b)$ is continuous, so can we imply that $G^{-1}$ is continuous because of zero-dimensional Hausdorff space (I can't see how to use this fact anywhere else:P)
 A: Since $A$ is a zero dimensional Hausdorff space (I assume you mean a zero dimensional manifold, which means that each point has an open neighborhood homeomorphic to $\mathbb{R}^{0}$, i.e. the one point set) it is comprised entirely of isolated points. So, given any function $f:A\to X$ and any open set $U\subseteq X$, we have $f^{-1}(U) = \bigcup_{a\in f^{-1}(U)}\{a\}$ which is open, i.e. any function from $A$ is continuous.
Much like $A$, if $T$ is finite (i.e. if $A$ is finite) $\{0,1\}^T$ also has  the discrete topology, and therefore any function from $\{0,1\}^T$ is also continuous. If, however $T$ is infinite, then the topology on $\{0,1\}^T$ is gonna depend on which topology you choose (and generally, the canonical topology is such that the basic open sets will be of the form $\prod_{\alpha} U_\alpha$, where $U_\alpha$ differs from $\{0,1\}$ only from finitely many indices, note that in this case $\{0,1\}$ is Hausdorff, and it is not zero-dimensional, since there is no open set comprising a single point).
From the two paragraphs above, we may have that $G$ is a homeomorphism only if $A$ is finite (because otherwise the $G(a) = (G^{-1})^{-1}(\{a\})$ has to have more than one point, since $\{a\}$ is open and we are assuming $G^{-1}$ is continuous).
However, note that $T$ is, the number of funcions $g_\alpha:A\to\{0,1\}$ is equal to $|\mathcal{P}(A)|$, i.e. the number of subsets of $A$, so $\{0,1\}^T$ has $2^{|\mathcal P(A)|} = 2^{2^{|A|}}$ elements, therefore, there can be no bijection between $A$ and $\{0,1\}^T$.
