basis for $\mathbb{R}^{\{0, 1\}^S}$ Let's say $S$ is a finite set.  If $R$ is any subset, then consider $f_R$ defined by $f_R(\eta)=\prod_{r \in R} (2\eta(r)-1)$ whenever $\eta \in X:= \{0, 1\}^S$.  Why is this a real basis for the real-valued functions on $X:= \{0, 1\}^S$?  I have tried mostly to show this collection is spanning, since it is of the right cardinality.  This I tried to do by induction.  I must be missing something easy, but no simple computation has worked so far and I have struggled to find even a nonsimple one that might work.
 A: A clever rewriting solves the problem. We can think the elements of $X$ as subsets of $S$, via $\eta\mapsto\{s\in S: \eta(s)=0\}$, and we denote this set by $\eta$ as well.
In this way, $2\eta(r)-1=1$ if $r\notin\eta$ and $2\eta(r)-1=-1$ if $r\in\eta$. Thus, $f_R(\eta)=(-1)^{|R\,\cap\,\eta\,|}$.
Suppose that $\sum_{R\subseteq S}a_Rf_R=0$, with $a_R\in\mathbb R$ for each $R\subseteq S$. Then $\sum_{R\subseteq S}a_R(-1)^{|R\,\cap\,\eta\,|}=0$ for all $\eta\subseteq S$. Let $s\in S$ be fixed and let $S'=S\setminus\{s\}$. Since every subset of $S$ is of the form $R\cup T$, with $R\subseteq S'$ and $T=\emptyset$ or $\{s\}$, then for each $\eta\subseteq S'$ we have
$$\begin{align*}
0=&\,\sum_{R\subseteq S'}a_R(-1)^{|R\,\cap\,\eta\,|}+\sum_{R\subseteq S'}a_R(-1)^{|(R\cup\{s\})\,\cap\,\eta\,|}\\
=&\,\sum_{R\subseteq S'}2a_R(-1)^{|R\,\cap\,\eta\,|}\,,
\end{align*}$$
because for $R\subseteq S'$ and $\eta\subseteq S'$ we have $(R\cup\{s\})\,\cap\,\eta=R\,\cap\,\eta$. By induction we have $2a_R=0$ for all $R\subseteq S'$. Since $s\in S$ was arbitrary, it follows that $a_R=0$ for all $R\subsetneqq S$, which in turn implies that the remaining coefficient, namely $a_S$, is equal to $0$ as well. This shows that the elements $f_R$ are $\mathbb R$-linearly independent.
