Prove $|W\cap P|\le 2^m$ Let $F$ be a field , let P be a subset of the linear space $\mathbb{F}^n$ :$$P= \{(x_1,\cdots,x_n)\in\mathbb{F}^n:x_i\in\{0,1\}\}$$
show that for any subspace $W$ of $\mathbb{F}^n$ generated by $m$ elements of $P$ , we have $|W\cap P|\le 2^m$
 A: Short answer : by grouping together coordinates behaving identically, one can assume $n\leq 2^m$ without loss of generality. 
More explicit answer : Let $W$ be  generated by $V\subseteq P$, with 
$|V|=m$. We may of course assume that $V$ is linearly independent ;
in particular $V$ does not contain the zero vector.
Denote by $p_i$ the $i$-th natural projection ${\mathbb F}^n \to {\mathbb F}$,
for $1\leq i \leq n$ : $p_i(x_1,x_2,\ldots,x_n)=x_i$. Also, denote by
$v_1,v_2,\ldots,v_m$ the vectors of $V$. 
Next, consider the map $q:\lbrace 1,2,\ldots, n \rbrace \to \lbrace 0,1 \rbrace^m$,
defined by $q(i)=(p_i(v_1),p_i(v_2),\ldots,p_i(v_m))$. Since the image set has cardinality
$2^m$, the fibers of $q$ (i.e. the inverse image sets $\lbrace x\in \lbrace 1,2,\ldots, n \rbrace \ | \ q(x)=y\rbrace$ for $y\in\lbrace 0,1 \rbrace^m$) are at most $2^m$
in number ; let us denote those fibers by $F_1,\ldots,F_r$ with $r\leq 2^m$. 
For each $k$ with $1\leq k \leq r$, select an index $i_k \in F_k$. For 
$v\in V$, define $\psi(v)\in {\mathbb F}^r$ by
$$
\psi(v)=(p_{i_1}(v),p_{i_2}(v),\ldots,p_{i_r}(v))
$$
Then because of the definition of the $F_k$, we have that
$\psi$ is injective and further that $\psi(V)$ is linearly independent just like $V$ is. 
So $|V| \leq |\psi(V)|$ because $\psi$ is injective, and $|\psi(V)| \leq r$ because
$\psi(V)$ is linearly independent in a $r$-dimensional space. Eventually
$|V| \leq r\leq 2^m$ which concludes the proof.
A: The cardinality of $P$, i.e. $|P|=2^n$ is a number of possible different combination of a binary (components $0$'s or $1$'s) vector of a length $n$. 
Note that $P$ isn't linear subspace of $\mathbb{F}$ since for $2\in\mathbb{F}$ and $v=(1,1,....1)$,  $2v \ne P$.
$W$ has at most $m$ basis vectors from $P$ (it is not stated that it generated by $m$ linearly independent elements from $P$, so it could be less). Therefore $W$ is equivalent to $\mathbb{R}^k$ with $k\leq m$. Since there is no more then 
$2^k$ binary vectors in $\mathbb{R}^k$ we get that $|W|\leq 2^m$.
Thus $W\cap P\leq 2^m$. 
