TITLE: $W$ be a $d$-dimensional subspace of $\mathbb R^n$. Then how to show that $|\{(x_1,...,x_n) \in W : x_i \in \{0,1\}\}| \le 2^d$?
QUESTION [6 upvotes]: Let $W$ be a $d$-dimensional subspace of $\mathbb R^n$. Then how to show that $|\{(x_1,...,x_n) \in W : x_i \in \{0,1\}\}| \le 2^d$ ?
I can see that we are done if we can find an injective function from $\{(x_1,...,x_n) \in W : x_i \in \{0,1\}\}$ to $\mathbb F_2^d $ , but I can't quite see any such natural injection.

REPLY [3 votes]: This is essentially the same argument as David C. Ullrich's, but explained without matrices:
We would like to show that elements of $W$ are determined by $d$ well-chosen, fixed, projections $p_i$ out of the $n$ natural projections $\mathbb R^n \to \mathbb R$. I.e. we want that the intersection of the $\ker p_i$ intersects $W$ trivially. Put differently, we want an $n-d$-dimensional subspace spanned by $n-d$ standard basis elements, which is complementary to $W$.
We can generalize the result we need as follows:
Proposition. Let $V$ be a finite-dimensional vector space, $\dim V = n$, and $W \subseteq V$ a subspace, $\dim W = d$. Let $U \subseteq V$ be a set that spans $V$ together with $W$. Then there exist $n-d$ elements in $U$ that span a subspace complementary to $W$.
Proof 1. By induction on $n-d$: For $d=n$ there is nothing to prove. Suppose $d < n$, then because $W \neq V$ there exists $u \in U - W$. Apply the induction hypothesis to $W' = W + \langle u \rangle$ and $U' = U - \{u\}$, and note that $\dim W' = d + 1$. $\square$
Proof 2. Quotienting everything by $W$, this simply says that if $U$ (actually: its image in the quotient) spans $V/W$, then it contains a basis of $V/W$. $\square$
It suffices to apply this for $V = \mathbb R^n$, $W=W$ and $U$ the standard basis.