Let $Z$ denote the set of points in $\mathbb{R}^{n}$ whose coordinates are $0$ or $1.$ (Thus $Z$ has $2^{n}$ elements, which are the vertices of a unit hypercube in $\mathbb{R}^{n}$.) Given a vector subspace $V$ of $\mathbb{R}^{n},$ let $Z(V)$ denote the number of members of $Z$ that lie in $V.$ Let $k$ be given, $0\le k\le n.$ Find the maximum, over all vector subspaces $V\subseteq\mathbb{R}^{n}$ of dimension $k,$ of the number of points in $V\cap Z.$
1. **Identify the problem and the goal**: We need to find the maximum number of points in the set \( Z \) that lie in a \( k \)-dimensional subspace \( V \subseteq \mathbb{R}^n \). The set \( Z \) consists of points in \(\mathbb{R}^n\) whose coordinates are either 0 or 1, making \( Z \) the set of vertices of a unit hypercube in \(\mathbb{R}^n\).

2. **Construct a specific subspace**: Consider the subspace \( V = \{(x_1, x_2, \dots, x_k, 0, \dots, 0) : x_i \in \mathbb{R}\} \). This subspace is \( k \)-dimensional and contains points where the last \( n-k \) coordinates are zero.

3. **Count the points in \( V \cap Z \)**: The points in \( V \cap Z \) are those in \( Z \) that have their last \( n-k \) coordinates equal to zero. The first \( k \) coordinates can be either 0 or 1, giving us \( 2^k \) such points. Therefore, \( |V \cap Z| = 2^k \).

4. **Show that \( 2^k \) is maximal**: Suppose \( S \) is a set of more than \( 2^k \) elements of \( Z \). Since each vector in \( Z \) has integer coordinates and all are distinct modulo 2, we can consider the elements of \( Z \) as elements of \( F_2^n \), where \( F_2 \) is the field of integers modulo 2.

5. **Properties of \( F_2^n \)**: In \( F_2^n \), every \( k \)-dimensional vector space contains exactly \( 2^k \) points. Therefore, if \( S \) has more than \( 2^k \) points, it cannot be contained in any \( k \)-dimensional \( F_2 \)-vector space. Consequently, there must be \( k+1 \) elements \( v_1, v_2, \dots, v_{k+1} \) in \( S \) that are linearly independent modulo 2.

6. **Linear independence in \( \mathbb{R} \)**: Write \( v_1, v_2, \dots, v_{k+1} \) as columns of a matrix. Since these columns are linearly independent modulo 2, some \( (k+1) \times (k+1) \) submatrix has a nonzero determinant modulo 2. This submatrix thus has an odd determinant, which is nonzero in \( \mathbb{R} \). Therefore, \( v_1, v_2, \dots, v_{k+1} \) are linearly independent over \( \mathbb{R} \), implying that \( S \) is not contained in any \( k \)-dimensional \( \mathbb{R} \)-vector space.

7. **Conclusion**: Since \( S \) cannot be contained in any \( k \)-dimensional subspace if it has more than \( 2^k \) points, the maximum number of points in \( V \cap Z \) for any \( k \)-dimensional subspace \( V \) is \( 2^k \).

\(\blacksquare\)

The final answer is \( \boxed{ 2^k } \)