We are given that \( x, y \in X_0 \), meaning that \( \|x\| = 1 \) and \( \|y\| = 1 \). This implies that both \( x \) and \( y \) are unit vectors.
Now, let's analyze the options:
(A) \( \|x + y\| = \|x\| + \|y\| \) and \( x, y \) are linearly independent:
This statement is correct. For two vectors \( x \) and \( y \) in a real normed linear space, the equality \( \|x + y\| = \|x\| + \|y\| \) holds if and only if \( x \) and \( y \) are linearly independent and they lie on the same line (i.e., they are not opposites). Since \( \|x\| = 1 \) and \( \|y\| = 1 \), the statement \( \|x + y\| = \|x\| + \|y\| \) is satisfied if and only if \( x \) and \( y \) are linearly independent.
(B) \( \|x + y\| = \|x\| + \|y\| \) and \( x, y \) are linearly dependent:
This is incorrect. If \( x \) and \( y \) are linearly dependent, \( \|x + y\| \) would not be equal to \( \|x\| + \|y\| \).
(C) \( \|x + y\|^2 = \|x\|^2 + \|y\|^2 \) and \( x, y \) are linearly independent:
This is incorrect because the norm of the sum of two vectors is not equal to the sum of the squares of the norms unless the vectors are orthogonal, which is not necessarily true here.
(D) \( \|x + y\| = 2\|x\|\|y\| \) and \( x, y \) are linearly dependent:
This is incorrect because the expression \( \|x + y\| = 2\|x\|\|y\| \) only holds when \( x \) and \( y \) are scalar multiples of each other (which implies linear dependence). However, this condition does not apply here, as it is not specified that \( x \) and \( y \) are scalar multiples.
Thus, the correct option is (A), as it correctly describes the relationship between the vectors \( x \) and \( y \).