91Ó°ÊÓ

A vector space is a set of elements, called vectors, equipped with two operations, addition and scalar multiplication, that satisfy certain axioms, such as associative and commutative properties of addition, distributive properties with respect to scalar multiplication, the existence of a zero vector, and the existence of additive inverses. A sequence of real numbers is convergent if it approaches some real number (its limit) as the index goes to infinity. In other words, given any positive number \(\epsilon\), there exists an integer \(N\) such that for all indices \(n > N\), the difference between the \(n\)-th term of the sequence and its limit is less than \(\epsilon\). Now, let's denote by \(C\) the set of all convergent sequences and by \(V\) the vector space of all sequences of real numbers. We want to show that \(C\) is an infinite-dimensional subspace of \(V\).

In order to show that \(C\) is a subspace of \(V\), we need to show that it contains the zero sequence, and it is closed under vector addition and scalar multiplication. 1. Containing the zero sequence: The zero sequence (\(0, 0, 0, ...\)) is a convergent sequence since it has a limit (which is 0). Therefore, the zero sequence belongs to the set of convergent sequences, \(C\). 2. Closed under vector addition: Let \(x = (x_1, x_2, x_3, ...)\) and \(y = (y_1, y_2, y_3, ...)\) be two convergent sequences in \(C\), with limits \(L_x\) and \(L_y\), respectively. We need to show that the sum of the sequences, \(x + y = (x_1 + y_1, x_2 + y_2, x_3 + y_3, ...)\), is also convergent. Given any positive number \(\epsilon\), since both sequences \(x\) and \(y\) are convergent, there exist integers \(N_x\) and \(N_y\), such that for all \(n > N_x\), \(|x_n - L_x| < \frac{\epsilon}{2}\), and for all \(n > N_y\), \(|y_n - L_y| < \frac{\epsilon}{2}\). Let \(N = \max\{N_x, N_y\}\), then for all \(n > N\), we have \(|(x_n + y_n) - (L_x + L_y)| \le |x_n - L_x| + |y_n - L_y| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\). Thus, the sum of the sequences is convergent, and \(x + y\) also belongs to \(C\). 3. Closed under scalar multiplication: Let \(x = (x_1, x_2, x_3, ...)\) be a convergent sequence in \(C\) with limit \(L_x\), and let \(c\) be a scalar (a real number). We need to show that the scalar multiplication, \(c \cdot x = (c \cdot x_1, c \cdot x_2, c \cdot x_3, ...)\), is also convergent. Given any positive number \(\epsilon\), since the sequence \(x\) is convergent, there exists an integer \(N\), such that for all \(n > N\), \(|x_n - L_x| < \frac{\epsilon}{|c|}\) if \(c \neq 0\), and the inequality is always true if \(c = 0\). Then, for all \(n>N\), we have \(|c \cdot x_n - c \cdot L_x| = |c| \cdot |x_n - L_x| < |c| \cdot \frac{\epsilon}{|c|} = \epsilon\). Thus, the scalar multiplication is convergent, and \(c \cdot x\) also belongs to \(C\). As a result of the three properties above, we can conclude that \(C\) is a subspace of \(V\).

To show that \(C\) is infinite-dimensional, we need to prove that there exists a basis for \(C\) containing an infinite number of vectors. Consider the set \(B = \{(1, 0, 0, ...), (0, 1, 0, ...), (0, 0, 1, ...), ...\}\). Each element in \(B\) is a convergent sequence with a limit of 0, so each element belongs to \(C\). Furthermore, the elements of \(B\) are linearly independent, since no finite linear combination of the vectors can equal another vector in the set. Thus, \(B\) is an infinite linearly independent set of vectors in \(C\). Since there exists an infinite linearly independent set in \(C\), we can conclude that \(C\) is infinite-dimensional. In summary, by showing that \(C\) is a subspace of \(V\) and it is infinite-dimensional, we have proven that the set of convergent sequences is an infinite-dimensional subspace of the vector space of all sequences of real numbers.