91Ó°ÊÓ

Consider the set of sequences \(S=\{s_1, s_2, s_3, \ldots \}\), where each sequence \(s_n\) is given by: \[ s_n = (0, 0, \ldots, 0, 1, 0, 0, \ldots) \] and the 1 is in the nth position, with all other elements being 0. In other words, the sequence \(s_n\) has a 1 in the nth position and zeros elsewhere.

To prove that the sequences in S are linearly independent, we need to show that for any finite set of distinct sequences \(\{s_{n_1}, s_{n_2}, \ldots, s_{n_k}\}\) in S, their linear combination is equal to the zero sequence if and only if all their coefficients are zero, i.e., if we have: \[ a_{n_1}s_{n_1} + a_{n_2}s_{n_2} + \cdots + a_{n_k}s_{n_k} = (0, 0, 0, \ldots) \] then the coefficients \(a_{n_1}, a_{n_2}, \ldots, a_{n_k}\) must all be equal to zero. Given the above linear combination, let's focus on the \(n_i\)th element of the linear combination for some \(i\). The only term in the sum that contributes a non-zero element at position \(n_i\) is the term \(a_{n_i}s_{n_i}\). Thus, for any \(i\), we have \(a_{n_i} = 0\). Since all other terms do not contribute anything at the \(n_i\)th position, we find that all the coefficients \(a_{n_1}, a_{n_2}, \ldots, a_{n_k}\) are equal to zero.

Since we have shown that the set of sequences S is made up of linearly independent sequences, and since S is an infinite set, we've shown there is an infinite set of linearly independent sequences in \(\mathbf{F}^{\infty}\). By definition, this means that \(\mathbf{F}^{\infty}\) is infinite dimensional.