Why does $\ell_p$ have infinitely many linearly independent vectors when it equals its double dual (and so is finite dimensional)?

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Let $\ell_p$ with $p \in (1,\infty)$ be the space of sequences which have finite $p$-norm, i.e. $(\sum\limits_{n=1}^\infty |x_n|^p)^\frac{1}{p}$ is finite. Let $(\ell_p)^*$ be the dual space of $\ell_p$. I have just proven that we have that $(\ell_p)^* = \ell_q$ where $\frac{1}{p}+\frac{1}{q} = 1$. This then implies $(\ell_p)^{**} = (\ell_q)^* = \ell_p$.

As the double dual space equals the primal space this implies (assuming AC) that $\ell_p$ is finite dimensional. However if we look at the sequences $e_i$ which are $1$ in their $i^{th}$ position and $0$ elsewhere we notice that they belong to $\ell_p$ and are linearly independent. But we have just found infinitely many linearly independent vectors in a finite dimensional space, a contradiction.

This means I must be making a mistake somewhere, however I can't quite see where, can someone please tell me where my reasoning goes wrong?

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There is no contradiction. One of the statements is about algebraic duals (the fact that a vector space is isomorphic to its double algebraic dual if and only if it is finite dimensional) whereas the other is about topological duals ($\ell^p$ is isomorphic to its double topological dual when $p\in (1,\infty)$).

Indeed the topological dual may be strictly smaller than the algebraic dual when the dimension is infinite, because not all linear functionals are continuous. Because of this, the topological dual (and the double topological dual) need not be strictly bigger than the initial space. In particular, it is possible for an infinite dimensional normed space to be isomorphic to its double topological dual.