It is shown that the algebraic dimension of an infinite dimensional Banach space is equal to it's cardinality. Using this fact, can we say that the algebraic dimension of $(\ell^{\infty}(X))^*$is equal to $2^{2^X}$. Here $X$ is an arbitrary set.
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If $X$ is finite, then $(\ell^\infty(X))^*$ has cardinality $\lvert X\rvert$. – May 04 '20 at 15:35
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1Basically you are asking what is the cardinality of $(l^\infty(X))^*$? – J. De Ro May 04 '20 at 15:37
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Here is the answer in the case that $X= \mathbb{N}$, i.e. if $X$ is countable: https://math.stackexchange.com/questions/2121069/does-ell-infty-have-the-cardinality-of-the-continuum – J. De Ro May 04 '20 at 15:38