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I'd like to prove this statement:

Let $X$ a normed space. If $X$ has a linearly independent subset of cardinal $n$, then so does $X'$.

I know that I have to use the Hahn-Banach theorem to prove the existence of functionals in $X'$ that hold the above mentioned property but I'm not sure how.

Any suggestions are welcome. Thanks in advance.

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Let $E=\{x_1, x_2, \cdots, x_n\}$ be the linearly independent set. Let $Y:=\textit{span}\{x_1, x_2, \cdots, x_n\}$. Then $E$ is a basis of $Y$. Let $p_i: Y \rightarrow \mathbb{K}$ ($\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$) be the co-ordinate functional, i.e., $p_i(\sum_{i=1}^nc_ix_i)=c_i$. Since $Y$ is finite dimensional each $p_i$ is continuous and $p_i(x_j)=1$ if $i=j$ and $p_i(x_j)=0$ if $i \neq j$. Now by the Hahn-Banach theorem, $\exists~~~f_1, f_2, \cdots, f_n \in X'$ such that $f_i|_Y=p_i~~~\forall~i$. $\therefore~~~f_i(x_j)=p_i(x_j)~~~\forall~i,j$. It can be easily shown that $f_i$'s are linearly independent.

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  • $\begingroup$ What if $E$ is infinite? $\endgroup$
    – Ruy
    Mar 19, 2021 at 12:27
  • $\begingroup$ See the question again, it said "$X$ has a linearly independent subset of cardinal n". I just answered the above question and not stating anything beyond that. $\endgroup$
    – absolute0
    Mar 19, 2021 at 14:26
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    $\begingroup$ Cardinals can be infinite. $\endgroup$
    – Ruy
    Mar 19, 2021 at 14:41

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