# Application of Hahn–Banach theorem [duplicate]

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.

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.
• What if $E$ is infinite?
• 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. Mar 19, 2021 at 14:26