# Question about functional analysis, bounded set

Let X be a normed space, Z⊆X, and for all $$f \in X^*$$ set $$\{ x: x \in \mathbb{R} ~,~ \exists z \in Z ~~with~~ x=f(z) \}$$ is bounded, prove that Z is bounded. I was thinking of using the Theorem (Bounded linear functionals) on page 223 of the book kreyszig introductory functional analysis with applications, but I could do the demonstration, if anyone has any ideas I would appreciate it.

For fixed $$z$$, let $$x_{z}^{\ast\ast}\in X^{\ast\ast}$$ be such that $$x_{z}^{\ast\ast}(f)=f(z)$$, so for fixed $$f\in X^{\ast}$$, $$|x_{z}^{\ast\ast}(f)|\leq M_{f}$$ for all $$z$$ and hence by Uniform Boundedness Principle we have $$\|x_{z}^{\ast\ast}\|\leq M$$ for all $$z$$, then for each $$z$$, choose an $$f$$ such that $$\|z\|=|f(z)|$$ and $$\|f\|=1$$, then $$\|z\|\leq\|x_{z}^{\ast\ast}\|\|f\|\leq M$$.