I'm reading lecture notes about analysis on infinite dimensional spaces and I ran into this exercise:
Every continuous linear functional $f\in (\mathbb{R}^\infty)^*$ is of the form
$$f(x)=\sum_n^N a_n x(n)$$
for some $(a_n)_{n=1}^N\in \mathbb{R}$. Thus the space can be identified with $c_{00}$, the space of real sequences that eventually are zero.
How do you prove such statements (All X are of the form Y) in general? if this implies the sets are bijective do I have to prove double sided inclusion or find an isomorphism?
Here $V=\mathbb{R}^{\infty}=\mathbb{R}^{\mathbb{N}}$ is equipped with the product topology. This means that as a topological vector space, $V$ is the locally convex space with topology defined by the seminorms $$ ||x||_n=|x_n| $$ with $n\in\mathbb{N}$. A linear form $f:V\rightarrow \mathbb{R}$ is continuous iff it is bounded by a finite positive linear combination of the above seminorms, i.e., iff there is exists a constant $K\ge 0$ and a set of indices $\{n_1,\ldots,n_p\}$ such that for all sequence $x\in V$, $$ |f(x)|\le K(||x||_{n_1}+\cdots||x||_{n_p})=K(|x_{n_1}|+\cdots+|x_{n_p}|)\ . $$ It is then immediate that $f(x)$ only depends on the entries of $x$ at the locations $n_1,\ldots,n_p$. One thus has $$ f(x)=a_1x_{n_1}+\cdots+ a_p x_{n_p} $$ for some constants $a_1,\ldots,a_p$.
BTW, $V$ is reflexive. The space $c_{00}=\oplus_{\mathbb{N}}\mathbb{R}$ seen as the strong dual of $V$ is such that $(c_{00})^{\ast}\simeq V$.