Show that the dual space of a subspace of $V$ can be identified with $V$

Question

Let $V$ be the vector space of all sequences of real numbers $ V=\{s=(a_1, a_2, . . .) | a_i ∈ \mathbb{R} \}$.

Also let $U$ be the vector space of all sequences of real numbers which are eventually $0$, that is $ U=\{s=(a_1,a_2, . . .) | \exists N$ such that $a_N = a_{N+1} = · · · = 0\}$. Show that the dual space $U'$ can be identified with $V$.

The hints I have been give is show that $V$ does not have a countable basis.

I have tried to should $V$ is uncountable, but it is not really useful, and I am not sure how to start this. Any helps are appreciated. Thanks a lot!

Answer 1

Hints. To show that $U'$ can be identified with $V$, we have to give an isomorphism $\Phi \colon V \to U'$. As $U'$ is the dual of $U$, to give $\Phi$, we have to explain, how an element $v \in V$ acts as a linear map on $U$. So to each $v \in V$ we have to associate a linear $\phi_v := \Phi(v) \colon U \to \mathbf R$. Thinking of the inner product on the finite dimensional analogues, we will give $$ \phi_v(u) := \sum_{n=1}^\infty u_n v_n $$ a try, note that the sum is actually finite for each $u \in U$, as all but finitely many $u_n$ are zero.

What remains?

1. Show that $\phi_v$ is linear
2. Show that $\Phi\colon v \mapsto \phi_v$ is linear, that is, the above expression is linear in $v$.
3. Show that for $v \ne v'$ we have $\phi_v \ne \phi_{v'}$ - that is $\Phi$ is one to one.
4. Show that $\Phi$ is onto, to do so, let $f \colon U \to \mathbf R$ an arbitrary linear functional. We want to write it in the above form. To do so, apply $f$ to the basic sequences $e^n = (0,\ldots, 0, 1, 0,\ldots)$ ($1$ in the $n$-th place), and let $v_n = f(e^n)$. Now show that $\phi_v = f$, using that the $e^n$ form a basis of $U$, by showing $f(e^n) = \phi_v(e^n)$ for each $n$.

Answer 2

For $n\in\mathbb{N}$, let $e_n$ denote the sequence whose $n$'th element is $1$, and all the other elements are $0$. The $e_n$'s form a basis of $U$. constructing a linear functional $U\to\mathbb{R}$ is equivalent to matching a real number to every element in the basis.