Real Analysis, Folland problem 5.5.65 Hilbert Spaces 
problem 5.5.65 - $l^2(A)$ is unitarily isomorphic to $l^2(B)$ if and only if $card(A) = card(B)$

Background Information: If $\mathcal{H}_1$ and $\mathcal{H_2}$ are Hilbert spaces with inner products $\langle \cdot, \cdot \rangle_1$ and $\langle \cdot, \cdot \rangle_2$, a unitary map from $\mathcal{H}_1$ to $\mathcal{H_2}$ is an invertible linear map $U:\mathcal{H}_1\rightarrow \mathcal{H}_2$ that preserves inner products $$\langle Ux, Uy\rangle = \langle x, y \rangle, \ \forall x\in\mathcal{H}_1$$
By taking $y = x$, we see that every unitary map is an isometry:$$\lVert Ux\rVert_2 = \|x\|_1$$
I am not really sure how to start with this proof, any suggestions is greatly appreciated. 
 A: Let $A$ be a set. The definition of $\mathscr l^2(A)$ is as the set of functions $f: A \to \mathbb C$ that are square summable, ie for which $\sum_{\alpha \in A} |f(\alpha)|^2 < \infty$ holds. One consequence of this is that $f$ can be non-zero only on a countable subset of $A$. The space naturally has the structure of a vector space and it also has an inner product:
$$\langle g,f \rangle := \sum_{\alpha \in A} \overline{g(\alpha)}\ f(\alpha)$$
Before continuing note that a way of writing elements as $f \equiv \sum_{\alpha \in A} f(\alpha)\ e_\alpha$ where $\{e_\alpha\}_{\alpha \in A}$ is an orthonormal basis may be a more familiar way of looking at the space.
If two sets $A, B$ have same cardinality then there exists a bijection $F: B \to A$ between them. This induces a map $F_*: \mathscr l^2(A) \to \mathscr l^2(B)$ via $F_*(f):=f\circ F$. Since $F$ is a bijection it is equivalent for $\sum_{\alpha \in A} |f(\alpha)|^2$ to be finite and for $\sum_{\beta \in B} |(f\circ F)(\beta)|^2$ to be finite.
So $F_*$ is actually well defined, furthermore $(F^{-1})_*:\mathscr l^2(B) \to \mathscr l^2(A)$ is the same as $(F_*)^{-1}$ as can easily be verified.
It is also easy to see that $F_*$ (and its inverse) are linear maps. It remains to show unitarity. First do this in an unrigorous manner:
$$\langle g,f\rangle_{\mathscr l^2(A)}=\sum_{\alpha \in A}\overline{g(\alpha)}\ f(\alpha)\overset{(*)}{=}\sum_{\beta \in B}\overline{g( F (\beta ) )}\ f(F(\beta))=\langle F_* (g), F_*(f)\rangle_{\mathscr l^2(B)}$$
Where $(*)$ holds because $F$ essentially just rearranges the terms of an absolutely converging sum, renaming the index set from $A$ to $B$. To do this part more cleanly note that the essential step is to show that if we have an absolutely summable expression $\sum_{\alpha \in A}E(\alpha)$, then $\sum_{\beta \in B} E(F(\beta))$ is also absolutely summable and evaluates to the same thing ($E(\alpha) \in \mathbb C$).
Note that $\mathcal A \subset A$ is a finite set iff $F(\mathcal A)=\mathcal B$ for some finite set $\mathcal B \subset B$ ($F$ is bijection), and for finite sets we have equality:
$$\sum_{\alpha \in \mathcal A} E(\alpha)=\sum_{\beta \in F(\mathcal A)}E(F(\beta))$$
But the definition of a sum over a general infinite set is given by
$$\sum_{\alpha \in A} E(\alpha)=\lim_{\mathcal A \to A}\sum_{\alpha \in \mathcal A}E(\alpha) = \lim_{\mathcal A \to A}\sum_{\beta \in F(\mathcal A)}E(F(\beta))=\lim_{\mathcal B \to B}\sum_{\beta \in \mathcal B}E(F(\beta))$$
Where the limit $\mathcal A \to A$ is understood as the limit of the expression in the sense of nets. The last inequality follows from the identification of finite sets of $A$ with finite sets of $B$ via $F$ (note summable in this sense is the same as absolutely summable).
So the two spaces are indeed unitarily equivalent.
