Construction of an explicit isomorphism Is there an explicit isomorphism between the Hilbert spaces: $L^2[0,1]$ and
$\ell^2$, if the answer is 'yes' how do I construct it ?
It is true to say that the Hilbert isomorphism must be in the same dimension ?
Thanks in advance
 A: For each $ n \in \mathbb{N} $, define $ \mathbf{e}_{n}: [0,1] \to \mathbb{C} $ by
$$
\forall x \in [0,1]: \quad {\mathbf{e}_{n}}(x) \stackrel{\text{def}}{=} e^{2 \pi inx}.
$$
Then $ \{ \mathbf{e}_{n} \}_{n \in \mathbb{N}} $ is an orthonormal basis (O.N.B.) of $ {L^{2}}([0,1]) $.

For each $ n \in \mathbb{N} $, define $ \mathbf{s}_{n}: \mathbb{N} \to \mathbb{C} $ by
$$
\mathbf{s}_{n} \stackrel{\text{def}}{=} (0,\ldots,0,1,0,\ldots),
$$
where the $ 1 $ occupies the $ n $-th position of the sequence. Then $ \{ \mathbf{s}_{n} \}_{n \in \mathbb{N}} $ is an O.N.B. of $ {\ell^{2}}(\mathbb{N}) $.

There exists an obvious Hilbert-space isomorphism $ \Psi: {L^{2}}([0,1]) \longrightarrow {\ell^{2}}(\mathbb{N}) $ that satisfies $ \Psi: \mathbf{e}_{n} \longmapsto \mathbf{s}_{n} $ for all $ n \in \mathbb{N} $.
To answer your final question, let me quote the following theorem.

Theorem Two Hilbert spaces $ \mathcal{H}_{1} $ and $ \mathcal{H}_{2} $ are isomorphic if and only if an O.N.B. of $ \mathcal{H}_{1} $ has the same cardinality as an O.N.B. of $ \mathcal{H}_{2} $.

