What is the 'Algebraic' Dimension of $l^2(\mathbb{N})$?

Question

Let $l^2(\mathbb{N})$ be the space of all complex sequences that are square-summable. Clearly this is a Hilbert space, and it has a maximal orthonormal system which is equipotent to $\mathbb{N}$. Hence, its 'Hilbert dimension' is $\aleph_0$ (please tell me if other termiology is used).

But what is the 'algebraic' dimension of this space? That is, what is the cardinality of its Hamel basis? Is there a way to compute it using only cardinal arithmetic, or does one have to know much more about its algebraic structure?

I was reading this article explaining Halmos' counterexample of an inner product space NOT having any complete, orthonormal subset. It is stated there (implicitly, I guess) that the algebraic dimension of $l^2(\mathbb{N})$ is the cardinality of the continuum, but I don't see why that has to be the case; this seems highly non-trivial to me. Could someone please explain this to me?

Answer 1

Yes, the dimension is $\mathfrak c = 2^{\aleph_0}$. You show the dimension is ${}\le \mathfrak c$ using cardinality, and ${}\ge \mathfrak c$ by explicitly exhibiting a linearly independent set of cardinal $\mathfrak c$.

For the linearly independent set, it is easier to use the isomorphic Hilbert space $L^2[0,1]$. Then the characteristic functions $\mathbf 1_{[0,t]}$ are linearly independent.

Answer 2

The dimension of $\ell^2(\mathbb{N})$ is equal to its cardinality, which is $\mathfrak c$.

It follows from this proposition:

Let $V$ be a complex vector space such that $\dim V \ge \mathfrak c$. Then $$\dim V = |V|$$ i.e the cardinality and dimension of $V$ are equal.

Proof:

Let $V$ be a Hamel basis for $V$. Every element of $V$ can be uniquely written as a finite linear combination of elements of $B$:

$$x = \sum_{b \in B} \alpha_b b = \sum_{i=1}^n \alpha_{b_i}b_i$$

Hence, it is uniquely determined by a finite sequence of pairs $(\alpha_{b_1}, b_1), (\alpha_{b_2}, b_2), \ldots, (\alpha_{b_n}, b_n)$ where $\alpha_{b_i}$ is a scalar, and $b_i \in B$.

Conversely, every such finite sequence of pairs determines an element of $V$, namely $\sum_{i=1}^n \alpha_{b_i}b_i \in V$.

Therefore:

$$|V| = \left|\bigcup_{n\in\mathbb{N}}(\mathbb{C} \times B)^n\right| = \aleph_0 \cdot |\mathbb{C} \times B| = \aleph_0 \cdot \mathfrak{c} \cdot |B| = |B|$$

In particular, this is applicable for $\ell^2(\mathbb{N})$ because the set of geometric sequences $$\left\{\left(1,t,t^2,t^3, \ldots\right) \in \ell^2(\mathbb{N}) : t \in \langle 0, 1\rangle\right\} \subseteq \ell^2(\mathbb{N})$$

has cardinality $\mathfrak c$ and it is linearly independent in $\ell^2(\mathbb{N})$. Hence, $\dim \ell^2(\mathbb{N}) \ge \mathfrak c$.

Also, the cardinality of $\ell^2(\mathbb{N})$ is $\mathfrak c$. It is at most $\mathfrak c$ because $\left|\mathbb{R}^{\mathbb{N}}\right| = {\mathfrak c}^{\aleph_0} = \mathfrak c$, and the subset above has cardinality $\mathfrak c$ so indeed $\left|\ell^2(\mathbb{N})\right| = \mathfrak c$.

Hence the above proposition implies $\dim \ell^2(\mathbb{N}) = \mathfrak c$.