It is known that not every separable Banach space has a Schauder basis, and certainly, non-separable Banach spaces cannot. This got me thinking about generalising Schauder bases to work for non-separable Hilbert spaces without them just winding up being orthogonal bases, and I came up with this, which we'll call an ordinal basis.
Let $\alpha$ be an ordinal number and $V$ a Banach space. (You could think about letting it be a more general ordered set but that doesn't make sense, you need well-ordering.) Let $f:\alpha\to V$. Define the sum $\Sigma f$ of $f$ by transfinite induction. $\Sigma f|_0=\mathbf{0}$. $\Sigma f|_{\beta+1}=\Sigma f|_\beta+f(\beta)$. For a limit ordinal $\beta$, $\Sigma f|_\beta=\lim_{\gamma\to\beta}\Sigma f|_\gamma$, meaning $v\in V$ such that, for all $\varepsilon>0$, there exists $\gamma<\beta$ such that for all $\beta>\delta\geq\gamma$, $\|v-\Sigma f|_\delta\|<\varepsilon$. An ordinal basis for $V$ is a function $f:\alpha\to V$ for some ordinal $\alpha$, such that for all $v\in V$, there exists a unique $g:\alpha\to\mathbb{R}$ such that $\Sigma gf=v$, where $gf$ is defined obviously. This does in fact generalise orthogonal bases for Hilbert spaces, and we can thus assume that $\alpha$ is a cardinal number for Hilbert spaces.
That got me wondering about general Banach spaces. Can we always assume $\alpha$ is a cardinal number for a Banach space with an ordinal basis. If the answer is no, this allows new possibilities. For example, could every separable Banach space possess an ordinal basis? After all, $\alpha$ could be a countable ordinal other than $\omega$. This is quite a technical question but it has a more straightforward corollary: every separable Banach space would have to have an infinite-dimensional subspace with a Schauder basis. In general, could every Banach space have an ordinal basis?