Uncountably infinite dimensioned vector space A countably infinite dimensioned vector space doesn't seem too far-fetched. An obvious basis exists and addition and scalar multiplication are trivial extensions.
Does an uncountably infinite (say, $\aleph_1$) dimensioned vector space make sense? Is it already a thing and, if so, what does it look like, how does it behave and what is it used for?
 A: There are lots of such vector spaces:


*

*The space of all functions $\mathbb{R}\rightarrow\mathbb{R}$ is an uncountable-dimensional vector space over $\mathbb{R}$.

*$\mathbb{R}$ is an uncountable-dimensional vector space over $\mathbb{Q}$.

*If $A$ is any set, the powerset $\mathcal{P}(A)$ of $A$ is a vector space over the field with two elements $\mathbb{Z}/2\mathbb{Z}$, with vector addition given by "bitwise XOR."
But let me say a bit more about why we should expect such things, even without examples.
It's useful in this context to go back to the usual (= finite-dimensional) vector spaces and abstract a bit. We often think of a finite-dimensional vector space as a set of finite tuples - e.g. $\mathbb{R}^n$ is the set of $n$-tuples of real numbers, and is the canonical $n$-dimensional vector space over $\mathbb{R}$. But a more abstract way to think of these vector spaces is as maps: "the canonical" $n$-dimensional vector space over $\mathbb{R}$ (if you're familiar with fields, we can replace $\mathbb{R}$ with any field $k$) is the set of functions $f: \{1, 2, ..., n\}\rightarrow \mathbb{R}$. This comes from thinking of a tuple as a function: a $3$-tuple of reals is a map $f$ from $\{1, 2, 3\}$ to $\mathbb{R}$, the idea being that the first element of the tuple is $f(1)$, the second is $f(2)$, and so forth.
With this in mind, we can now generalize to arbitrary sets. Given a set $X$, we can consider the set $Fn(X)$ of all functions from $X$ to $\mathbb{R}$ (previously $X$ was $\{1, 2, ..., n\}$ for some $n$, or more generally a finite set). It turns out that $Fn(X)$ is a vector space! Vector addition is given by $$f+g: x\mapsto f(x)+g(x),$$ and scalar multiplication is given by $$af: x\mapsto af(x).$$ If $X$ is uncountable, $Fn(X)$ an uncountable vector space. So by reframing the usual examples in a slightly more abstract way, we see that there's no obstacle to there being really big vector spaces!
Actually the construction I've described above is often not the one we actually want; rather, we usually care more about the set $Fn_{fin}(X)$ of maps $X\rightarrow \mathbb{R}$ with finite support, that is, which are equal to $0$ on all but finitely many elements of $X$. This doesn't make a difference when $X$ is finite, obviously, but makes a huge difference when $X$ is infinite: for example, $Fn_{fin}(\mathbb{N})$ is countable-dimensional, but $Fn(\mathbb{N})$ is not. This is a good exercise; for one half, show that $Fn_{fin}(\mathbb{N})$ can be thought of as the set of finite sequences of real numbers, and for the other half use a diagonal argument.
A: For any cardinality $\alpha$ and any field $F$, there exists an $F$-vector space whose dimension is at least as big as that cardinality. To show this, let $X$ be any set such that $\vert X \vert = \alpha$. Then let $V = F^{X<\omega}$ (that is, the set of all functions from $X$ to $F$ with finite support). Given $u,v\in V$ and $c\in F$, define vector addition and scalar multiplication such that
$$(u+v)(x) = u(x)+v(x)$$
$$(c u)(x) = c(u(x))$$
holds for all $x\in X$. Then $V$ is a vector space. 
For each $x\in X$, define $v_x$ to be the function which sends $x$ to $1$, and everything else to $0$. Then the set $\{v_x\, \vert \, x\in X\}$ is a basis of $V$ with cardinality equal to $\vert X \vert$, and so $\dim V \geq \vert X \vert$.
A: An important example is the set of all continuous functions over the real numbers.
