Are there vector spaces with uncountable basis?

For any set $X$, consider maps $f:X \rightarrow \mathbb R$ such that $f(x)=0$ for all but a finite number of $x$. These form a vector space, with basis $\{\delta_x\}$, where $\delta_x(x)=1$ and $\delta_x(y)=0$ when $x \neq y$. So, the number of basis elements is the same as the cardinality of $X$. Take $X$ uncountable, and this space will have uncountable basis.

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Some confusion may arise from trying to sum an infinite (e.g. uncountable) number of vectors. However, we don't do that! A basis allows any vector be decomposed into a linear combination of basis vectors, and linear combinations are finite by definition (or, equivalently, infinite, but having only a finite number of non-zero coefficients).

Might be worth noting that there are two sorts of bases that are commonly used.

A Hamel basis is one in which any element can be written as a finite linear combination of elements of the basis. A Schauder basis is similar, but one takes the closure of the linear span and hence a topology is needed.

I presume you are looking for an uncountable Hamel basis.

Let $V = \{ f: \mathbb{R} \to \mathbb{R} | f^{-1} (\{ 0\}^c) \text { is finite}\}$ with the usual function addition and scalar multiplication.

It is not hard to see that $B = \{ 1_{\{x\}} \}_{x \in \mathbb{R}}$ is a basis and is uncountable.

Take the space $F$ of all functions from $\mathbb R$ into itself which take non-zero values at finitely many points only. For each $x\in\mathbb R$, let$$e_x(y)=\begin{cases}1&\text{ if }y=x\\0&\text{ otherwise.}\end{cases}$$Then the $e_x$'s form an uncountable basis of $F$.