TITLE: Uncountability of basis of $\mathbb R^{\mathbb N}$
QUESTION [5 upvotes]: Given vector space $V$ over $\mathbb R$ such that the elements of $V$ are infinite-tuples. How to show that any basis of it is uncountable?

REPLY [3 votes]: I like the following solution (a colleague of mine told me about it):
For any $a\in\mathbb R$ we have a sequence $\widehat a=(1,a,a^2,\dots,a^k,\dots)$.
The set $\{\widehat a; a\in\mathbb R\}$ is linearly independent. (To see this just notice that if you choose $n$ sequences from this set then the first $n$ coordinates of these sequences form Vandermonde matrix.)
Thus we have a linearly independent of cardinality $\mathfrak c$. Hence cardinality of any Hamel basis is at least $\mathfrak c$.
At the same time, the cardinality of the whole space is $|\mathbb R^{\mathbb N}|=\mathfrak c^{\aleph_0}=\mathfrak c$, so the basis cannot have more than $\mathfrak c$ elements. Thus the Hamel dimension of this space is $\mathfrak c$.