Uncountability of basis of $\mathbb R^{\mathbb N}$ 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?
 A: Take any almost disjoint family $\mathcal A$ of infinite subsets of $\mathbb N$ with cardinality $2^{\aleph_0}$.
Construction of such set is given here.
I.e. for any two set $A,B\in\mathcal A$ the intersection $A\cap B$ is finite.
Notice that 
$$\{\chi_A; A\in\mathcal A\}$$
is a subset of $\mathbb R^{\mathbb N}$ which has cardinality $2^{\aleph_0}$.
We will show that this set is linearly independent. This implies that the base must have cardinality at least $2^{\aleph_0}$. (Since every independent set is contained in a basis - this can be shown using Zorn lemma. You can find the proof in many places, for example these notes on applications of Zorn lemma by Keith Conrad.)
Suppose that, on the contrary,
$$\chi_A=\sum_{i\in F} c_i\chi_{A_i}$$
for some finite set $F$ and $A,A_i\in\mathcal A$ (where  $A_i\ne A$ for $i\in F$). The set $P=A\setminus \bigcup\limits_{i\in F}(A\cap A_i)$ is infinite.
For any $n\in P$ we have $\chi_A(n)=1$ and $\sum\limits_{i\in F} c_i\chi_{A_i}(n)=0$. So the above equality cannot hold. 

You can find a proof about dimension of the space $\mathbb R^{\mathbb R}$ (together with some basic facts about Hamel bases) here: Does $\mathbb{R}^\mathbb{R}$ have a basis?
In fact, it can be shown that already smaller spaces must have dimension $2^{\aleph_0}$, see Cardinality of a Hamel basis of $\ell_1(\mathbb{R})$.
A: 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 $\mathfrak c$.
