Show a vector space has no countable basis The question is: Prove that there is no isomorphism $T:W\rightarrow V=\mathbb{R}^\mathbb{N}$, hence deduce that $V$ has no countable basis.
I think I know how to show $V$ has no countable basis:
Suppose there is, we can just see them as $(1,0,...),(0,1,0,..)...$ by a change of basis, and that is clearly not a basis. (correct me if the argument is wrong)
I was wondering how I am supposed to show the non-existence of isomorphism and link that to the non-existence of a countable basis?
Thank you!
 A: I suppose that you are talking about $\mathbb{R}$-vector spaces. $W$ has a countable basis $B=(e_i)_{i\in \mathbb{N}}$ then it means that $W$ is a set of cardinality at most $\mathbb{R}$ since it is a countable union of sets of cardinality $\mathbb{R}$. Indeed, you can write
$$W= \bigcup_{k\geq 1}V_k$$ where $V_k= \text{Span}(e_1,...,e_k)$. Since $V_k$ has finite dimension $k$, it has cardinality $\mathbb{R}$.
But this implies in particular that you cannot find a bijection between $W$ and $\mathbb{R}^{\mathbb{N}}$ since the latter is a set of cardinality $2^{\mathbb{R}}$.
A: Here is my solution from discussing this problem with my friends:
Lemma: Let $V$ be a vector space with countable basis. If $S\subset V$ is an independent set, then $S$ is at most countably infinite.
Proof: Let $(v_k)_{k\ge 1}$ be a basis of $S$. Note $$S = \bigcup_{n\ge 1} S\cap span(v_1,\cdots,v_n)$$
Since $S\cap span(v_1,\cdots,v_n)$ is a subspace of $V$ with dimension at most $n$, it follows that if $S$ has more than $n$ vectors in $span(v_1,\cdots,v_n)$, these vectors are linearly dependent. Thus, each $S\cap span(v_1,\cdots,v_n)$ is finite, so $S$ is countable.
Claim: $\mathbb{R}^{\mathbb{N}}$ has an uncountable independent set.
Proof: Consider $\{ f(n) = \exp(rn), r\in \mathbb{R}\}$. This set is uncountable because $\mathbb{R}$ is uncountable. Given any finite set $\{r_1,\cdots,r_m\}$, by Vandermonde matrix or taking $n\to\infty$ suggests there doesn't exist reals $a_1,\cdots,a_m$ such that $\sum\limits_{j=1}^m a_j \exp(r_jn) = 0\forall n\in \mathbb{N}$.
These two claims suggest $\mathbb{R}^{\mathbb{N}}$ doesn't have a countable basis.
