Extension of $\mathbb{C}$ 
I want to prove that $\mathbb{C} \subset \mathbb{k}$ and $\dim_{\mathbb{C}}\mathbb{k} \leq \aleph_{0} \implies \mathbb{C = k}$($\mathbb{k}$ - field).

If $\mathbb{k} \ni e, e^2,\dots,e^n$ are linearly dependent then $\mathbb{C}(e) = \mathbb{C}$ (since the minimal polynomial of $e$ have degree $1$). So now we can consider $\mathbb{k}$ with basis without elements such as $e$ linearly dependent with its powers or powers of another basis elements. So now we have countable basis such as for any basis element $e$ span of its powers has dimension $\aleph_0 \implies$ his potency equal $\aleph_1$.
It sounds terrible and I'm afraid that this is not true.
Can u please indicate my mistakes, but not tell me a solution.
 A: The statement you want to prove can be reformulated, perhaps more clearly, as follows:

Let $\mathbb{k}$ be an extension field of $\mathbb{C}$; if $\mathbb{k}$ is a proper extension, then $\dim_{\mathbb{C}}\mathbb{k}>\aleph_0$.

The hint is to observe that


*

*no proper extension of $\mathbb{C}$ can be algebraic, because $\mathbb{C}$ is algebraically closed;

*a transcendental extension of $\mathbb{C}$ contains a subfield isomorphic to $\mathbb{C}(X)$ (field of rational functions on the indeterminate $X$).
So it's sufficient to prove that
$$
\dim_{\mathbb{C}}\mathbb{C}(X)>\aleph_0
$$
by finding an uncountable set of rational functions such that no finite subset is linearly dependent over $\mathbb{C}$. Try your hand with
$$
\left\{\frac{1}{X-c}:c\in\mathbb{C}\right\}
$$
that is suggested by partial fraction decompositions.

Note. Your original statement can be rewritten as

If $\mathbb{k}$ is an extension field of $\mathbb{C}$ and $\dim_{\mathbb{C}}\mathbb{k}=\aleph_0$ then $\mathbb{k}=\mathbb{C}$.

I commented (wrongly) that this is false and was corrected by Georges Elencwajg. Indeed, this is a statement of the form “if $A$ then $B$”, where $A$ is false (because we're able to prove the statement above). As part of my comment I suggested to change $=\aleph_0$ into $\le\aleph_0$.
I find that the original formulation is, at the least, misleading.
