$\text{codim}_EE_1=\text{codim}_EE_2=1$ implies $\text{codim}_E{(E_1 \cap E_2)} \leq 2$ I want to prove that if $E_1, E_2 \subset E$ are subspaces of a linear space $E$ then $\text{codim}_EE_1=\text{codim}_EE_2=1$ implies $\text{codim}_E{(E_1 \cap E_2)} \leq 2$ .
I tried the following: since $\text{codim}_EE_1 =1$ assume $[e_1]$ is a basis element of $E/E_1$, then for any $x \in E$ there is a unique $a_1$ s.t. $x - a_1e_1 \in E_1$. The following I'm not sure of: Since  $\text{codim}_EE_2 =1$ , that is all $[e_2] \in E/E_2$ are equivalent, we can choose any $[e_2]$ to be our basis, including some $e_2$ that is linear dependent on $x - a_1e_1$ (might be zero). In that case we have unique $a_2$ s.t. $x - a_1e_1 - a_2e_2 \in E_2$ and also $x - a_1e_1 - a_2e_2 \in E_1$ and the rest follows.
Is that correct? Is there a more elegant way to prove it?
 A: $E_1$ and $E_2$ are just kernels of functionals $f_1$, $f_2:E\to k$
(where $k$ is the field you are working over).
Then $E_1\cap E_2$ is the kernel of the linear map $F:E\to k^2$ defined
by $F(e)=(f(e_1),f(e_2))$. What could the dimension of the image of $F$ be?
A: A problem with your proof: the statement

Since  $\text{codim}_EE_2 =1$ , that is all $[e_2] \in E/E_2$ are equivalent, we can choose any $[e_2]$ to be our basis, including some $e_2$ that is linear dependent on $x - a_1e_1$ (might be zero)

Requires clarification and proof.
Here's a proof I like:

Proposition: $\operatorname{codim}(E_1 \cap E_2) \leq \operatorname{codim}(E_1) + \operatorname{codim}(E_2)$

Proof: Consider the map $\phi:E/(E_1 \cap E_2) \to (E/E_1) \times (E/E_2)$ given by 
$$
\phi([x]_{E_1 \cap E_2}) = ([x]_{E_1},[x]_{E_2})
$$
Show that this map is well-defined and linear.  Then, show that this map has a trivial kernel.  Since there is an injective map from $E/(E_1 \cap E_2)$ to $(E/E_1) \times (E/E_2)$, it must be that
$$
\dim[E/(E_1 \cap E_2)] \leq \dim[(E/E_1) \times (E/E_2)] = 
\dim(E/E_1) + \dim (E/E_2)
$$
Which is the desired conclusion.
