Intersection of a Hyperplane and a Subspace Let $Y$ be a dense linear subspace of a normed space $X$, and let $M$ be a closed hyperplane in $X$. I'm trying to show that $M \cap Y$ is a hyperplane in $Y$ and dense in $M$.
I've been trying to show that the codimension of $M \cap Y$ in $Y$ is $1$, and relying on the identity:
$$\dim(A + B) + \dim(A \cap B) = \dim(A) + \dim(B)$$
and keeping in mind that $\dim(X) - \dim(M) = 1$, but this just keeps bringing me round in circles.
 A: The result that you are trying to use will not help as it is valid only for finite-dimensional spaces and clearly the setup in your problem is for infinite-dimensional spaces because any proper subspace of a finite-dimensional normed space is closed and hence it can not be dense.
For showing the first part what you can do is as follows. We need to show that $\dim\left(\dfrac{Y}{M\cap Y}\right) = 1$. This you can do by the following simple computation
\begin{align*}
\dim\left(\dfrac{Y}{M\cap Y}\right)&=\dim\left(\dfrac{M+Y}{M}\right)& & \text{(Second Isomorphism Theorem)}\\
&\leq \dim(X/M) =  1.
\end{align*}
The above dimension can not be $0$ because then it will mean that $M+Y = M$, i.e., $Y \subset M$ and hence $\overline{Y} \subset \overline{M} = M$, which is not true as $\overline{Y} = X$ and $M$ is a hyperplane. 
Second part I have still not figured out how to prove. I'll post it as soon as I get it. On the bright side there are others to help us!!!
A: Yes,  $Y\cap M$ is dense in $M$.  Here is why.
Write $Y=(Y\cap M) \oplus Y'$ (algebraic direct sum) and notice that the projection $\pi :X\to X/M$ is injective on $Y'$ because $Y'\cap M=\{0\}$.  Since
$\text{dim}(X/M)=1$ we deduce that $\text{dim}(Y')\leq 1$.
If $\text{dim}(Y')=0$ then $Y\subseteq M$, and hence $Y$ is obviously dense in $M$, so let us suppose that  $\text{dim}(Y')=1$.  We
then have that $X=M\oplus Y'$ as a topological direct sum (the projections are continuous).
Setting $M'=\overline {Y\cap M}$, suppose that $M'\neq M$.  Then $M'\oplus Y'$ is a proper closed  subspace but this is a
contradiction because $Y\subseteq M'\oplus Y'$.
