Linear subspace of codimension one in infinite dimensional Banach space Let $Y$ be a linear subspace of codimension 1 in an infinite dimensional Banach space $X$ i.e. $\dim (X/Y)=1$. Then how to prove that $X\setminus Y$ is path connected if and only if $Y$ is dense in $X$ ?
I am unable to prove either direction. If $Y$ is not dense and of codimension 1, then we can conclude $Y$ is closed subset. Also, $Y$ is not dense in $X$ iff $\exists y \in X \setminus Y$ such that $B(y,1) \subseteq X \setminus Y$. I am unable to see how this could be equivalent with $X \setminus Y$ being not path connected. Please help. 
 A: If $Y$ is not dense and hence closed, then $X/Y$ has a natural norm and hence is homeomorphic to $\mathbb{R}$.  The image of $X\setminus Y$ under the quotient map $X\to X/Y\cong\mathbb{R}$ is $\mathbb{R}\setminus\{0\}$ which is disconnected, so $X\setminus Y$ cannot be connected.
Conversely, suppose $Y$ is dense in $X$.  Fix two points $a,b\in X\setminus Y$. Since $Y$ is dense in $X$, we can choose a sequence $(y_n)$ in $Y$ whose limit is $b-a$.  Now define $f:[0,1]\to X$ to take the linear path from $a$ to $a+y_0$ on $[0,1/2]$, the linear path from $a+y_0$ to $a+y_1$ on $[1/2,2/3]$, the linear path from $a+y_1$ to $a+y_2$ on $[2/3,3/4]$, and so on.  I claim that if we define $f(1)=b$ then $f$ is continuous.  Indeed, it is clear that $f$ is continuous everywhere except possibly at $1$.  Moreover, for any $\epsilon>0$, there exists $N$ such that $y_n$ is within $\epsilon$ of $b-a$ for all $n>N$.  It follows that every point on the path from $a+y_n$ to $a+y_{n+1}$ is within $\epsilon$ of $b$ for all $n>N$, and so $f(t)$ is within $\epsilon$ of $f(1)=b$ for $t$ sufficiently close to $1$.  Thus $f$ is continuous.
Finally, $f(t)\in X\setminus Y$ for all $t$.  For $t\in [0,1)$, $f(t)$ has the form $a+y$ for some $y\in Y$ (namely, $y$ is some linear combination of at most two of the $y_n$), so $f(t)\not\in Y$ since $a\not\in Y$.  Also, $f(1)=b\not\in Y$.
