# 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.

• if $Y$ is not dense (hence closed, like you said), you can find a continuous linear form $f: X \to \mathbb R$ such that $Y=\ker f$. then prove that you cannot find a path in $X-Y$ joining some $x_1$ and $x_2$ if $f(x_1)>0$ and $f(x_2)<0$ – Glougloubarbaki Jun 22 '18 at 14:46
• @Glougloubarbaki, you mean a "non-zero" continuous linear form. – fourierwho Jun 22 '18 at 14:47
• @Glougloubarbaki : Ah I see ... thanks... yes that part is done then ... what about the converse ? – user495643 Jun 22 '18 at 14:49
• @fourierwho if $Y=\ker f$ and $Y$ has codimension 1, then $f$ cannot be the zero linear form – Glougloubarbaki Jun 22 '18 at 14:57

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, $f(1)=b\not\in Y$.