$W$ be an $m$-dimensional linear subspace of $\mathbb R^n$ such that $m\le n-2$ , then is it true that $\mathbb R^n \setminus W$ is connected?

Question

Let $W$ be an $m$-dimensional linear subspace of $\mathbb R^n$ such that $m\le n-2$ , then is it true that $\mathbb R^n \setminus W$ is connected (hence path-connected as it is open in $\mathbb R^n$ ) ? My motivation comes from what I can intuitively feel is that if we remove a straight line from 3 dimensional space then the resulting space still remains path connected . But I have a hard time going about proving it ( believing it is true ) rigorously . I though I writing $\mathbb R^n=W \oplus \hat W$ if I could show that every point in $\mathbb R^n \setminus W$ can be connected with some point in $\hat W \setminus \{0\}$ I would be done ( as $\hat W \setminus \{0\}$ is path connected and subset of $ \mathbb R^n \setminus W $ ) ; but couldn't prove it . Please help .Thanks in advance

Answer 1

Start with a base for $W$ and extend it to one of $\mathbb{R}^n$: $w_1, \ldots, w_m, y_1, \ldots y_{n-m}$. The assumption is that $n-m \ge 2$.

If $x_1 = \sum_{i \le m} \alpha_i w_i + \sum_{j \le n-m} \beta_i y_i$ and $x_2 = \sum_{i \le m} \alpha'_i w_i + \sum_{j \le n-m} \beta'_i y_i$ are points of $\mathbb{R}^n\setminus W$. we know that $\sum_{j \le n-m} \beta_i y_i$ and $\sum_{j \le n-m} \beta'_i y_i$ are both non-zero. As $n-m \ge 2$, $\mathbb{R}^{n-m} \setminus \{0\}$ is path connected. Combine that path in that space with a line segment in the first $m$ coordinates to define a path between $x_1$ and $x_2$.

Answer 2

Yes it is true. Firstly suppose that $dim W=n-2$ and remark that $R^2-0$ is connected. Write $(e_1,...,e_n)$ a basis of $R^n$ such that $(e_1,...,e_{n-2})$ is a basis of $W$.

Consider $(x_1,...,x_n), (y_1,...,y_n)\in R^n-W$. You can connect find a path $c:[0,1]\rightarrow R^2-\{0\}$ such that $c(0)=(x_{n-1},x_n)$ and $c(1)=(y_{n-1},y_n)$. You can find a path $d$ in $W$ such that $d(0)=(x_1,...,x_{n-2})$ and $d(1)=(y_1,...,y_{n-2})$. Write $e(t)=(d(t),c(t))$ it is a path between $(x_1,...,x_n)$ and $(y_1,...,y_n)$ in $R^n-W$ since $c(t)\neq 0, t\in [0,1]$.

For the general case, consider $(x_1,...,x_n)$ and $(y_1,..,y_n)$ in $R^n-W$. Let $(e_1,..,e_n)$ a basis of $R^n$ such that $(e_1,..,e_m)$ is a basis of $W$. Since $(x_1,..,x_n)$ and $(y_1,..,y_n)$ are in $R^n-W$, there exists $i,j>m$ such that $x_i,y_j\neq 0$, if $i\neq j$ we can suppose that $i=m+1, j=m+2$, if $i=j$, we suppose that $i=m+1$. write $R^n=W\oplus U\oplus V$ where $ U=Vect(e_{m+1},e_{m+2})$ and $V=Vect(e_{m+3},..,e_n)$. You can find a path $c(t):[0,1]\rightarrow W\oplus U-W$ between $(x_1,..,x_{m+2})$ and $(y_1,..,y_{m+2})$ and a path $d(t)$ between $(x_{m+3},..,x_n)$ and $(y_{m+3},..,y_n)$ in $V$. The path $(c(t),d(t))$ is a path between $(x_1,..,x_n)$ and $(y_1,..,y_n)$ in $R^n$.

Answer 3

Hint

In a vector space of dimension at least equal to $2$, you can join two points $P_1,P_2$ distinct to the origin by a path not passing through the origin. If the origin doesn't belong to the segment joining the two points, you're done. If the line $L$ passing through the two points contains the origin, consider a point $Q \notin L$ then the path $P_1 Q P_2$ doesn't pass through the origin.

Based on this, you'll be able to prove that $\mathbb R^n \setminus W$ is path connected. For two points $x_i=w_i+v_i \in V=\mathbb R^n$, $i=1,2$ with $w_1,w_2 \in W$, the path $$x(t) =(1-t)w_1 + t w_2 +v(t)$$ where $v(t)$ is a path joining $v_1$ to $v_2$ as in previous paragraph is not passing in $W$.

Answer 4

Let $N$ be a complement to $W$, so $\mathbb R^n = W \oplus N$. We then have the map $$ \mathbb R^n \setminus W \to W \times (N \setminus \{ 0\}), \quad x \mapsto (x_W, x_N) $$ of topological spaces, where $(x_W, x_N)$ is the orthogonal decomposition of $x$ according to the direct sum above. This map is continuous and has a continuous inverse, so it is a homeomorphism.

Now both of the spaces on the right-hand side are connected if $\dim N > 1$, and the product of connected spaces is connected.