4
$\begingroup$

This is a homework question. $d,n\ge 2$.

Let $L=\{(x_1,...,x_n)\in (\mathbb{R}^d)^n: x_i\in \mathbb{R}^d, x_i\ne x_j \forall i\ne j\}$.

I tend to think it is not path connected because if you assume $d=1$, this is the case. But I cannot prove for $d\ge 2$.

Any help will be appreciated! :)

$\endgroup$
2
  • $\begingroup$ It's connected for $d\ge 2$: I just checked. $L$ is basically complement of finite number of vector subspaces of dimension $dn-d$, which is connected precisely when $d\ge 2$. To visualize this consider deleting a $2-$ dimensional subspace from $\mathbb{R}^4$, which and the complement is still connected, much like deleting a line from $\mathbb{R}^3$. $\endgroup$ Mar 20, 2015 at 17:05
  • $\begingroup$ The cod$\ge2$ argument works, but to make it formal it's not that easy (I think). Of course, it's a general fact from dimension theory of metric spaces, but in this case it's quite easy to prove directly path-connectedness. I post that (too long for this comment). $\endgroup$
    – Jesus RS
    Mar 21, 2015 at 19:13

1 Answer 1

3
$\begingroup$

Given two points $a=(a_1,\dots,a_n),b=(b_1,\dots,b_n)\in L$, there is always another $c=(c_1,\dots,c_n)\in L$ with all $c_k\in\mathbb R^d$ different from all $a_i,b_j\in\mathbb R^d$.

One can draw a path in $L$ from $a$ to $(c_1,a_2,\dots,a_n)$, then another to $(c_1,c_2,a_3,\dots)$, and so on till one reaches $c=(c_1,\dots,c_n)$. Note that all intermediate joint points are in $L$ by the choice of $c$.

From $a=(a_1,\dots,a_n)$ to $(c_1,a_2,\dots,a_n)$. Consider in $\mathbb R^d$ the complement $U$ of the points $a_2,\dots,a_n$. Since $d\ge2$, the open set $U$ is path-connected (even by polygonals) and $a_1,c_1\in U$ by the choice of $c$. Hence, there is a path $x_1:[0,1]\to U$ with $x_1(0)=a_1,x_1(1)=c_1$. For every $t$ we have $x(t)\in U$, so that $x_1(t)\ne a_2,\dots,a_n$ and, consequently, $(x_1(t),a_2,\dots,a_n)$ is a path in $L$ as we were looking for.

From $a=(c_1,a_2,\dots,a_n)$ to $(c_1,c_2,a_3,\dots,a_n)$. Argue the same now with the complement $U$ of the points $c_1,a_3,\dots,a_n$, complement that contains $a_2,c_2$, to get a path $(c_1,x_2(t),a_3,\dots,a_n)$ in $L$ with $x_2(0)=a_2,x_2(1)=c_2$.

Thus we end up with a path in $L$ from $a$ to $c$. Similarly, we construct another from $b$ to $c$, and put toghether we connect $a$ with $b$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .