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Let $k$ be an infinite field. There is a claim in the article Remark 2.16, page 1155, that if $U\subset \mathbb{A}^n_k$ is an open subset such that the complement of $U$ in $\mathbb{A}^n_k$ is of codimension $\geq 2$, then $U$ is $\mathbb{A}^1$-chain connected.

Is there an elementary way to see this? Comments are most appreciated!

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    $\begingroup$ Let $p,q$ be points in $U$, and let $r$ be a generic point of $U$. Then the line through $p,r$ and the line through $q,r$ both lie entirely inside $U$ (this is where the codimension $\geq 2$ gets used.) $\endgroup$
    – dhy
    Commented May 7, 2021 at 13:25
  • $\begingroup$ @dhy: Thanks for your comment. Could you elaborate on why the line passing through $p$ and $r$ doesn't meet the complement of $U$? My reasoning was as follows: The line is defined over the function field $F=k(\mathbb{A}^n_k)$ of $\mathbb{A}^n_k$. After base changing to $F$, the line and $Z:=\mathbb{A}^n_F-U_F$ are both closed subsets of $\mathbb{A}^n_F$ of dimensions 1 and atmost $n-2$ respectively. So their intersection has to be empty. Is that correct? $\endgroup$
    – Evans Gambit
    Commented May 8, 2021 at 7:49
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    $\begingroup$ Maybe try the following. The join of a point and a subvariety of dimension d (the closure of the union of all the lines containing both the subvariety and the point) is closed of dimension d+1. If Z is the complement of U and J(p,Z) and J(q,Z) are the joins of p to Z and q to Z then the union has dimension < n. Any point not in the union will work. $\endgroup$
    – Yosemite Stan
    Commented May 8, 2021 at 8:51
  • $\begingroup$ @Yosemite Stan: Thanks for your comment. So for a point $r$ not in the union, the line joining $p$ and $r$ has to be disjoint from $Z$. Otherwise, if the line $l$ meets $Z$ at a point $s$, then $l=\vec{ps}$, so must be contained in the above union. This is much clearer. Thanks again! Slight tidbit: Though we know the claim in the post is false for finite fields, I am unclear as to where we are using $k$ is infinite in your argument. $\endgroup$
    – Evans Gambit
    Commented May 8, 2021 at 9:41
  • $\begingroup$ @EvansGambit To conclude that there is a point in $\mathbb{A}^n$ that does not lie in a given proper subvariety (in this case the union of Yosemite Stan's comment) you need $k$ to be infinite. (Think about the case of the subvariety $x(x-1)\cdots(x-(p-1))=0$ in $\mathbb{A}^1_{\mathbb{F}_p}$.) $\endgroup$
    – dhy
    Commented May 9, 2021 at 17:45

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