Consider the lattice $\mathbb Z^d$ in $\mathbb R^d$ and the following two statements:
Let $(P)$ be a property on vectors in $\mathbb R^d$ and let $1\le k \le d$.
(1) For any $k$ linearly independent vectors $v_1, ... , v_k$ in $\mathbb Z^d$, there exists $v_i$ ($1 \le i \le k$) satisfying the property (P).
(2) For any rank-$k$ subgroup $L$ of $\mathbb Z^d$, there exists a nonzero vector $v$ in $L$ satisfying the property (P) that can be extended to a basis of $L$.
I wonder if they are equivalent. Clearly (1) implies (2) for any property (P). But does (2) imply (1)? Can anyone prove it or show there is a property (P) such that this may not happen?
For all $k$, (1) implies (2):
Proof. Suppose (1) holds. Then let $L$ be a rank-$k$ subgroup of $\mathbb{Z}^d$. Let $v_1, \dots, v_k$ be a basis for $L$. Then by (1), WLOG, $v_1$ satisfies $P$. Now $v_1$ is a nonzero vector in $L$ which satisfies $P$ and can be extended to a basis for $L$. Since $L$ was arbitrary, we conclude that (2) holds. $\square$
However, (2) does not always imply (1).
Let $d = k = 1$, and let $P(v)$ be the property "$v$ is positive".
All rank-$1$ subgroups of $\mathbb{Z}$ have a basis of the form $\{n\}$ where $n$ is a positive integer, and thus (2) holds in this case. However, not all nonzero vectors have this property, so (1) does not hold.