Lattices in vector space over local field contained in a common lattice Let $A=\mathbb{F}_q[t]$, let $v_\infty$ be the usual valuation on $k=\mathbb{F}_q(t)$ such that $ v_\infty(1/t)=1$, and let $k_\infty=\mathbb{F}_q((t^{-1}))$ the completion of $k$ with respect to this valuation. Let $V$ be a finite dimensional $k_\infty$- vector space. I found the following claim which I don't know how to prove:
Take a $\mathbb{F}_q[t]$-lattice $\Lambda_1\subset V$ (since $A$ is PID, $\Lambda_1 $is of the form $\Lambda_1= v_1A+...+v_rA$, where $r=\dim_{k_\infty} V$). Then the set of lattices $\Lambda_2$ such that there is a lattice $\Lambda$ containing both $\Lambda_1$ and $\Lambda_2$ is dense in the set of all latices $\mathcal{L}(V)$.
Any help towards proving this will be greatly appreciated
 A: What is a $\mathbb{F}_q[t]$-lattice in $k_\infty{}^n$, a free discrete $\mathbb{F}_q[t]$-submodule such that $k_\infty{}^n/\Lambda$ is compact ?
If so then (being a PID) any finitely generated $\mathbb{F}_q[t]$-submodule is free, and $k_\infty{}^n/\Lambda$ is compact iff $k_\infty \Lambda=k_\infty^n$.
(if $k_\infty \Lambda$ is smaller than $k_\infty^n$ then the quotient contains a copy of $k_\infty$, it is not compact, if $k_\infty \Lambda=k_\infty^n$ then up to a $k_\infty$-linear transformation it reduces to a quotient of $k_\infty{}^n / \Bbb{F}_q[t]^n$ which is $\cong t^{-1} \Bbb{F}_q[[t^{-1}]]^n$ as topological groups)
So it remains to check the discreteness.
Let $\Lambda_1$ be a lattice which is of rank $n$ as a $\Bbb{F}_q[t]$-module. With a $k_\infty$-linear transformation we can assume that $\Lambda_1=\Bbb{F}_q[t]^n$.
Then $\Lambda_1+ \Bbb{F}_q[t] u$ is discrete iff $u\in \Bbb{F}_q(t)^n$: look at the image of $\Bbb{F}_q[t] u$ in the compact set $k_\infty{}^n/\Lambda_1$, for it to have no accumulation point we need that the image of $\{ t^au, a\ge 0\}$ is a finite set whence $(t^a-t^b) u \in \Lambda_1$ for some $a>b$.
Conversely if $u\in \Bbb{F}_q(t)^n$ then the image of $\Bbb{F}_q[t] u$ in $k_\infty{}^n/\Lambda_1$ is a $\Bbb{F}_q[t]/(f(t))$ module, where $f$ is the denominator of $u$, so it is a finite set.
Therefore take any lattice $\Lambda_2$ and approximate it by a lattice $\Lambda_2'$ in $\Bbb{F}_q(t)^n$ to obtain that $\Lambda_1+\Lambda_2'$ is a lattice.
