TITLE: Is Bruhat cell dense in p-adic topology?
QUESTION [5 upvotes]: I've seen in literature a statement like 'there exists an open and dense Bruhat cell'. In $GL(2,F)$ for example, where $F$ is a p-adic field, let $\omega=\begin{pmatrix} & 1 \\ 1 & \end{pmatrix}$, then $A_2N_2\omega N_2$ is the big cell, here $A_2$ is the diagonal matrices, $N_2$ is the upper triangular unipotents.
Question: Is $A_2N_2\omega N_2$ an open and dense subset of $GL(2)$ in p-adic topology, or in Zariski topology?
I asked this because, if we want to find a sequence of elements in this cell approaching the identity, compute 
$$\begin{pmatrix} a & \\  & b\end{pmatrix}\begin{pmatrix} 1 & x \\  & 1 \end{pmatrix}\begin{pmatrix} & 1 \\ 1 & \end{pmatrix}\begin{pmatrix} 1 & y \\  & 1 \end{pmatrix}=\begin{pmatrix} ax & a+axy \\ b & by \end{pmatrix}$$
If this product goes to the identity matrix in the p-adic topology, we will get $b\to 0$, $ax\to 1$, $by\to 1$, and $a+axy\to 0$, from which we will have a contradiction.
Does this mean the big cell is not dense in the p-adic topology? Or I'm misunderstanding things.
Thank you very much for any comment.

REPLY [5 votes]: The big Bruhat cell is exactly the complement of the Borel subgroup, so it is defined by the non-vanishing of an algebraic function on $G$ (the bottom left entry); thus it is certainly open and dense in both the Zariski and $p$-adic topologies.
PS. You might find the identity
$$ \begin{pmatrix} r & s \\ t & u \end{pmatrix} = \begin{pmatrix} s - ru/t &  \\  & t \end{pmatrix}\begin{pmatrix} 1& -rt/(ru-st) \\  & 1 \end{pmatrix}\begin{pmatrix}  & 1 \\ 1 & \end{pmatrix}\begin{pmatrix} 1 & u/t \\  & 1 \end{pmatrix} $$
helpful. Note that as $t \to 0$ then lots of things diverge.