The set of all invertible upper triangular matrices is open or not?

Question

Let $M_2(\mathbb R)$ be the set of all $2\times 2$ real matrices and $U$ be the set of all invertible upper triangular matrices ,i.e.,$ U=\{\begin{bmatrix} {x_1}&{x_2}\\{x_3}&{x_4}\end{bmatrix}\in M_2(\mathbb R): x_3=0 \ and\ x_1x_4\neq0\}. $

Now the question is 'Is $U$ open or not w.r.t. usual topology?'

Here if we suppose $U\subseteq\mathbb R^3$ then $U=\{(x_1,x_2,x_4)\in \mathbb R^3:x_1x_4\neq0\}$ and this is an open set as $U=f^{-1}((-\infty,0)\cup(0,\infty))$ where $f:\mathbb R^3\to \mathbb R $ such that $f(x_1,x_2,x_4)=x_1x_4$ and $f$ is a continuous function. So in this case $U$is an open set.

Now if we suppose $U\subseteq\mathbb R^4$ then $U=\{(x_1,x_2,x_3,x_4):x_3=0\ and\ x_1x_4\neq0\}\subset \mathbb R^3\times \{0\}.$ So in this case $U$ can't be open,right?

So what will be the right answer?

Any help is appreciated. Thank you.

Answer 1

As you essentially showed, $U$ is an open subset of the space of upper triangular matrices, but not an open subset of the space of all matrices.

A set is never open per se. You always have to mention the ambient topological space you want to consider ("is open in" or "is an open subset of").

Answer 2

Short answer: the identity matrix $I=\begin{bmatrix} {1}&{0}\\{0}&{1}\end{bmatrix}$ is in $U$ but for any $t\neq 0$, $I_t=\begin{bmatrix} {1}&{0}\\{t}&{1}\end{bmatrix}$ is not in $U$ (not upper triangular). Therefore, $U$ is not open.

Details: the distance between $I_t$ and $I$ is $|t|$, since $I_t-I=\begin{bmatrix} {0}&{0}\\{t}&{0}\end{bmatrix}$. Therefore, any ball with center $I$ and radius $r>0$ will contain $I_{r/2}$, hence not be wholly contained in $U$.