Is $SL_{2} (\Bbb R)$ compact in $GL_{2} (\Bbb R)$

Question

Let, $GL_{2}(\Bbb R) = \{\begin{pmatrix} x_{1} & x_{2} \\ x_{3} & x_{4} \end{pmatrix} :x_{i} \in \Bbb R \forall 1 \leq i \leq 4, x_{1} x_{4} - x_{2}x_{3} \neq 0 \}$

And $SL_{2}(\Bbb R) = \{\begin{pmatrix} x_{1} & x_{2} \\ x_{3} & x_{4} \end{pmatrix} :x_{i} \in \Bbb R \forall 1 \leq i \leq 4, x_{1} x_{4} - x_{2}x_{3} = 1 \}$ .

Clearly,$SL_{2}(\Bbb R)$ is closed in $GL_{2}(\Bbb R)$ but is it bounded in ${\Bbb R}^4$ so that due to Heien-Borel theorem it becomes compact in $GL_{2}(\Bbb R)$.

My attempt:

Choosing $x_{1} , x_{4} \in \Bbb R$ so that $x_{1} x_{4} = 1$ , we can consider arbitrary $x_{2} \in \Bbb R$ by fixing $x_{3} = 0$ . Hence the set does not remain bounded. Is my answer okay?

Answer 1

The sequence of matrices $A_n$ with $x_1 = x_4 = 1$, $x_3 = 0$ and $x_2 = n$, where $n$ is an arbitrary natural number.