A connected group which is disconnected

Question

It is known that $G = \mathrm{SL}(2, \mathbb{R})$ is a connected group. Consequently $G$ cannot have any open subgroups. (Any such subgroup would be closed as well, contradicting connectedness.)

How can one reconcile the above fact with the following phenomenon?

Let

\begin{align*} A = \left\lbrace \left( \begin{smallmatrix} t & 0 \\0 & 1/t \end{smallmatrix} \right) : t \neq 0 \right \rbrace, \quad N = \Big\lbrace \left( \begin{smallmatrix} 1 & x \\0 & 1 \end{smallmatrix} \right) : x \in \mathbb{R} \Big\rbrace. \end{align*}

$A$ and $N$ are closed subgroups of $G$. Then,

\begin{align*} AN = \left\lbrace \left( \begin{smallmatrix} t & tx \\0 & 1/t \end{smallmatrix} \right) : x \in \mathbb{R}, \, t \neq 0 \right \rbrace \end{align*} is a group (since $AN = NA$) which contains an open neighbourhood $O$ of $I = \left( \begin{smallmatrix} 1 & 0 \\0 & 1 \end{smallmatrix} \right)$, e.g.

\begin{align*} O_\varepsilon := \left\lbrace \left( \begin{smallmatrix} t & tx \\0 & 1/t \end{smallmatrix} \right) : |x| < \varepsilon, \, |t-1| < \varepsilon \right \rbrace, \quad \varepsilon > 0. \end{align*} This implies that $AN \leqslant G$ is an open subgroup, since $AN \supset a_0 O_\varepsilon n_0$, which is an open neighbourhood of an arbitrary point $a_0 n_0 \in AN$.

Answer 1

I think this is not an open neighbourhood in $\mathrm{SL} (2,\mathbb{R})$. For example, for each given $\varepsilon >0$, we have the element $$ K=\begin{pmatrix} 1&\sqrt{\varepsilon}\\ \sqrt{\varepsilon}&1+\varepsilon \end{pmatrix} $$ that satisfies $\Vert K-\mathrm{Id}_2\Vert_2\leq\sqrt{3}\varepsilon $ but $K$ does not lie in the neighbourhood described in your question.

Answer 2

It turns out that $AN$ is not open. Your error lies in the assertion that $O_\varepsilon$ is an open subset of $SL(2,\Bbb R)$. It is not.