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Let $G$ be the group of all $2\times2$ non singular upper triangular matrices (with matrix multiplication) with entries in $\mathbb{R}$.View G as a topological subspace of $\mathbb{R^4}$.Let U be a subgroup of G which is an open subset in $G$.Prove that

$U$$\supseteq${$\begin{pmatrix}a& b\\0& d\end{pmatrix},\;\;$$a,d$ $\in$$\mathbb{R}_{>0}$, $b \in \mathbb{R}$}

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Since $U$ is a subgroup of $G$ it follows that

$\begin{pmatrix}1& 0\\0& 1\end{pmatrix} \in U$.

Since the topology of $G$ is induced by the topology of $\mathbb R^4$ and $U$ is open in $G$ it follows that there exists $r > 0$ such that

$\begin{pmatrix}1 + \delta & \mu \\0& 1 + \nu \end{pmatrix} \in U$

for all $\delta, \mu, \nu \in ]-r,r[$. Since

$\begin{pmatrix}1 + \delta & 0 \\0& 1 + \nu \end{pmatrix}^2 = \begin{pmatrix} (1 + \delta)^2 & 0 \\0& (1 + \nu)^2 \end{pmatrix}$

it follows that

$\begin{pmatrix}a & 0 \\0& d \end{pmatrix} \in U$

for all $a,d \in \mathbb R_{> 0}$, since for given $a,d \in \mathbb R_{> 0}$ there exists $k \in \mathbb N$ and $\delta, \nu \in ]-r,r[$ such that $a = (1 + \delta)^k$ and $b = (1 + \nu)^k$. Similarly it follows that

$\begin{pmatrix} 1 & b \\0& 1 \end{pmatrix} \in U$

for all $b \in \mathbb R$ by solving $\begin{pmatrix}1 & \mu \\0& 1 \end{pmatrix}^k = \begin{pmatrix}1 & b \\0& 1 \end{pmatrix}$, for $\mu \in ]-r,r[$.

Since $\begin{pmatrix}1 & b \\0& 1 \end{pmatrix}\begin{pmatrix}a & 0 \\0& d \end{pmatrix} = \begin{pmatrix}a & bd \\0&d \end{pmatrix}$ the claim follows.

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  • $\begingroup$ *since for given a,d∈R>0 there exists k∈N and δ,ν∈]r,r[ such that a=(1+δ)^k and b=(1+ν)^k*\,why is it so?? $\endgroup$ – Kayoken Sep 29 '14 at 6:34
  • $\begingroup$ @Prayagdeep : Let me discuss the simplest case that $a > d > 1$: Choose some $k \in \mathbb N$ such that $(1 + r)^k > a$, which is possible since $(1 + r)^n \to \infty$ for $n \to \infty$. Then, by continuity, there exists some $0 < \delta < r$ with $(1 + \delta)^k = a$. Since $1 < b < a$ there exists some $0 < \nu < \delta$ with $(1 + \nu)^k = b$, again by continuity. The other cases work with the same idea. $\endgroup$ – wspin Sep 29 '14 at 14:23

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