I have lie group question :
Let $G$ be a linear group and $A$ and $B$ closed subset of $G$ show that $AB$ need not to be closed.
Hint: Consider upper and lower triangular matrices in $SL(2,\mathbb{R})$
I think we should think topology on G as topology which is coming from $\exp$ map i.e a neighborhood of $a \in G $ is any subset of $G$ which contains $\{a\exp X\,|\,X\in \mathfrak{g},||X|| < \epsilon \} $ where $\mathfrak{g} $ is the Lie algebra of $G $.
I can not show at this topology upper matrices are closed and I can not figure out their product what looks like I will be appriciate if somebody can help .
I don't understand what you mean about the topology on $SL(2,\mathbb{R})$ and I will assume that we are dealing with the usual topology here, that is, that topology induced by the distance$$D\left(\begin{pmatrix}a&b\\c&d\end{pmatrix},\begin{pmatrix}e&f\\g&h\end{pmatrix}\right)=\sqrt{(a-e)^2+(b-f)^2+(c-g)^2+(d-h)^2}.$$
It is clear that the set $U$ of upper triangular matrices is closed in $SL(2,\mathbb{R})$, since it is the zero set of the continuous function $f\colon SL(2,\mathbb{R})\longrightarrow\mathbb{R}$ defined by $f\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)=c$. For the same reason, the set $L$ of lower triangular matrices in $SL(2,\mathbb{R})$ is closed. It is also clear that the matrix $\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)\in SL(2,\mathbb{R})$ does not belong to $UL$, since the lower right entry of an element of $UL$ is the product of two real numbers distinct from $0$, and therefore it cannot be $0$.
But$$\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\lim_{n\in\mathbb{N}}\begin{pmatrix}\frac1n&-\frac1{n^2}-1\\1&-\frac1n\end{pmatrix}$$and, for each $n\in\mathbb N$,$$\begin{pmatrix}\frac1n&-\frac1{n^2}-1\\1&-\frac1n\end{pmatrix}=\begin{pmatrix}n&\frac1{n^2}+1\\0&\frac1n\end{pmatrix}\begin{pmatrix}-1&0\\n&-1\end{pmatrix}\in UL.$$Therefore, $UL$ is not closed.