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I have read the following sentece on a lecture notes:

If $G\subset\mathrm{GL}(n,\mathbb{R})$ is a subgroup then its topological closure $\overline{G}$ is a topologically closed subgroup. $\overline{G}$ is obviously topologically closed, and I'm claiming that it's still a subgroup.

Now, I guess the topology is a metric topology (on $\mathrm{GL}(n,\mathbb{R})$ all metrics are equivalent), but I cannot see how it is true: as it is, it seems a false statement to me.

For example, if $G=\mathrm{GL}(n,\mathbb{R})$ then its closure is all the space of $n\times n$ matrixes, which is not a group.

Or if I consider the subgroup of diagonal matrix of the form $\lambda\mathrm{Id}_n$, $0\neq\lambda\in\mathbb{R}$, then the zero matrix is in the closure so it is not a group (with respect to the matrix multiplication).

I would need the assumption that $\overline{G}\subset\mathrm{GL}(n,\mathbb{R})$ to prove that it is a subgroup.

Am I missing something?

malklera kwezibalo
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  • If $G=\mathrm{GL}(n, \mathbb{R})$ then $\overline{G}=G$. – Randall May 25 '21 at 21:01
  • The closure is understood inside $GL(n, {\mathbb R})$. The claim that the closure of a subgroup is again a subgroup holds for general topological groups. – Moishe Kohan May 25 '21 at 21:05
  • I disagree. On $\mathrm{GL}(1,\mathbb{R})={x\in\mathbb{R}:x\neq0}$ you have that the closure is $\mathbb{R}$ but $0$ has no inverse. On $\mathrm{GL}(2,\mathbb{R})$ you can take a diagonal matrix with 1 and a $\epsilon$ on the other entry, which converge to a singular matrix. – malklera kwezibalo May 25 '21 at 21:06
  • Closed, as defined on that notes, is if any convergence sequence (of element of the set) has limit in the set. And closure is defined as the set plus the limit points. – malklera kwezibalo May 25 '21 at 21:10
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    You can disagree, but that is what is meant: If you take the closure of $\mathbb R \setminus {0}$ in $\mathbb R \setminus {0}$, you get $\mathbb R \setminus {0}$ again. (If you take the closure in $\mathbb R$, you do get $\mathbb R$; but again, that is not what is meant by “closure” here.) – Eike Schulte May 25 '21 at 21:10
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    Without looking at your notes, I have no idea what was said there, but there is always a chance that the notes were sloppily written. In this case, complain to the instructor. – Moishe Kohan May 25 '21 at 21:12
  • Yes, I got the point. There is some discrepancy on that notes. – malklera kwezibalo May 25 '21 at 21:13
  • That definition of closure is fine, but it (implicitly) depends on which things are points that can be limits. In this setting, we work purely in $\operatorname{GL}(n)$ and non-invertible matrices simply don’t exist as potential limit points. (Similar to how a sequence in the reals can converge to infinity, but usually infinity is not in the closure of $\mathbb R$ because we don’t consider it a valid point. Which points are valid as limits depends on the context.) – Eike Schulte May 25 '21 at 21:17

1 Answers1

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You are missing the fact that, when the author of those notes wrote about $\overline G$, what it was meant was its closure in $GL(n,\Bbb R)$.

Besides, it is not true that, in $GL(n,\Bbb R)$ all metrics are equivalent. The usual metric that it is used on $GL(n,\Bbb R)$ is the one for which$$d\bigl((a_{ij})_{1\leqslant i,j\leqslant n},(b_{ij})_{1\leqslant i,j\leqslant n}\bigr)=\sqrt{\sum_{i,j=1}^n(a_{ij}-b_{ij})^2}.$$

José Carlos Santos
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  • Good point. I meant norm. For closed, as reported on the other comment, is not specified it is in $\mathrm{GL}$. That exactly the assumption I want to use – malklera kwezibalo May 25 '21 at 21:10