Why a Lie Group is closed in GL(n,C)?

[BiPi]

The Brian Hall's book reads: A Lie group is any subgroup G of GL(n,C) with the following property: If Am is a secuence of matrices in G, and Am converges to some matrix A then either A belongs to G, or A is not invertible. Then He concludes G is closed en GL(n,C), ¿How can this be possible, if Am can converge to A, out of GL(n,C), why is closed?

 

[George Jones]

Forget Lie groups for a bit.

Let $A$ be the subset of $\mathbb{R}$ that is the interval $\left(0,2\right)$, and let $B$ be the subset of $A$ that is the interval $\left(0,1\right]$. Is $B$ closed in $A$?

 

[BiPi]

I think it is, am I right?

 

[George Jones]

I think it is, am I right?

Yes.

Changing the notation in my example so as to avoid notional clash with Hall:

Let $U$ be the subset of $\mathbb{R}$ that is the interval $\left(0,2\right)$, and let $V$ be the subset of $U$ that is the interval $\left(0,1\right]$. Then, $V$ closed in $U$.

Consider the sequence of real numbers $x_m =1/m$ for $m$ a positive integer. This is a sequence of numbers in $V$ that converges in $\mathbb{R}$ to the real number zero, which in not in either $V$ or $U$.

Hall writes

$A_m$ is any sequence of matrices in $G$, and $A_m$ converges to some matrix $A$, then either $A$ is in $G$ or $A$ is not in $\mathrm{GL}\left(n;\mathbb{C}\right)$.

In my example, $\mathbb{R}$ plays the role of the set of matrices $M_n \left(\mathbb{C}\right)$, $U$ plays the role of $\mathrm{GL}\left(n;\mathbb{C}\right)$, and $V$ plays the role of group $G$.

 

[BiPi]

In your example, the limit xm=1/m" role="presentation">xm=1/m converges to zero, out of U and V, but You claim V is closed, as far as I know, a closed interval contains all its limits, what am I missing?

 

[George Jones]

In your example, the limit xm=1/m" role="presentation">xm=1/m converges to zero, out of U and V, but You claim V is closed, as far as I know, a closed interval contains all its limits, what am I missing?

Are you familiar with the concept of subspace topology (sometimes called the relative topology, or the induced topology)?

$V$ is closed in $U$, but $V$ is not closed in $\mathbb{R}$. In order for $V$ to be closed in $U$, when a sequence in $V$ converges to a point $U$, then that point must be $V$. In order for $V$ to be closed in $\mathbb{R}$, when a sequence in $V$ converges to a point in $\mathbb{R}$, then that point must be $U$.

My sequence shows that $V = \left(0,1\right]$ is not closed in $\mathbb{R}$, but it does not show that $V$ is not closed in $U$.

 

[BiPi]

Does Not Showing V is not closed in U, implies V is closed in U?

 

[George Jones]

Does Not Showing V is not closed in U, implies V is closed in U?

No, it doesn't.

One method to show that $V$ is (sequentially) closed in $U$ is the following.

Suppose that $\left\{y_m\right\} \subset V$, and that $y_m \rightarrow u$, with $u \in U$ and $u \not\in V$. Establish a contradiction.