# confused in the term “closed” in closed subgroup

well, In Brian C Halls Book, I am not getting the definition of Matrix Lie group, as he says : A matrix Lie Group is any subgroup $G$ of $GL_n(\mathbb{C})$ with the following property: If $A_m$ is any sequence of matrix in $G$ and and $A_m$ converges to $A$ then either $A\in G$ or $A$ is not invertible. well, That may happen true, but after that page he gave an example of a subgroupof $GL_n(\mathbb{C})$ which is not closed hence not a matrix lie group! could any one tell me clearly what is the definition?

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I don't understand what the question is. Hall defined a matrix Lie group and then gave an example of something that was not a matrix Lie group. – Qiaochu Yuan Oct 8 '12 at 8:14
The first example he gave , he demanded that the limit of the sequence of matrix should be in the group but in the definition he did not claimed that isn't it? I understand clearly!' – La Belle Noiseuse Oct 8 '12 at 8:28
@Flute: Can you rewrite that comment in English, please? – Chris Eagle Oct 8 '12 at 8:30
well, just tell me do I need my subgroup closed (topologically) to be a matrix lie group or not? – La Belle Noiseuse Oct 8 '12 at 8:39

The essence of that definition is that the subgroup be closed as a subset of the topological space $GL_n(\Bbb C)$. It will usually not be closed as a subset of the space of all matrices (although it might in some cases be) because $GL_n(\Bbb C)$ itself is not. This is why the exception made is for sequences converging to a non-invertible matrix: within the space $GL_n(\Bbb C)$ such sequences are not convergent at all, so need not be considered for the question of closure.
$GL_n(\mathbb{R})$ is not a matrix Lie Group then right? – La Belle Noiseuse Oct 8 '12 at 9:32
@Flute: well no, to the contrary, it is a matrix Lie group. Consider a sequence of real invertible matrices that converge to an invertible matrix $A$ (converging to a singular matrix being uninteresting). Then $A$ is also a real (invertible) matrix, as you can easily see, so $GL_n(\Bbb R)$ is a closed subgroup of $GL_n(\Bbb C)$. – Marc van Leeuwen Oct 8 '12 at 9:48
well, $GL_n(\mathbb{R})$ is not a closed subgroup rather it is dense! – La Belle Noiseuse Oct 8 '12 at 9:54
@Flute: You are totally confused. How would you approximate a complex (non-real) invertible matrix with real matrices? It cannot be done, so the real matrices are not dense in $GL_n(\Bbb C)$. (Of course $GL_n(\Bbb R)$ is dense in something, in itself for instance, but I cannot suppose you meant that.) – Marc van Leeuwen Oct 8 '12 at 9:59
I finally have understood dear sir, It will be a matrix Lie group if either it is inside $GL_n(\mathbb{C}$ or outside of $GL_n(\mathbb{C}$ – La Belle Noiseuse Oct 10 '12 at 8:25