So, in spite of an obvious lack of enthusiasm for this thread, I shall continue to the bitter end (already in sight!)

We have defined the General Linear Group $$GL(n, \mathbb{F})$$ of all invertible $$n \times n$$ matrices. We now want to show that this group is also a manifold. I proceed as follows.

Suppose there is a map $$ \mathbb{Z} \to GL(n,\mathbb{F})$$ such that, for some subset $$\mathbf{A}_n$$ of matrices in $$GL(n,\mathbb{F})$$, that $$\mathbf{A_1,A_2,....,A_n}$$ is a sequence of matrices.

I will say that the sequence $$\bf{A_n}$$ converges uniformly on the matrix $$\mathbf{A}$$ iff, as $$ n \to \infty$$, each entry $$ \mathcal{A}^i_j$$ in the the n-th matrix$$\bf{A_n}$$ converges on the corresponding entry in $$\bf{A}$$. That is, $$(\mathcal{A}_n)^i_j \to \mathcal{A}^i_j$$.

Writing this in the usual way, I say that, for any $$ n \in \mathbb{Z}$$ and some $$\epsilon \gt 0$$, the criterion for uniform convergence is

$$|(\mathcal{A}_n)^i_j - \mathcal{A}^i_j| \lt \epsilon$$. As I may take $$\epsilon$$ as small as I like, I can ensure that, provided $$\bf{A}$$ has non-vanishing determinant, then so do all $$\bf{A}_n \in \epsilon \subset GL(n, \mathbb{F})$$

Under this circumstance, I will say that $$\epsilon$$ is a neighbourhood of the matrix$$\bf{A}$$ whose coordinates are given by the numbers $$(\mathcal{A}_n)^i_j - \mathcal{A}^i_j = x^i_j$$, that is $$GL(n,\mathbb{F})$$ has the structure of a manifold.

Quite obviously there are $$n \times n = n^2$$ of the coordinates $$x^i_j$$, so our manifold is of dimension $$n^2$$. It is therefore a submanifold of $$\mathbb{F}^{n^2}$$, and therefore inherits the subspace topology.

(Hmm, did I say what this is? Do I need to?)

We have defined the General Linear Group $$GL(n, \mathbb{F})$$ of all invertible $$n \times n$$ matrices. We now want to show that this group is also a manifold. I proceed as follows.

Suppose there is a map $$ \mathbb{Z} \to GL(n,\mathbb{F})$$ such that, for some subset $$\mathbf{A}_n$$ of matrices in $$GL(n,\mathbb{F})$$, that $$\mathbf{A_1,A_2,....,A_n}$$ is a sequence of matrices.

I will say that the sequence $$\bf{A_n}$$ converges uniformly on the matrix $$\mathbf{A}$$ iff, as $$ n \to \infty$$, each entry $$ \mathcal{A}^i_j$$ in the the n-th matrix$$\bf{A_n}$$ converges on the corresponding entry in $$\bf{A}$$. That is, $$(\mathcal{A}_n)^i_j \to \mathcal{A}^i_j$$.

Writing this in the usual way, I say that, for any $$ n \in \mathbb{Z}$$ and some $$\epsilon \gt 0$$, the criterion for uniform convergence is

$$|(\mathcal{A}_n)^i_j - \mathcal{A}^i_j| \lt \epsilon$$. As I may take $$\epsilon$$ as small as I like, I can ensure that, provided $$\bf{A}$$ has non-vanishing determinant, then so do all $$\bf{A}_n \in \epsilon \subset GL(n, \mathbb{F})$$

Under this circumstance, I will say that $$\epsilon$$ is a neighbourhood of the matrix$$\bf{A}$$ whose coordinates are given by the numbers $$(\mathcal{A}_n)^i_j - \mathcal{A}^i_j = x^i_j$$, that is $$GL(n,\mathbb{F})$$ has the structure of a manifold.

Quite obviously there are $$n \times n = n^2$$ of the coordinates $$x^i_j$$, so our manifold is of dimension $$n^2$$. It is therefore a submanifold of $$\mathbb{F}^{n^2}$$, and therefore inherits the subspace topology.

(Hmm, did I say what this is? Do I need to?)

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