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Why do the $n \times n$ non-singular matrices form an “open” set?

Like the title says how would you show that the set of matrices such that $\det A \neq 0$ is open?

I can't even see where to start! As I can't envisage how I would find a matrix 'ball' of diameter $\epsilon$ for every element of the set?

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    Hint: The determinant function is continuous.2012-05-13
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    @ThomasE. Ah excellent hint, but what metric do you use on the set of $n \times n$ matrices? As I'm trying to show that $|\det A - \det C| < \epsilon$ right?2012-05-13
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    Ah actually on second thoughts it's obviously continuous as it's a polynomial function of the entries of the matrix2012-05-13
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    Another hint: It is easier in this case to use the open-set definition of continuity than the $\epsilon$-$\delta$ definition.2012-05-13
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    Well, you'll have to make the set M(n,n) of all nxn matrices a metric space (or a topological space) first, to talk of openness. What I suggest is that, you use the metric: d(A,B)= rank(A-B). It is a metric because rank(A-B)>=0 and rank(A-B)=0 => A=B; rank(A-B)=rank(B-A) and rank(A-C)=rank(A-B+B-C)<=rank(A-B)+rank(B-C); Now, all that you'll have to show is that the set of all non-singular matrices of order n is open, in the metric space (M(n,n),rank).2012-05-13
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    You can identify real $n\times n$-matrices with $\mathbb{R}^{n\times n}$, which has natural topology that makes the determinant continuous.2012-05-13
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    It's important to think of the matrices as having a given topology. It's quite easy to cook up a topology on $\operatorname{Mat}_{n \times n}(\mathbb R)$ that makes this set open, but that's not really the question!2012-05-13
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    Summing up, the set is open and the question is closed.2012-05-13

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