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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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    Summing up, the set is open and the question is closed.2012-05-13

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Make the set $M(n,n)$ of all $n\times n$ matrices a metric space first, by taking $d(A,B)=\text{rank}(A-B)$. This is a metric, since $\begin{align} \text{rank}(A-B)&\geq 0, \text{rank}(A-B)=0 \Rightarrow\\ &A=B ; \end{align}$ $ \begin{align} \text{rank}(A-B) &= \text{rank}(B-A) \text{ and } \text{rank}(A-B) \\ &=\text{rank}((A-C)+(C-B)) \\ &\leq \text{rank}(A-C)+\text{rank}(C-B), \Rightarrow \\ &(M(n,n),\text{rank}) \text{ is a metric space.} \end{align} $ Now, you'll have to show that the set of all non-singular i.e. full-rank matrices is open. Because, if $A$ is a full-rank matrix, taking $\epsilon=1/2$, we have: $\text{rank}(A-X)<1/2 \Rightarrow \text{rank}(A-X)=0 \Rightarrow A=X$ i.e. $X$ is of full-rank. Thus any point(matrix) of the set of all nonsingular matrices, is an interior point of it. So, unless the metric is specified, the problem is meaningless!

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    @SomabhaMukherjee It is standard when a topology on a Euclidean space is not specified to give it the Euclidean topology. In this case, I think the question is clearly asking for a proof in the case that you use the topology on $\mathbb{R}^{n \times n}$2012-05-13