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

Consider the space of all nxn matrices with real entries with the standard metric, i.e.,view the matrix as an element of $R^{n^2} $and use the usual Euclidean metric on $R^{n^2} $. I need to prove that the subset of all invertible matrices is open. Please any idea?

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marked as duplicate by Zev Chonoles Feb 18 '12 at 6:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Another related question: math.stackexchange.com/questions/18964/… –  Jonas Meyer Feb 18 '12 at 8:54

1 Answer 1

up vote 5 down vote accepted

$\bf Hint:$ $A$ is invertible iff the determinant is different from zero.

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