# Open condition for groups

## Main Question or Discussion Point

What does it mean for a condition to be "open"? E.g. it is said that det(A)≠0 is an open condition for a matrix group.

Furthermore, this implies that GL(n) has the same dimensions as the group of all nxn matrices as, and I quote, "the subgroup of matrices with det(A)=0 is a subset of measure zero".

1. Could you explain why this is true?
2. Could you shed light on what kind of measure we are talking about here (Lebesgue, Dirac, or else)?

Thank you.

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Office_Shredder
Staff Emeritus
Gold Member
mxn matrices are trivially isomorphic to [itex] \mathbb{R}^{m+n}[/tex] and have the Lebesgue measure induced from this. det(A) is a polynomial, so all they are using is that the set defined by p(x) = 0 for some polynomial p is a measure zero subset of Euclidean space.

I don't really understand the explanation that Wikipedia gives of a Lebesgue measure-zero set. I understand intuitively that subsets consisting of single points are of measure zero, but the proof evades me. =(

any intuitive explanations, or a short proof perhaps?

jbunniii
Homework Helper
Gold Member
Intuitively, a set has measure zero if it can be covered by a collection of open "intervals" whose total "size" is arbitrarily small. In ##n## dimensions, "interval" means ##n##-dimensional box, and "size" means the ##n##-dimensional volume of that box, i.e. the product of its ##n## side lengths. We specify open boxes simply to avoid degenerate boxes where one or more sides have length zero.

If we have a set ##A## consisting of one point ##x##, then surely it can be covered by a single open box centered at ##x##, with all of its sides having length ##\epsilon##. Thus the box has volume ##\epsilon^n##. We can do this for arbitrarily small ##\epsilon##, so ##A## has measure zero.

The same argument works if ##A## consists of a finite number of points, and it can be modified to work if ##A## is countably infinite.

Of course, there are uncountably many matrices with determinant zero, so a different argument is needed, but the intuitive idea is still the same.

Thank you, very clear. However, do we need to establish a comparison between the subset of matrices with determinant zero and that (complementary) of matrices with determinant not equal zero, in order to say that the first has, comparatively, measure-zero? Or is the measure absolutely true?

jbunniii