Closed set of rational numbers

[Andy_ToK]

Hi,
here is the question, if A is a closed set that contains every rational number r: [0,1], show that [0,1] is a subset of A.

But, how could A be closed? If A is closed, R^n-A is open, so any point in R^n-A would have a open sphere around it and this open sphere wouldn't intersect A. apparently, this is not true. eg. sqrt(0.5) has no open sphere around it that is disjoint from A.

 

[Hurkyl]

[0, 1] is closed set containing every rational number between 0 and 1, is it not? Ponder that.

 

[Andy_ToK]

[0,1] is but A isn't, i think. because A doesn't contain those irrational numbers between 0 and 1.

 

[Hurkyl]

Why not? The problem you stated doesn't assert otherwise.

Incidentally, A is not uniquely specified -- there are lots of sets that have the property of both being closed and of containing every rational number in [0, 1], and the hypotheses is merely that A is one such set.

 

[HallsofIvy]

I think your title shows a misunderstanding. You titled this "closed set of rational numbers", which implies it contains only rational numbers, but the question is about a closed set that contains rational numbers- it doesn't say only rational numbers and clearly that cannot be true. If A is to contain [0,1] then clearly it contains irrational numbers as well. The point of the exercise is to show that any closed set that contains all rational numbers in [0,1] must also contain all irrational numbers in [0,1].

 

[arildno]

Note that in order for A to be closed (by premise!), A must include all accumulation points for sequences in A.

In particular, it means that A must contain all accumulation points for all sequences whose terms are rational numbers in the unit interval.

What you then need to show is that any irrational number within the unit interval is an accumulation point for at least one such sequence of rational numbers in the unit interval.

Remember that the rationals are dense in the reals..

 

[Andy_ToK]

Thank you all.