What does it mean when rational numbers are dense?

Then it must be defined differently: it means that every open set in the plane intersects the set of all rational points. No matter how small you make an open disk in the plane, it cannot avoid containing some rational points; so the set of all rational points is dense in the plane.

Why is the measure of the rationals 0?

The inner measure is always less than or equal to the outer measure, so it must also be 0. Therefore, although the set of rational numbers is infinite, their measure is 0.

Is rationals nowhere dense?

No they are not: Wikipedia and Wolfram MathWorld indicate that a “nowhere dense set” is one whose closure has empty interior. Since ˉQ=R in this case, the rationals are not nowhere dense.

Why Q is dense in R?

A subset S ⊂ X S \subset X S⊂X is called dense in X if any real number can be arbitrarily well-approximated by elements of S. For example, the rational numbers Q are dense in R, since every real number has rational numbers that are arbitrarily close to it.

Are the rationals Jordan measurable?

For example, the set of rational numbers contained in the interval [0,1] is then not Jordan measurable, as its boundary is [0,1] which is not of Jordan measure zero. Intuitively however, the set of rational numbers is a “small” set, as it is countable, and it should have “size” zero.

Are rational numbers measurable?

But then, this set of rationals in the unit interval is the countable union of point sets so it MUST BE measurable.

Is the set 1 n dense in R?

Every other answer tells you that in the standard topology of R, N is not dense.

Is Empty set dense in itself?

The empty set is nowhere dense. In a discrete space, the empty set is the only such subset. In a T1 space, any singleton set that is not an isolated point is nowhere dense. The boundary of every open set and of every closed set is nowhere dense.

Are integers dense?

The integers, for example, are not dense in the reals because one can find two reals with no integers between them.