Why is the boundary of the rationals (Q) equal to R?

[filter54321]

I was reading a website that said the boundary of a set's boundary is equal to the first boundary. Visually, this makes sense for subsets of R1 and R2 because the first boundary will not have an interior (no ball about the points will fall into the boundary).

However, the reading went on to say that the boundary of the rationals Q is R. This seems wrong to me so I am questioning the entire site.

Wouldn't the boundary of Q be Q? A ball of positive radius about any point in Q would contain both points from Q as well as irrationals from P. ddQ = dQ = Q (where d is boundary operator). The theorem at top does appear to hold but the example is messed up...no?

 

[quasar987]

If A is a subset of R^n, then a boundary point of A is, by definition, a point x of R^n such that every open ball about x contains both points of A and of R^n\A.

So for instance, in the case of A=Q, yes, every point of Q is a boundary point, but also every point of R\Q because every irrational admits rationals arbitrarily close to it. So in the end, dQ=R.

 

[Tinyboss]

The rationals in the reals are good for all kinds of examples and counterexamples.

Here's a cool one: there is a proper open subset $$U\subset\mathbb{R}$$ with $$\mathbb{Q}\subset U$$. That is, U is open and contains all of Q, but not all of R.

 

[quasar987]

Isn't it obvious? Take U=R minus any irrational point.