# Boundary of the rationals

1. Aug 5, 2011

### alexfloo

I've read in several places that the boundary of the rational numbers is the empty set. I feel I must be misinterpreting the definition of a boundary, because this doesn't seem right to me.

My understanding of the boundary of a set S is that it is the set of all elements which can be approached from both the inside and the outside. That is, the set of all r such that r is the limit of a sequence in S and also the limit of a sequence outside of S.

We know of course that every real number is the limit of a sequence of rational numbers. We know also that every real number r is the limit of the constant sequence (r). So shouldn't the boundary of the rationals be the set of all irrational numbers?

2. Aug 5, 2011

### HallsofIvy

What you say is correct up to the last sentence:
Therefore the boundary of the rational numbers, as a set of real numbers with the usual topology, is the set of all real numbers, both rational and irrational.

The boundary of the rational numbers, as a subset of the rational numbers with the usual topology, is empty. Perhaps that is what you saw? In any topology, the entire space has empty boundary.

3. Aug 5, 2011

### alexfloo

Okay, I think that makes perfect sense, but just to clarify:

The discrepancy is that the the rationals can have multiple algebraic structures, and therefore multiple boundaries, depending on whether we consider them in isolation, or as a subset of the reals. (Or, presumably, as a subset of some other completion that I know less about, like the p-adic numbers).

4. Aug 5, 2011

### HallsofIvy

Yes, many of the topological properties of sets depend upon whether the set is a subset of some larger topology. Those that do not (compactness for example) are called "intrinsic".

5. Aug 5, 2011

### Bacle

But I don't know if it would make sense to talk about, e.g., the boundary of the rationals a stand-alone space; I assume you always talk about the boundary of a subset A embedded in a space X; usually A is a subspace of X, I think.

6. Aug 6, 2011

### SteveL27

HallsofIvy already pointed out that the boundary of $\mathbb{Q}$ considered as a subset of $\mathbb{R}$ is all of $\mathbb{R}$. I just wanted to expand on that a little.

If $X$ is a topological space and $A \subset X$, the boundary of $A$ is the set of points with this property: each neighborhood of the point intersects both $A$ and $X\setminus{A}$.

$[X\setminus{A}$ is the set difference: the set of elements of $X$ that are not elements of $A]$.

This definition of boundary is equivalent to the one you gave. (Needs proof, of course).

Now we can see that if $A = X$ then $X\setminus{A}$ is empty, so there can't be any elements in the boundary. That's why the boundary of $\mathbb{Q}$ in itself is empty.

The more interesting case is when $A$ is a proper subset of $X$; for example $\mathbb{Q}$ and $\mathbb{R}$. If $x$ is a rational number, then any neighborhood about $x$ contains both rationals and irrationals. So all of $\mathbb{Q}$ is in the boundary. But if $x$ is an irrational number, it also has the property that each of its neighborhoods contain both rationals an irrationals. So the irrationals are in the boundary too.

So, the boundary of $\mathbb{Q}$ is all of $\mathbb{R}$.

Hope this helps.