# Why the set of all boundary points of irrational numbers are real numbers?

Hello nice mathematicians, thanks for reading my question, I have a question. Why the set of all boundary points of the irrational is the set of real numbers?

# Definitions:

## Boundary of a Set

Let $A$ $\subset$ $\Bbb R$ be a set of real numbers and $x$ $\in$ $\Bbb R$. We say that $x$ is a boundary point of $A$ if every neighborhood of $x$ contains at lest one point of $A$ and at least one point of $A^C$. The set of all boundary points of $A$ is called the boundary of $A$, and is denoted $A^b$.

## $\epsilon$-Neighborhood

Let $x$ $\in$ $\Bbb R$ and $\epsilon$>$0$. The interval $(x-\epsilon,x+\epsilon)$ will be called $\epsilon$-Neighborhood of $N_{\epsilon}(x)$ is the set of all points that are within a distance of $\epsilon$ from $x$.

Thank you!

Quote of the day:

"Therefore, O students, study mathematics and do not build without foundations".

Leonardo di ser Piero da Vinci
1452-1519

• There are two parts here: 1. the irrationals are dense; 2. the irrationals have empty interior. To prove the latter it is sufficient to show that the rationals are also dense. The choice of a convenient proof for these two statements depends a bit on your choice of definition of the reals. (For example, if the reals are by definition the Cauchy completion of $\mathbb{Q}$, then density of the rationals is immediate.) – Ian Nov 10 '16 at 23:04
• Useful fact: $\partial A$ is the set of points $x\in \mathbb R$ such that $x$ is the limit of a sequence in $A$ as well as the limit of a sequence in $\mathbb R \setminus A$ – zhw. Nov 14 '16 at 0:43
• Thank you, i'll study your proof. – Oromion Nov 14 '16 at 0:49