Can $\pi$ be defined using Dedekind cuts? I have read that Dedekind cuts allow you to define the real numbers from the rationals. For example, $\sqrt{2}$ can be defined in the following way:


*

*Partition the rational numbers into two sets $A$ and $B$, such that all of the elements of $A$ are less than the elements of $B$

*$A=\{a \in \mathbb{Q}:x<0 \text{ or } x^2 < 2$}

*$B=\{b \in \mathbb{Q}: x > 0  \text{ and } x^2 \geq 2$}

*Because $B$ has no lower bound, there is a 'gap' in the number line

*We define $\sqrt{2}$ to fill this gap: $\sqrt{2}$ is the unique number such that $x^2=2, x>0$. $\sqrt{2}$ sits on the partition line that we previously drew (hence why we say $\sqrt{2}$ is that cut)


Hopefully, I understand Dedekind cuts well enough to ask this question. It seems to me that because $\pi$ is transcendental, it cannot fill a gap in the same way that $\sqrt{2}$ does. There is no polynomial equation that $\pi$ helps solve. By contrast, $\sqrt{2}$ solves the equation $x^2=2, x \geq 0$. Nevertheless, $\pi$ is a real number, and Dedekind cuts define the real numbers. So, can $\pi$ be defined using Dedekind cuts, or is more work needed? Moreover, does the usual geometric definiton of $\pi$ have a well-defined mathematical meaning if you have not constructed the real numbers from the rationals?
 A: We know that
$$\pi=4\int_0^1\frac{dx}{1+x^2}.$$
A lower Riemann sum of this integral is
$$L_n=\frac4n\sum_{k=1}^{n}\frac1{1+k^2/n^2}.$$
An upper Riemann sum of this integral is
$$U_n=\frac4n\sum_{k=0}^{n-1}\frac1{1+k^2/n^2}.$$
We can define $A=\{a\in\Bbb Q:a<L_n\text{ for some }n\in\Bbb N\}$
and $B=\{b\in\Bbb Q:b>U_n\text{ for some }n\in\Bbb N\}$.
A: Any sequence $c_n$ that converges to $\pi$ can be used to define such a cut, with
$$
A=\{a\in\mathbb Q\mid a\lt c_n\text{ infinitely often}\}\hphantom{\;.}
$$
and
$$
B=\{b\in\mathbb Q\mid b\gt c_n\text{ infinitely often}\}\;.
$$
A: There are many ways to do this, but my preference is to use Euler's solution to the Basel problem (https://en.wikipedia.org/wiki/Basel_problem).  Here's how I would define the lower set of the Dedekind cut:
$$\left\{ a \in \Bbb Q | (a \lt 0) \lor \exists n \in \Bbb N : a^2 < \sum_{i=1}^n\frac{6}{i^2} \right\}$$
The upper set can be defined this way:
$$\left\{ a \in \Bbb Q | (a > 0) \land \forall n \in \Bbb N : a^2 > \sum_{i=1}^n\frac{6}{i^2} \right\}$$
