Inverse Completion Less Zero of Integral Domain is Closed

Theorem
Let $$\left({D, +, \circ}\right)$$ be an integral domain whose zero is $$0_D$$ and whose unity is $$1_D$$.

Let $$\left({K, \circ}\right)$$ be the inverse completion of $$\left({D, \circ}\right)$$.

Then $$\left({K^*, \circ}\right)$$ is closed, where $$K^* = K - \left\{{0_K}\right\}$$.

Proof
Let $$\left({K, \circ}\right)$$ be the inverse completion of $$\left({D, \circ}\right)$$.

We define $$\left({K, \circ}\right)$$ of $$\left({D, \circ}\right)$$ by Inverse Completion of Integral Domain.

The structure of $$\left({K, \circ}\right)$$ is such that element of $$\left({K, \circ}\right)$$ is of the form $$x \circ y^{-1}$$, where $$x \in D$$ and $$y \in D^*$$.

From Zero of Inverse Completion of Integral Domain, $$0_K$$ is all elements of $$D \times D^*$$ of the form $$\frac {0_D} x$$.

Therefore the elements of $$K^*$$ are those of the form $$\frac x \circ y^{-1}$$ where $$x, y \in D^*$$.

By Product of Division Products, $$\frac a b \circ \frac c d = \frac {a \circ c} {b \circ d}$$.

As $$\left({D, +, \circ}\right)$$ is an integral domain, none of its non-zero elements are zero divisors.

Therefore $$\forall x, y \in D^*: x \circ y \ne 0_D$$.

So $$\forall \frac a b, \frac c d \in K^*: \frac {a \circ c} {b \circ d} \in K^*$$.

Hence the result.