Ok, so if $x$ is a real radius that simply means $x$ is a real number, correct?:Consider the definition of a real number: it is a cut in the real number line. For any real radius around the origin, the real number line leaves that radius.

$$x \in (0,\infty) \implies \exists y \in \mathbb{R} \text{ such that } y > x$$

Ok, so:Cuts in the part beyond the radius are real numbers in the neighborhood of infinity.

$$y > x \implies y \in \text{ 'the neighborhood of infinity'}$$

$$x = 1 \implies (1,\infty) = \text{ 'the neighborhood of infinity'}$$

This seems nonstandard. I believe the standard would be that $\infty - b = \infty \neq y \in (1,\infty)$.I have defined these types of real numbers to have the form "infinity minus b."

The real numbers are already defined. One can define a number in any fashion they see fit, but that doesn't make it a real number. Have you proven that $\infty - b \neq \infty$ and instead that $\infty - b \in \mathbb{R}$ or something? What am I missing...?Since it is known that one can make definitions, I have proven that a real number can have this form.