I know that this isn't a normal constraint but let's say we are working with the standard Von Neumann Naturals $\mathbb{N}$ within ZFC and not some arbitrary system. So no results from $\mathbb{R},\mathbb{Z}, \mathbb{Q}$ are applicable. Now lets get on with the question. First, lets define our extended Naturals set.
$\overline{\mathbb{N}}=S(\mathbb{N})=\mathbb{N}\cup\{\mathbb{N}\}$ where we can use $\mathbb{N}=\infty$ within the set and while we are at it lets also define one other set.
$\mathbb{N}^{+}=\mathbb{N}\setminus\{0\}$
Now lets define addition, multiplication and order over the extended naturals as we'd expect:
For all n and m in the naturals,
$n+_{\overline{\mathbb{N}}} m:=n+m\,\,\mathrm{and}\\ n+_{\overline{\mathbb{N}}} \infty=\infty+_{\overline{\mathbb{N}}} n= \infty+_{\overline{\mathbb{N}}} \infty:=\infty$
$n\cdot_{\overline{\mathbb{N}}} m:=n\cdot m\,\,\mathrm{and}\\n \neq 0 \implies n \cdot_{\overline{\mathbb{N}}} \infty=\infty\cdot_{\overline{\mathbb{N}}}n=\infty\\ \infty\cdot_{\overline{\mathbb{N}}} \infty:= \infty$
$n >_{\overline{\mathbb{N}}} m \iff n>m\,\,\mathrm{and}\\ \infty >_{\overline{\mathbb{N}}} n$
$n \geqslant_{\overline{\mathbb{N}}} m \iff n >_{\overline{\mathbb{N}}} m \lor n=m$
Next let us define the summation and products:
$\bigg(\sum_{k=b}^{S(n)}a_k \bigg):=\bigg(\sum_{k=b}^n a_k\bigg)+_{\overline{\mathbb{N}}}\big(a_{n+1}\big)$ where b<n and $a_n$ is defined from b to $S(n)$
$\bigg(\prod_{k=b}^{S(n)}a_k\bigg):=\bigg(\prod_{k=b}^n a_k \bigg)\cdot_{\overline{\mathbb{N}}}\big(a_{n+1}\big)$ where b<n and $a_n$ is defined from b to $S(n)$
So here are my actual questions:
- With just these tools is it possible to define $\sum_{n=0}^{\infty} a_n$ and $\prod_{n=0}^{\infty} a_n$? or would we have to use something similar to limits, metrics or order to define it here?
- If it is defineable without those tools, is it possible to prove that $\sum_{n=0}^{\infty} a_n=\infty$ and $\prod_{n=0}^{\infty} a_n=\infty$ given that for all n, $a_n \in \mathbb{N}^{+}$?
My only idea is to somehow prove that for all naturals n, $a >_{\overline{\mathbb{N}}} n$ iff $a=\infty$ and somehow attempting to show that $\sum_{n=0}^{\infty} a_n$ and $\prod_{n=0}^{\infty} a_n$ would be greater than any natural and thus $\sum_{n=0}^{\infty} a_n=\prod_{n=0}^{\infty} a_n=\infty$
