In Axler's Linear Algebra Done Right, the following problem (1.B.6) was posed:
Let $\infty$ and $-\infty$ denote two distinct objects, neither of which is in $\mathbb{R}$. Define an addition and scalar multiplication on $\mathbb{R} \cup \{\infty\} \cup \{-\infty\}$ as you could guess from the notation. Specifically, the sum and product of two real numbers is as usual, and for $t \in \mathbb{R}$ define \begin{align*} & t \infty = \begin{cases} -\infty & \text{ if $t < 0$} \\ 0 & \text{ if $t = 0$} \\ \infty & \text{ if $t > 0$} \end{cases} \; \; \; \; \; t(-\infty) = \begin{cases} \infty & \text{ if $t < 0$} \\ 0 & \text{ if $t = 0$} \\ -\infty & \text{ if $t > 0$} \end{cases} \\ & t + \infty = \infty + t = \infty, \; \; \; \; \; t + (-\infty) = (-\infty) + t = -\infty, \\ & \infty + \infty = \infty, \; \; \; \; (-\infty) + (-\infty) = -\infty, \; \; \; \; \infty + (-\infty) = 0. \end{align*} Is $\mathbb{R} \cup \{\infty\} \cup \{-\infty\}$ a vector space over $\mathbb{R}$? Explain.
Now, it came out that this is not a vector space because it fails on the associative and distributive properties. And, looking into why it is that associativity fails, I wondered whether associativity is a property that stipulates the retention of previous operations in some basic sense. Here is what I mean.
Take $x$, $y$, and $z$ to be elements of this set where $x$ $\in$ $\mathbb{R}$, $y = ∞$ and $z = −∞$. The one grouping leaves $x$ as the result of the addition, and the other leaves $0$ as the result. The discrepancy, from what I can see, is due to the fact that addition with either $∞$ or $−∞$ (on one hand) and a real number (on the other) completely "loses the information" that a real number was involved in the operation. Perhaps this type of information loss is allowed only for the additive identity?
So, I understand that one way to frame associativity is that grouping should not matter, but in what ways is it also a stipulation that no information loss must occur? Are those the same? What is wrong with looking at it this way?
Here’s another reason why it’s not even a group:
We know that $ab=a$, implies $b=e_{G}$ in group $G$ (Cancellation law). Now as you mentioned $\infty +t = \infty$, which implies that $t=0$ for any $t\in \mathbb{R}$. Impossible! You’re absolutely right we lose some information in addition if we talk a a little bit friendly.