Axler Exercise 1.B.6 I am trying to solve this following exercise in Axler.

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.

Here is what I have been able to put together so far.
Solution. Define $M = \mathbb{R} \cup \{\infty\} \cup \{-\infty\}$. Iterating through the axioms, we run into the following issues:
(1) $(-\infty) + \infty$ is not defined, even though $\infty + (-\infty) = 0$. Hence, not only is addition on $M$ not defined for every element of $M$, but the fact that $-\infty + \infty \neq 0$ implies that $M$ doesn't satisfy commutativity of addition.
(2) Multiplication fails since $\infty \times \infty$, $-\infty \times \infty$, $\infty \times -\infty$, and $-\infty \times -\infty$ are not defined.
(3) Commutativity of multiplication fails because $\infty \times t$ is not defined for any $t \in \mathbb{R}$.
It's possible that I am missing some things. But how do these look? Am I correct that these are problems with the described space?
 A: *

*is indeed a problem, but it's easily patched by also defining $(-\infty) + \infty = 0$ (and I imagine this is just an oversight by the author)


*Be careful. We aren't disproving that $M$ has multiplication, only that it isn't a vector space over $\mathbb{R}$. it doesn't make sense to say "oh but $\infty \times \infty$ isn't defined!" because $\infty$ is an element of the (supposed) vector space, and we aren't required to have multiplication of vectors.


*Again, here $\infty$ is a "vector", and $t$ is a scalar, so we don't need $\infty \times t$ to be defined, only $t\infty$.
In linear algebra, there are two types of "multiplication": multiplication of elements of our field (in this case $\mathbb{R}$), and scalar multiplication, i.e. multiplication of a scalar and a vector (written $\lambda v$, scalar first then vector). You seem to be confusing the two.

As for actual problems with $M$, consider
$$ (-\infty + \infty) + \infty = 0 + \infty = \infty$$
while
$$ -\infty + (\infty + \infty) = -\infty + \infty = 0$$
So $M$ fails to be associative.
A: Regarding points $(2)$ and $(3)$: We are considering $\mathbb R\cup\{\infty\}\cup\{-\infty\}$ as a vector space over plain old $\mathbb R$. So we don't need both products of the form $t\times \infty$ and $\infty \times t$ to be defined. It is enough that all elements of $\mathbb R$ have a well-defined multiplication on either just the left or just the right of all elements in $\mathbb R\cup\{\infty\}\cup\{-\infty\}$. 
Also, since the field of scalars is just $\mathbb R$, quantities like $\infty\times-\infty$, etc. don't need to be defined (in general, there is no requirement that you should be able to multiply vectors with each other - only scalars with scalars and scalars with vectors).  
Regarding $(1)$: While Axler may not have explicitly stated what $(-\infty)+\infty$ was, Axler's rules imply that $-1\times -\infty=\infty$
 and $-1\times \infty=-\infty$, so, multiplying both sides of $\infty+(-\infty)=0$ by $-1$ gives us $(-\infty)+\infty=0$. Of course, here we have assumed distributivity. You could raise an objection to this, as Axler does not seem to have made explicit that multiplication must be distributive, but if you assume it isn't distributive, the problem becomes trivial. Note also that we are guaranteed $a(b+c)=ab+ac$ in the 'vector space' that Axler has defined, as long as $(b,c)\neq (-\infty,\infty)$ (this is easy to check). So really, once $(-\infty)+\infty=0$ has been defined, distributivity will also hold in general. So you can look at this as a failure of commutativity of distributivity. But we should reasonably expect any vector space to have these, so this answer is not very satisfying.
If you've had any experience with naive notions of infinity, you know that it is easy to run into inconsistencies, and even attempts to patch up those problems make it so that mathematical structures with infinities behaving strangely and lack many 'nice' properties we find in more familiar structures. This kind of experience should lead you to expect, intuitively, that the structure Axler has defined just cannot be a vector space. 
The spirit of Axler's question, then, is to show that, given the information Axler has given you, regardless of how you 'complete' the rules of multiplication and addition for the structure, it still cannot be a vector space.  
That is to say, there's a deeper reason to why the structure cannot be a vector space other than that Axler has not explicitly said its multiplication is distributive or its addition commutative. 
Look at the last two equations Axler has outlined: $(-\infty)+(-\infty)=-\infty$ and $\infty+(-\infty)=0$ - is there any possible way you could define a vector space over $\mathbb R\cup\{\infty\}\cup\{-\infty\}$ with $\mathbb R$ so that those two rules (along with the others Axler listed) and all the vector space axioms are satisfied?
