Let $\infty$ and $- \infty$ denote objects neither of which in $\mathbb R$. The sum and product of two numbers is as usual and for $t \in \mathbb R$ define

$$t\infty = \begin{cases} -\infty, & \text{if $t < 0$} \\ 0, & \text{if $t = 0$} \\ \infty & \text{if $t > 0$} \end{cases}$$

$ t + \infty = \infty + t = \infty$ and $t + (-\infty) = (-\infty) + t = (-\infty)$

$\infty + \infty = \infty$ and $(-\infty) + (-\infty) = (-\infty)$ and $\infty + (-\infty) = 0$.

$(-\infty) + (-\infty) = -\infty$ implies $-\infty = 0$ which doesn't make sense so that the operations on the given vector spaces are contradictory.

Is is correct but missing something else? Is it wrong?

Here's a capture for context

  • $\begingroup$ That's the way it's defined in my textbook. Everything in the quotes is given. $\endgroup$ – vasya Jan 7 '16 at 20:04
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    $\begingroup$ $\infty=3\infty-2\infty=\infty-\infty=0$, On the other hand, with different definitions, $\mathbb{R}\cup \infty$ can be a vector space, of course. $\endgroup$ – Dietrich Burde Jan 7 '16 at 20:09
  • $\begingroup$ Unless you make the extended real line a field. Is $\infty$ + (-$\infty$) well-defined? Usually it's not a vector space. $\endgroup$ – ZTransformer Jan 7 '16 at 20:11
  • $\begingroup$ I think the problem might be the rule $\infty + (-\infty) = 0$. It will cause all kinds of trouble in expressions that involve $\infty$ or $-\infty$. Maybe it's better to consider the expression $\infty + (-\infty)$ to be indeterminate, notwithstanding what your book says. Does the book make use of the idea that this expression equals zero? $\endgroup$ – David K Jan 7 '16 at 20:11
  • $\begingroup$ @vasya It's given in your textbook in what context? Is it asking you the question in the title? Does it suggest that it is a vector space? $\endgroup$ – rschwieb Jan 7 '16 at 20:12

Your first line of equalities says it all: if you want to enforce a vector space structure on your extended real line but maintain the intuition of infinity, $t+\infty=\infty$ for any real $t$ implies $t=0$ (in the sense of being the additive identity), which certainly isn't true. Moreover $t\infty=\infty$ for all $t$ also implies $\infty=0$.

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    $\begingroup$ Is $t \neq 0$ because otherwise it would mean $\mathbb R = \{0\}$? $\endgroup$ – vasya Jan 7 '16 at 20:11

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