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I have always wondered, is negative infinity less than positive infinity? Can I compare them?

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You can declare $ - \infty < x < \infty$ for all real numbers $x$. This gives a total order on $[-\infty,\infty]$. –  Mark Oct 5 '11 at 10:13
    
$+\infty>0 \Rightarrow \ln(+\infty)>\ln(0) \Rightarrow +\infty>-\infty$ .. end of proof :) –  pedja Oct 5 '11 at 10:42
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up vote 4 down vote accepted

It depends what you mean by "infinity".

In standard analysis, $\infty$ appears mainly as a placeholder for "perform that otherwise limited computation as if the limit $+\infty$ was a large positive number, but don't stop at any point." For instance, $$ \sum_{i=1}^\infty \frac1{i^2} $$ means (naïvely) "sum the numbers $\frac11,\frac14,\frac19,\ldots$ and just don't stop". In much the same way, we can stretch such a computation simply in both directions: $$ \sum_{k=-\infty}^\infty 2^{-k^2} = \ldots + 2^{-4} + 2^{-1} + 2^0 + 2^{-1} + 2^{-4} + 2^{-9} + \ldots $$ Both of these sums are well-defined, while $\sum_{k=\infty}^\infty$ wouldn't make any sense.
When just using $\infty$ as such a placeholder, it's not really meaningful to compare anything to it, still it is common to write $$ \sum_{i=1}^\infty \frac1{i^2} < \infty $$ but this just means that the sum is well behaved, does not diverge to infinity like, for instance, $$ \sum_{i=1}^\infty 5 = 5+5+\ldots \not< \infty. $$ But this is not a comparison of mathematical objects.


On the other hand, it is possible to have $\infty$ as an actual mathematical object on its own, an element of some set. Such a set is called a compactification of the real numbers. It can be defined either with one single infinity representing both $-\infty$ and $+\infty$ (as one equivalence class). In this case the result is homeomorphic to the unit circle, which does not have an ordering at all. Or with distinct elements $-\infty$ and $+\infty$. Then you do, in fact, have $-\infty<+\infty$ if you declare it to be so. That's what Mark Schwarzmann said in his comment.

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