Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
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
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.