Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If you have an infinite set $X$, how do you prove that $|$ $X$ $\times$ $X$ $|$=$|$ $X$ $|$. I believe AC is needed but I don't quite see how to write the proof. Thanks.

share|improve this question
    
While this is a great question that everyone should see at some point of their mathematical career, I somewhat doubt this is close enough to research level to warrant a post here. As a hint, note two sets have the same cardinality if you can find a bijection between them. –  mathic Dec 11 '10 at 10:09
    
Please read the FAQ. The question is not appropriate for MO. Just two remarks: 1) For infinite cardinals $\kappa$ there is a bijection $\kappa \to \kappa \times \kappa$ which is given by the type of a well-ordering of the product. This works without AC. 2) The statement you cited is equivalent to AC in ZF. –  Martin Brandenburg Dec 11 '10 at 10:11
1  
Yes, fair enough. I am a maths research student (though not in set theory) and I just wanted some clarification on this point made in a paper I'm reading. Since it's not anywhere near the area I'm working on I didn't realise it was not a high enough level to post here. Guess I'll go somewhere else! Sorry! –  G-Unit Dec 11 '10 at 10:21
    
No problem. Check out math.stackexchange.com. –  Martin Brandenburg Dec 11 '10 at 11:16
add comment

closed as too localized by Martin Brandenburg, Chandan Singh Dalawat, Robin Chapman, Simon Thomas, Ryan Budney Dec 11 '10 at 14:33

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Browse other questions tagged or ask your own question.