Timeline for Bijection from the plane to itself that sends circles to squares
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 11, 2017 at 14:53 | comment | added | amakelov | further, for example, we have a bijection $\mathbb{R}\to\mathbb{R}$ given by $x\mapsto 4x$ for which it is true that the image of any integer is an even integer, but yet it is not true that the inverse image of any even integer is an integer | |
| Apr 11, 2017 at 14:38 | comment | added | Timothy Chow | @MattiVirkkunen : A bijection has to be invertible, but just because a bijection maps every circle to a square doesn't mean that its inverse maps every square to a circle. "Bijection" just means that points are put into one-to-one correspondence, and it doesn't automatically mean that the set of circles is put into bijection with the set of squares. The bijection could inject the set of circles into the set of squares. | |
| Apr 11, 2017 at 14:20 | comment | added | Matti Virkkunen | @WillBrian: Doesn't a bijection have to be invertible by definition though? If A maps into B one way, B has to map back to A the other. So showing that a mapping doesn't exist one way should be enough to prove that there can be no such bijection. | |
| Apr 10, 2017 at 21:10 | comment | added | amakelov | sorry - you're right! | |
| Apr 10, 2017 at 19:32 | comment | added | Will Brian | The question was whether there is a bijection sending circles to squares. As far as I can tell, your argument only shows that there is no bijection sending squares to circles. (It is not obvious to me that if a bijection makes every circle a square, then its inverse must make every square a circle.) Am I missing something? | |
| Apr 10, 2017 at 19:29 | history | answered | amakelov | CC BY-SA 3.0 |