bloodhound Posted November 18, 2005 Posted November 18, 2005 Can someone give me an example of a bijection between [math]\mathbb{R}[/math] and [math][0,1][/math]
shmoe Posted November 18, 2005 Posted November 18, 2005 Do you need to explicitly exhibit a bijection or is it enough to prove one exists?
matt grime Posted November 18, 2005 Posted November 18, 2005 Evidently no continuous function will do as any continuous image of a compact set is compact, and I think that is your problem with trying to figure this out. Existence proof: consider the maps i:[0,1] to R the inclusion and j:R to [0,1] which is the composition of the inverse tan function, which maps R bijectively with (-pi/2,pi/2) followed by adding pi/2 to map it to (0,pi) bijectively, followed by dividing by pi. j is an injection from R to [0,1] thus we have injections between the two sets and hence there exists some bijection by cantor-schroeder bernstein. writing one directly. notice that we can easily get a bijection between (0,1) and R using the tan trick. so we just need one from [0,1] to (0,1) and that is the standard trick: map 0 to 1/3, 1/3 to 1/5, 1/5 to 1/7 etc ..... map 1 to 1/2, 1/2 to 1/4, 1/4 to 1/6...etc leave x fixed if x has not been mentioned this is a bijeciton from [0,1] to (0,1)
bloodhound Posted November 18, 2005 Author Posted November 18, 2005 Evidently no continuous function will do as any continuous image of a compact set is compact' date=' and I think that is your problem with trying to figure this out. Existence proof: consider the maps i:[0,1'] to R the inclusion and j:R to [0,1] which is the composition of the inverse tan function, which maps R bijectively with (-pi/2,pi/2) followed by adding pi/2 to map it to (0,pi) bijectively, followed by dividing by pi. j is an injection from R to [0,1] thus we have injections between the two sets and hence there exists some bijection by cantor-schroeder bernstein. writing one directly. notice that we can easily get a bijection between (0,1) and R using the tan trick. so we just need one from [0,1] to (0,1) and that is the standard trick: map 0 to 1/3, 1/3 to 1/5, 1/5 to 1/7 etc ..... map 1 to 1/2, 1/2 to 1/4, 1/4 to 1/6...etc leave x fixed if x has not been mentioned this is a bijeciton from [0,1] to (0,1) Thanks a lot! I must warn you that I will probably swamp this board with questions on Algebra and Topological spaces in the coming days. Meanwhile, I'll go read up on the cantor schroeder bernstein theorem. I just hope it's not too complicated.
matt grime Posted November 19, 2005 Posted November 19, 2005 cantor-scroeder bernstein, or whatever the correct spelling is, is an obviously true theorem, though that is different from a trivially true theorem. it states that if we partially order sets by X <= Y if X injects to Y then X <=Y and Y<=X implies X=Y, which we all agree is 'obviously' true. the proof is not hard to understand but it is very clever. however, it is easy to write down a bijection directly, as i did, as long as you remember the tricks and that you've got to stop thinking of 'nice' functions.
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