bloodhound Posted November 18, 2005 Share Posted November 18, 2005 Can someone give me an example of a bijection between [math]\mathbb{R}[/math] and [math][0,1][/math] Link to comment Share on other sites More sharing options...

shmoe Posted November 18, 2005 Share Posted November 18, 2005 Do you need to explicitly exhibit a bijection or is it enough to prove one exists? Link to comment Share on other sites More sharing options...

bloodhound Posted November 18, 2005 Author Share Posted November 18, 2005 I'll take any for now. Link to comment Share on other sites More sharing options...

matt grime Posted November 18, 2005 Share 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) Link to comment Share on other sites More sharing options...

bloodhound Posted November 18, 2005 Author Share 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. Link to comment Share on other sites More sharing options...

matt grime Posted November 19, 2005 Share 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. Link to comment Share on other sites More sharing options...

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