
#37
Dec1108, 09:33 AM

P: 179

kathrynag already stated a bijection between (0, 1) and (0, 1] in the very first post of this thread. (It doesn't take too much effort from that to conclude there is a bijection between (0, 1) and [0, 1].) Her problem is showing that it is onetoone and onto.
I suggest practicing showing that simpler functions are bijections, such as f : R > R defined by f(x) = 5  x^3, so that you understand the form of the proof. 



#38
Dec1108, 01:00 PM

P: 607

f(b)5b^3 5x^3=5b^3 a^3=b^3 a=b Thus 11 y=5x^3 y+5=x^3 x=(y+5)^1/3 x includes all real numbers. Thus onto. Then how do i work around proving [0,1) is equivalent to (0,1]? Like I thought about using f(x)=1x because f(0)=1 and f(1)=0 but then that would be a mpping and that's not quite right. 



#39
Dec1108, 11:06 PM

P: 607

Ok could I still use C=[0,1]
Then A=Cx element of C B=C but B=Cnull set null set is an element of C so A=C some elemnt of C=B? 



#40
Dec1208, 10:19 PM

P: 607

Ok I know [0,1) is equivalent to (0,1] by f(x)=1x
Now I'm having trouble finding a similar function for (0,1) equivalent to [0,1] 



#41
Dec1308, 12:25 PM

P: 607

Ok, I found something else. 2 sets A and B are equivalent iff there is a 11 function from A onto B.




#42
Dec1308, 12:34 PM

P: 420

After reading this topic, I am completely speechless.




#43
Dec1308, 12:49 PM

P: 179

You already have f(x) = 1  x, a bijection between [0, 1) and (0, 1], which is easier to show. If you also have a bijection between [0, 1) and [0, 1], then that implies (0, 1) and [0, 1] are cardinally equivalent. If you want a single bijection between (0, 1) and [0, 1], think about how you can "combine" the 3 functions. 



#44
Dec1308, 01:41 PM

P: 607

Ok, so (0,1) and (0,1] For f(x)=x, we get 0 and 1 For f(1/n)=1/(n1), we get {1/1/2,1/1,1/4....} Thus, (0,1] [0, 1) and [0, 1] For this I use {0}U(0,1) because I already know (0,1) is a bijection to (0,1]. Thus we have a bijection to [0,1]. Ok, for (0,1) equivalent to [0,1]. Let f(x)=x for all elemnt not of the form 1/n Then we get the points (0,1) Thne for all other points let f(x)={0}U 1/(n1) 



#45
Dec1308, 01:59 PM

P: 179

Also, what about all other points not in {0, 1, 1/2, 1/3, 1/4, ...}? 



#46
Dec1308, 02:34 PM

P: 607

All other points not in {0,1,1/2,1/3,1/4....} are in f(x)=x Yeah, i guess setting g(0)=0 and g(x)=f(x) would work. So then let f:(0,1)>(0,1] and let g:(0,1)>[0,1] Let f(x)=g(x) for all x in (0,1) and let g(1)=0? 



#47
Dec1308, 03:01 PM

P: 179





#48
Dec1308, 03:18 PM

P: 607

So then let f:(0,1)>(0,1] and let g:(0,1)>[0,1] Let f(x)=g(x) for all x in (0,1) Is this part correct? I know there needs to be some number to equal 0 right? Let g(x)=f(x)=? 



#49
Dec1308, 03:39 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

If If f:(0,1)>(0,1] then there is NO "x" left over to give f(x)= 0. That's the whole point.




#50
Dec1308, 03:44 PM

P: 607

Ok I understand that.




#51
Dec1308, 06:12 PM

P: 607

I just am unsure of what x value to plug into g(x)=0 



#52
Dec1308, 07:12 PM

P: 607

Ok, I'm gonna lay out what I've done thus far.
proving f is a 11 function from (0,1) into (0,1]. Let a=1/m b=1/n f(a)=1/(m1) f(b)=1/(n1) f(a)=f(b) m=n 1/m=1/n a=b f(a)=a f(b)=b f(a)=f(b) a=b proving f is a function from (0,1) onto (0,1]: To prove f:A>B is onto, let B be arbitrary and show there exists a in A such that f(a)=b. I need to show that im f=B. y=1/(n1) n=1/(y1) y1=1/n f inverse=1/n+1 f inverse =(1+n)/n f inverse of b=(1+b)/b=a f(a)=1/[(1+b)/b1]=1/1/b=b y=x x=y f inverse=x f inverse (b)=b=a f(a)=a=b Finding a 11 function from [0,1) onto [0,1] [0, 1) and [0, 1] given a bijection f:(0,1)>(0,1] , define g:[0,1)>[0,1] by g(0) =0, and g(x) = f(x) for x in (0,1) Proving [0,1) is equivalent to (0,1]: Use the function f(x)=1x Proving (0,1) is equivalent to [0,1]: So then let f:(0,1)>(0,1] and let g:(0,1)>[0,1] Let f(x)=g(x) for all x in (0,1) Is this part correct? I know there needs to be some number to equal 0 right? Let g(x)=f(x)=? 



#53
Dec1408, 10:40 AM

P: 607

Ok, do I use a composition?
Let f:(0,1)>(0,1] Let g:=[0,1)>[0,1] Let g(x)=1x Then h(x)=g(f(x)) 



#54
Dec1408, 11:52 AM

P: 607

Ok then I would also need to define g(0) as 0 right because we get:
h(x)=g([0,1]). No never mind because I should be doing h(x)=f(g(x) =f([0,1] f(0)=1 and f(1)=0 Please let me know if this is right! 


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