# Prove (0,1)=[0,1]

Ok, 0 is not even in f.

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)=?
Is that the right idea or not?
I just am unsure of what x value to plug into g(x)=0

Ok, I'm gonna lay out what I've done thus far.

proving f is a 1-1 function from (0,1) into (0,1].
Let a=1/m b=1/n
f(a)=1/(m-1) f(b)=1/(n-1)
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/(n-1)
n=1/(y-1)
y-1=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)/b-1]=1/1/b=b

y=x
x=y
f inverse=x
f inverse (b)=b=a
f(a)=a=b

Finding a 1-1 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)=1-x

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)=?

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

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!

Also f(x)=1-x
and g:[0,1)-->[0,1]

Ok it's all figured out. They key was letting [0,1) was f(x)=x in (0,1) and f(0)=1. Then I just used transitivity.