set theory problem, open and close intervals equivalance

hello there,

i am trying to prove a set theory problem relating open and half open intervals. help needed.

1- [0,1] ~ (a,b)

2- [0,1] ~ [a,b)

3- [0,1] ~ (a,b]

where a,b belongs to R and ~ is equivalent sign.

for question 2, [0,1] ~ [a,b)

i have defined a function such that f(x) = (bx+a/2) / 2 for all x belongs to [0,1]. is that correct??

Re: set theory problem, open and close intervals equivalance

Two sets are "equivalent" if and only if there exist an invertible function that maps every point in the first set to every point in the other set. Your first problem requires that you find an invertible function that maps (0, 1) to (a, b). You want to use f(x)= (bx+ a/2)/2 which **is** invertible. But f(0)= a/4 and f(1)= b/2+ a/4. That does NOT map (0, 1) to (a, b).

But there is a more serious problem. To do this problem you should know that a **continuous** invertible function must map open sets to open sets and closed sets to closed sets. Since [0, 1] is closed but none of (a, b), [a, b), or (a, b] are closed, no continuous, and certainly no **linear** function will work. I would recommend something like: map every irrational number to itself and then use the fact that the set of rational numbers in both [0, 1], (a, b), [a, b), (a, b] are countable.

Re: set theory problem, open and close intervals equivalance

Quote:

Originally Posted by

**MathGeni04** where a,b belongs to R and ~ is equivalent sign.

The name of the sign is irrelevant; it's what it *denotes* that matters.

See also this thread.

Re: set theory problem, open and close intervals equivalance

Quote:

Originally Posted by

**HallsofIvy** Two sets are "equivalent" if and only if there exist an invertible function that maps every point in the first set to every point in the other set. Your first problem requires that you find an invertible function that maps (0, 1) to (a, b). You want to use f(x)= (bx+ a/2)/2 which **is** invertible. But f(0)= a/4 and f(1)= b/2+ a/4. That does NOT map (0, 1) to (a, b).

But there is a more serious problem. To do this problem you should know that a **continuous** invertible function must map open sets to open sets and closed sets to closed sets. Since [0, 1] is closed but none of (a, b), [a, b), or (a, b] are closed, no continuous, and certainly no **linear** function will work. I would recommend something like: map every irrational number to itself and then use the fact that the set of rational numbers in both [0, 1], (a, b), [a, b), (a, b] are countable.

would you mind giving any examples of f(x) which can be used in these problem. i am new to this stuff, dnt have much grip. i'll much obliged.

Re: set theory problem, open and close intervals equivalance

Quote:

Originally Posted by

**MathGeni04** would you mind giving any examples of f(x) which can be used in these problem.

Have you seen the tread I linked to? Do you think I kept in my bookmarks for 1.5 years in vain? :)

Re: set theory problem, open and close intervals equivalance

Quote:

Originally Posted by

**emakarov** Have you seen the tread I linked to? Do you think I kept in my bookmarks for 1.5 years in vain? :)

i hv seen it. thnx :p

Re: set theory problem, open and close intervals equivalance

Quote:

Originally Posted by

**MathGeni04** i am trying to prove a set theory problem relating open and half open intervals. help needed.

1- [0,1] ~ (a,b)

2- [0,1] ~ [a,b)

3- [0,1] ~ (a,b]

I see that you have solved this question. Here is some advice for future questions like this.

You can show that $\displaystyle [0,1]\sim (0,1]\sim [0,1)\sim (0,1)$.

We do that by point shifting of a countable set. Define a sequence $\displaystyle f_0=0,~f_n=n^{-1},~n\in\mathbb{Z}^+$.

Now if $\displaystyle x\in [0,1]$ and if $\displaystyle (\exists k)[x=f_k]$ then define $\displaystyle \pi:~x\mapsto f_{k+1}$ or else $\displaystyle \pi:~x\mapsto x$.

That shows that $\displaystyle [0,1]\overset{\pi}\sim(0.1]$. Similar functions made with minor adjustments in $\displaystyle \pi$.

Note that if $\displaystyle a<b$ then $\displaystyle \sigma: x\mapsto a+(b-a)x$ makes $\displaystyle [0,1]\overset{\sigma}\sim [a,b]$.