[Math] Proving intervals are equinumerous

elementary-set-theoryproof-writing

a.) Show that (0, 1] is equinermous to the interval (0, 1) by giving an example of a bijection from (0, 1] to (0, 1).

My attempt: 

Define f: (0, 1] -> (0, 1)
    by f(x) { x, x != 1/2^n
        { 1/2^(n+1), x = 1/2^n for n = 0, 1...
Let S = { 1/2n | n = 0, 1 ... } = {1, 1/2, 1/4...}
Range(f) = (0, 1) 
        ie 1 --> 1/2
            x => x

One can easily see range(f) = (0, 1)

Is f 1-1?
    x, x' element of (0, 1]
    if x =! x' => f(x) != f(x')

    Case 1: 
        Suppose both x, x' are element of S...
        **I DUNNO WHAT TO SAY REALLY.**
    Case 2:
        x' element s, x not element of S...
        **I DUNNO WHAT TO SAY REALLY.**

    Case 3:
        x, x' not element of S
        Then f(x) = x != x' = f(x')

b.) Show [0, 1] equinumerous to [0, 1).

My attempt: 

    Define f: [0, 1] -> [0, 1)
    by f(x) { x if 0 <= x < 1, and x != 1/2^n, for all n in N
        { 1/2^(n+1), x = 1/2^n for n all n in N
    Let S = { 1/2n | n = 0, 1 ... } = {1, 1/2, 1/4...}
    Is f 1-1?
        x, x' element of [0, 1]
        if x =! x' => f(x) != f(x')
        Case 1: 
            Suppose both x, x' are element of S
            **I DUNNO WHAT TO SAY REALLY.**
        Case 2:
            x' element s, x not element of S
            **I DUNNO WHAT TO SAY REALLY.**
        Case 3:
            x, x' not element of S
            Then f(x) = x != x' = f(x')

c.) Explain how it follows from (a) and (b) that all four inteverals [0, 1], [0, 1), (0, 1] and (0,1) have the same number of elements.

    You can find a function that is a bijection between all of these intervals. // I don't really know what to say here? Should I just point out it's possible to find a bijection between each of the intervals, or would I need to do four more proofs?

d.) For good measure, show directly that the interval [0, 1] is equinumerous to (0, 1) by giving an example of a bijection from [0,1] to (0,1).

    Define f: [0, 1] -> (0, 1)
    by f(x) { x if 0 < x < 1, and x != 1/2^n, for all n in N
        { 1/2^(n+1), x = 1/2^n for n all n in N
    Let S = { 1/2n | n = 0, 1 ... } = {1, 1/2, 1/4...}
    Is f 1-1?
        x, x' element of [0, 1]
        if x =! x' => f(x) != f(x')
        Case 1: 
            Suppose both x, x' are element of S
            **I DUNNO WHAT TO SAY REALLY.**
        Case 2:
            x' element s, x not element of S
            **I DUNNO WHAT TO SAY REALLY.**
        Case 3:
            x, x' not element of S
            Then f(x) = x != x' = f(x')

I think I have the functions correct, because I copied them out of the book/my notes. However I have no idea how to derive these functions. I understand the general idea behind it, finding a function to map one to the other, but unless it's a really easy function I don't know how I'm supposed to think of these on my own.

Best Answer

On the question of how to think of a bijection $f:(0,1] \to (0,1)$ for part (a): If you defined $f$ so that $f(x)=x$ for all $x \in (0,1)$, your only problem would be where to map $1$. Well, it has to be mapped somewhere, say at $a_1=f(1)\in(0,1)$. Then everything would be ok except that two points are mapped to $a_1$: $1$ and $a_1$ itself. You have to modify $f$ a bit so that $a_1$ gets mapped somewhere else, say to $a_2=f(a_1) \in (0,1)\setminus\{a_1\}$. But now there are two points being mapped to $a_2$. Again, you have to modify $f$ a bit so that $f(a_2)=a_3 \in (0,1)\setminus\{a_1,a_2\}$. And so on... So you could select any sequence $1=a_0, a_1, a_2, \dots$ of pairwise distinct numbers in $(0,1]$ and let $$f(x)=\begin{cases} a_{i+1} & \text{, if } x = a_i \text{ for some } i \in \mathbb{N}\\ x & \text{, otherwise} \end{cases}$$

About part (d): You don't seem to have defined f(0) in your solution. See if the idea I described above can be used to find a bijection between $[0,1]$ and $(0,1)$.

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