7

The problem is stated as follows:

Show that there is a one-to-one correspondence between the points of the closed interval $[0,1]$ and the points of the open interval $(0,1)$. Give an explicit description of such a correspondence.

Now, I think I can prove the first part of the problem by demonstrating the following:

Define $f: (0,1) \to \mathbb{R}$ as follows.

For $n \in \mathbb{N}$, $n \geq 2$, $\space{ }f(\frac{1}{n}) = \frac{1}{n-1}$ and for all other $x \in (0,1)$, $\space{}f(x) = x$

  1. Prove that $f$ is a $1-1$ function from $(0,1)$ onto $(0,1]$

  2. Slightly modify the above function to prove that $[0,1)$ is equivalent to $[0,1]$

  3. Prove that $[0,1)$ is equivalent to $(0,1]$

Since the "equivalent to" relation is both symmetric and transitive, it should follow that $[0,1]$ is equivalent to $(0,1)$. Hence, there does exist a one-to-one correspondence between $[0,1]$ and $(0,1)$.

I have no trouble with the above. My problem is in "finding an explicit description of such a correspondence." Can I modify the above function, or will that not suffice?

David Smith
  • 530
  • 1
  • 4
  • 10

3 Answers3

6

Steps 2 and 3 are not necessary. The function $g:(0,1] \to [0,1]$ defined by $g(1) = 0$ and $g(x) = f(x)$ if $x \neq 1$ is a bijection. This shows that $(0,1]$ is equivalent to $[0,1]$ and, by transitivity, that $(0,1)$ is equivalent to $[0,1]$. Furthermore, the function $g \circ f$ is a one-to-one correspondence between $(0,1)$ and $[0,1]$ that you can describe explicitly.

Math536
  • 1,898
2

I would just let $a_n=2^{-n}$ for $n=0,1,2,\ldots$., and $I_n=(a_{n+1},a_{n})$. Then $$(0,1)=\bigcup_{n=0}^{\infty}I_n \cup \bigcup_{n=1}^{\infty}\{a_{n}\}$$ and $$ [0,1]=\bigcup_{n=0}^{\infty}I_n \cup \bigcup_{n=0}^{\infty}\{a_{n}\}\cup \{0\}.$$ The function that maps $0$ to $a_1$, $a_n$ to $a_{n+2}$ for all $n$, and is the identity on each interval $I_n$, is a bijective mapping from $[0,1]$ to $(0,1)$. That is, $$ f(x)=\begin{cases} 1/2 & \text{if} \; x=0 \\ x/4 & \text{if} \; x=2^{-n} \; \text{for} \; n\in\mathbb{N}\\ x & \text{otherwise}. \end{cases} $$ Its inverse is simply $$ f^{-1}(x)=\begin{cases} 0 & \text{if} \; x=1/2 \\ 4x & \text{if} \; x=2^{-n} \; \text{for} \; n\ge 2\\ x & \text{otherwise}. \end{cases} $$

mjqxxxx
  • 41,358
-1

It's far easier than step 1.

Step 2 : same definition than step 1, but with: $\forall x∈[0,1), \space f(x)=x$ instead of $\forall x∈(0,1), \space f(x)=x$.

Step 3 : $\forall x, \space f(x)=1-x.$

daOnlyBG
  • 2,711
boris
  • 1