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I have proved this but my teacher wants me to put more but I have no idea what to add. He says he wants a proof that they are explicitly in fact a bijection.

For the first one this is what I did $g(x)=x$ if $x$ is not contained in $A$ Otherwise $g(x) = f(x)$ Then, $g$ is a required bijection from $(0,1]$ to $(0,1)$

For the second one I said $$A=\{0,1,\frac{1}{2},\frac{1}{3},\frac{1}{4}, \dots, \frac{1}{n-2}\}$$ $$B=\{\frac{1}{2},\frac{1}{3},\frac{1}{4}, \dots, \frac{1}{n}\}$$ $$A \to B$$ such that $$f\left(\frac{1}{n-2}\right) \to \frac{1}{n}$$

dantopa
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Ayoshna
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    What is $f(x)$? What is $A$? – Thomas Apr 19 '14 at 19:11
  • I don't know... I don't remember what I was doing. I did this problems months ago... I think I'm suppose to let g(x) be equal to 1/n+1 is x=1/n and it equals x if x does not equal 1/2 for any n – Ayoshna Apr 19 '14 at 19:17
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    It is hard to see what you are doing when you don't remember it yourself... so indeed it can be a good thing to write things down carefully. – Thomas Apr 19 '14 at 19:19
  • That was exactly what I wrote down... sorry if its not making sense – Ayoshna Apr 19 '14 at 19:20

3 Answers3

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We try to construct a mapping $f \colon (0,1] \to (0,1)$ as follows:

Let $A \colon= \{ 1, 1/2, 1/3, \ldots \}$. Then let $$ f(x) \colon= \frac{x}{x+1}$$ if $x \in A$,

while $$ f(x) \colon= x$$ if $x \not \in A$.

In other words, let $$f(x) \colon= \frac{1}{n+1} $$ if $x = 1/n$ for some natural number $n$,

while $$f(x) \colon= x$$ for all other $x \in (0,1]$.

Now one can easily see that $f$ thus defined is a bijection.

Now we try to define a bijection $g \colon [0,1] \to (0,1)$ as follows:

Let $$g(0) \colon = \frac{1}{2},$$ $$ g(x) \colon= \frac{1}{n+2} $$ if $x = 1/n$ for some natural number $n$, and $$g(x) \colon= x$$ if $x \in (0,1)$ such that $x \ne 1/n$ for any natural number $n$.

The map $g$ is clearly a bijection of $[0,1]$ with $(0,1)$.

Saaqib Mahmood
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Ok, here is one way to do it. It might not be as explicit as what you teacher requires, but I think that it is pretty explicit and it looks like what you were trying to do... I will just do the first case and let you think more about the second case.

Note that $$ (0,1) \cap \mathbb{Q} \quad \text{and}\quad (0,1] \cap \mathbb{Q} $$ are countable. Write $$ (0,1) \cap \mathbb{Q} = \{x_1, x_2, x_3 , \dots \}= \{\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{3}{4}. \dots\}\\ (0,1] \cap \mathbb{Q} = \{y_1, y_2, y_3 , \dots \}= \{1, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{3}{4}. \dots\} $$ That is, for example, $x_3 = \frac{2}{3}$, $y_3 = \frac{1}{3}$.

Then $$ g(y) = \begin{cases} y & \text{if } y\text{ is irrational} \\ x_i &\text{if }y = y_i\text{ is rational}. \end{cases} $$ will be a bijection that you want.

Thomas
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Here is my take.

1.

Let $A=\left\{\dfrac{1}{2^n} \mid n\in\Bbb N \text{ and } n>0 \right\}$

We define $f:(0,1] \to (0,1)$ by $f(1)=\dfrac{1}{2}$ and $f(x)=\dfrac{x}{2}$ for all $x\in A$ and $f(x)=x$ for all $x\in (0,1)\setminus A$.

It's easy to verify that $f$ is bijective.

2.

Let $A=\left\{\dfrac{1}{2^n} \mid n\in\Bbb N \text{ and } n>0 \right\}$

We define $f:[0,1] \to (0,1)$ by $f(0)=\dfrac{1}{2}$ and $f(1)=\dfrac{1}{2^2}$ and $f(x)=\dfrac{x}{2^2}$ for all $x\in A$ and $f(x)=x$ for all $x\in (0,1)\setminus A$.

It's easy to verify that $f$ is bijective.

Akira
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