3

I want to find a bijection between $[1,2)$ and $(1,2)$ and prove it.

My attempt:

$[1,2) = \{x \in \mathbb R | 1 \leq x <2\}$

$(1,2) = \{x \in \mathbb R | 1 < x < 2\}$

$f(x) = x$ if $x \ne 1\frac{1}{n}$ for $n = 1,2,3,...$ and $f(x) = 1\frac{1}{x+1}$, if $x =1\frac{1}{n}$ for $n = 1,2,3,...$

Proof - Injective - Prove $x_1 = x_2$ for $f(x) = x$ \begin{align*} f(x_1) = f(x_2) &\implies x_1 = x_2.\\ \end{align*}

Proof - Injective - Prove $x_1 = x_2$ for $f(x) = 1\frac{1}{x+1}$ \begin{align*} f(x_1) = f(x_2) &\implies 1\frac{1}{x_1+1} = 1\frac{1}{x_2+1}\\ &\implies \frac{1}{x_1+1} = \frac{1}{x_2+1}\\ &\implies x_2+1 = x_1+1\\ &\implies x_2 = x_1.\\ \end{align*}

Therefore $f$ is injective.

Any help will be appreciated.

Thanks.

MJD
  • 65,394
  • 39
  • 298
  • 580
Moz
  • 103
  • https://math.stackexchange.com/questions/213391/how-to-construct-a-bijection-from-0-1-to-0-1 This could help – thewatcher Sep 20 '21 at 07:47
  • You don’t need an explicit function definition btw. You could always just shift everything and make 1 unique. I.e. map every point except one to 1.0 + unique non terminating decimal expansion and make 1 map to 1.1 – Jake Freeman Sep 20 '21 at 07:47
  • To see that the question found by José Carlos Santos indeed provides the answer to your question, note that there is an obvious bijection between $(0,1)$ and $(1,2)$ and similarly for $(0,1]$ and $[1,2)$ (can you write them?). Compose these bijections and the bijection in the linked answer to get your answer. – Ennar Sep 20 '21 at 16:56

2 Answers2

1

This is the right idea, but the details are not quite right. For example, you say

$$f(x) = 1\frac{1}{x+1}\text{ if }x =1\frac{1}{n}\text{ for }n = 1,2,3,...$$

I think you meant to say $f(x) = 1\frac1{n+1}$ there on the left. I will suppose that you meant that. if you really meant $f(x) = 1\frac1{x+1}$, that is an additional problem.

Supposing that you meant $f(x)=1\frac1{n+1}$, then when $n=1$ you've said that $f(2)=1\frac12=\frac32$. But you shouldn't be defining $f(2)$ at all, because $2$ is not an element of $[1,2)$. At the other end of the interval, $1$ does not have the form $1\frac1n$, so you've defined $f(1)=1$. But $1$ is not in the desired range $(1,2)$.

Also it's not enough to prove that $f$ is injective; you must also prove that $f$ is surjective: for each $y$ in $(1,2)$, there is some $x$ in $[1,2)$ for which $f(x)=y$.

Check the details a little more carefully.

Also you should know that nobody writes $1\frac1n$ to mean $1+\frac1n$.

MJD
  • 65,394
  • 39
  • 298
  • 580
-2

In your question, you noted that card((1,2)) <= card([1,2)) Let $f: [1,2)\to(1,2)$

Fix $x \in (1,2)$. Let $\{A_i\}_{1}^{\infty}$ represent the non-terminating decimal expansion right of the decimal place (this is unique to each number). Let $f(x) := 1 + \sum_{i=1}^{\infty}A_i*10^{-i-1}$. All values of f are of the form 1.0... Therefore, let $f(1):=1.1$. As each decimal expansion is unique the resulting function f is injective. Therefore, there exists a bijection between (1,2) and [1,2)

Jake Freeman
  • 351