Constructing a bijection between $S=[1,2]$ and $T=[1,2)$

Question

I am having a bit of trouble with constructing the biijection here. I know that the bijection has to be a piecewise function with one component being $f(x) = x$ when $x$ does not equal $1/n$. But the other component is a bit more tricky, none pass the horizontal line test. Could the other component be $2/n+2$ or how would I go about figuring that component out?

Your question title obscures your actual question—$S$ and $T$ are out of context. Are you trying to find a bijection between $[1,2]$ and $[1,2)$? — Matthew Leingang, Sep 26 '16 at 12:47
Yes, that is exactly what I am trying to do. Sorry, english isnt my native language. — Aggrawal Puja, Sep 26 '16 at 12:50
No problem. I took the liberty of editing the question title to declare the variables. — Matthew Leingang, Sep 26 '16 at 12:58

5xum · Answer 1 · 2016-09-26T12:54:27.557

2

You can construct a bijection from $S$ to $T$ by saying that

$$2\mapsto 1+\frac12\\ 1+\frac12\mapsto 1+\frac13\\ 1+\frac13\mapsto 1+\frac14\\ \vdots$$

and that $x$ maps to $x$ if $x\notin \{1+\frac1n, n\in\mathbb N\}$

edited Sep 26 '16 at 12:54

answered Sep 26 '16 at 12:52

5xum

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1/2 is not in $S$ – Nex Sep 26 '16 at 12:53
Perhaps $1+\frac{1}{n} \mapsto 1+\frac{1}{n+1}$? – Nex Sep 26 '16 at 12:54
@Nex Fixed it. Thanks. The idea is the same, anyway. – 5xum Sep 26 '16 at 12:54
1

Map every irrational number to itself. The rational numbers are countable so we can list the rational numbers in [1, 2], starting with 2. Then map the nth rational on that list to the n+1 rational. – user247327 Sep 26 '16 at 13:00
@user247327 Sure, that's another way of doing it. – 5xum Sep 26 '16 at 13:00
I like to call this the Infinite Hotel bijection; room $n$ corresponds to $1/n$ ($1+1/n$ here) and the bijection leaves only room 1 vacant. – Parcly Taxel Sep 26 '16 at 13:03
So would this be a piecewise function if im writing in function notation? Where one function is f(x) = x and another function is 1+1/n where x=1/n? – Aggrawal Puja Sep 26 '16 at 13:08
@AggrawalPuja Almost. You're sloppy, though. $x$ is never $\frac1n$. – 5xum Sep 26 '16 at 13:08

score 0 · Answer 2 · answered Sep 26 '16 at 13:30

The usual Hilbert hotel argument works this way: split $[1,2]$ as $[1,2]\cap\mathbb{Q}$ and its complement. Do nothing on the complement, just leave the elements where they are. Consider that $[1,2]\cap\mathbb{Q}$ is a countable set, and enumerate its elements as $2=q_0, 1=q_1, q_2,q_3,\ldots$. On $[1,2]\cap\mathbb{Q}$, consider the map $\varphi$ that sends $q_k$ into $q_{k+1}$. This is an invertible map between $[1,2]\cap\mathbb{Q}$ and $[1,2\color{red}{)}\cap\mathbb{Q}$.

Constructing a bijection between $S=[1,2]$ and $T=[1,2)$

2 Answers2