Bijection between (0,1) and [0,1) in R?

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Homework Help Overview

The discussion revolves around finding a bijection between the intervals (0,1) and [0,1) in the real numbers. Participants explore various approaches and considerations related to this mapping.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants suggest different functions and mappings, including a piecewise function and considerations for mapping rational and irrational numbers. Questions arise regarding the handling of zero and the direction of the bijection.

Discussion Status

There are multiple attempts to define the bijection, with some participants providing specific mappings and others questioning the implications of these mappings, particularly concerning the treatment of zero. The discussion is ongoing, with no clear consensus yet.

Contextual Notes

Participants note issues related to the definition of the function at zero and the nature of the rational numbers within the intervals. There is an acknowledgment of the countability of rational numbers and the need to address discontinuities in the proposed functions.

Colleen G
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Homework Statement


I need to find a bijection between (0,1) and [0,1) in R. It can go in either direction since it is a bijection.

Homework Equations


I can't think of any equations at all!

The Attempt at a Solution


Something like f(x) = 1/[(1/x)+1] for x in A
x for x not in A
where A={1/2, 1/3, 1/4, ...}

Having an issue with zero though.
 
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Colleen G said:

Homework Statement


I need to find a bijection between (0,1) and [0,1) in R. It can go in either direction since it is a bijection.

Homework Equations


I can't think of any equations at all!

The Attempt at a Solution


Something like f(x) = 1/[(1/x)+1] for x in A
x for x not in A
where A={1/2, 1/3, 1/4, ...}

Having an issue with zero though.
Not sure which direction you are trying to do. For ##[0,1)\rightarrow (0,1)## try starting with ##0\to 1/2##, ##1/2\to 1/3## etc.
 
The crucial point here is that the set of rational numbers, between 0 and 1, is countable. Write the rational numbers as "a_1, a_2, a_3, ..." and map a_1 to 0 and a_{n+1} to a_n. Map each irrational number to itself.
 
Colleen G said:
Having an issue with zero though.
Simplify ##\displaystyle\ \frac{1}{\displaystyle\frac{1}{x}+1}\ ## to get x out of the denominator.

Although f(0) is undefined, ##\displaystyle\ \lim_{x\to 0}\,f(x)=0\ ##, so f(x) has a removable discontinuity at x=0 .
 

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