Equinumerous

Dec 2009
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Can anyone help me solve this?
Show that the interval (0,1] is equinumerous to the interval (0,1) by giving an example of a bijection from (0,1] to (0,1).
Thanks.
 

undefined

MHF Hall of Honor
Mar 2010
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Chicago
Can anyone help me solve this?
Show that the interval (0,1] is equinumerous to the interval (0,1) by giving an example of a bijection from (0,1] to (0,1).
Thanks.
Let \(\displaystyle A=(0,1]\subseteq\mathbb{R}\) and \(\displaystyle B=(0,1)\subseteq\mathbb{R}\) and define \(\displaystyle f:A\to B\) as follows:

\(\displaystyle f(x)=\begin{cases}\frac{1}{2^{n+1}}\quad \exists\ n \in \mathbb{Z}: x=\frac{1}{2^n}\\ x\quad\ \ \not \exists\ n \in \mathbb{Z}: x=\frac{1}{2^n} \end{cases}\)

This is very similar to the question in this thread.
 
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Drexel28

MHF Hall of Honor
Nov 2009
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Berkeley, California
Can anyone help me solve this?
Show that the interval (0,1] is equinumerous to the interval (0,1) by giving an example of a bijection from (0,1] to (0,1).
Thanks.
I don't understand the point of having to exhibit the bijection. Clearly one exists since the countable union of countable sets is countable, and using this (decompose your set into the disjoint union of a countable set and an uncountable set) one can prove that if \(\displaystyle E\) is uncountable then \(\displaystyle E\cup N\simeq E\) for any \(\displaystyle N\simeq\mathbb{N}\)
 
Dec 2009
8
0
Thank you undefined, it rings the bell. Then I find there are many ways to show such bijection.

Let \(\displaystyle A=(0,1]\subseteq\mathbb{R}\) and \(\displaystyle B=(0,1)\subseteq\mathbb{R}\) and define \(\displaystyle f:A\to B\) as follows:

\(\displaystyle f(x)=\begin{cases}\frac{1}{2^{n+1}}\quad \exists\ n \in \mathbb{Z}: x=\frac{1}{2^n}\\ x\quad\ \ \not \exists\ n \in \mathbb{Z}: x=\frac{1}{2^n} \end{cases}\)

This is very similar to the question in this thread.