How do you prove that [3,4)~[3,4)\{pi} ?

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March 22nd 2012, 10:47 AM
jfk

How do you prove that [3,4)~[3,4)\{pi} ?

Can anyone provide a proof, plz? I need to understand how the proof works, 10x in advance...(Happy)
March 22nd 2012, 12:05 PM
emakarov

Re: How do you prove that [3,4)~[3,4)\{pi} ?

Let $a = \pi-3$ and let $A = \left\{3+\frac{a}{2^n}\;\middle\vert\; n\in\mathbb{N}\right\}$ where $\mathbb{N}=\{0,1,2,\dots\}$ . In particular, $\pi\in A$ . Then we can construct $f:A\to A$ such that $\pi$ is not in the image of $f$ . Namely, define $f\left(3+\frac{a}{2^n}\right)=3+\frac{a}{2^{n+1}}$ . Then it is easy to extend $f$ to $[3, 4)$ so that $\pi$ is still not in the image of $f$ .
March 22nd 2012, 12:47 PM
jfk

Re: How do you prove that [3,4)~[3,4)\{pi} ?

thank you very much, can I ask you what is the procedure to deal with this kind of questions?
March 22nd 2012, 12:57 PM
Plato

Re: How do you prove that [3,4)~[3,4)\{pi} ?

Quote:

Originally Posted by jfk

can I ask you what is the procedure to deal with this kind of questions?

That is known as subset shift. You are proving an uncountable set equivalent to a subset of itself formed by removing one point.
March 22nd 2012, 01:48 PM
emakarov

Re: How do you prove that [3,4)~[3,4)\{pi} ?

See also Hilbert's paradox of the Grand Hotel.

In general, when we are dealing with infinite sets, the cardinality of many set operations depends only on the cardinality of the larger set. In particular, finite set don't matter compared to countably infinite sets, and those don't matter compared to continuum. So, if B is infinite and $|A|\le |B|$ (the cardinality of A is less than or equal to the cardinality of B), then $|A\cup B| = |B|$ and $|A\times B|=|B|$ .
March 22nd 2012, 02:33 PM
jfk

Re: How do you prove that [3,4)~[3,4)\{pi} ?

Thank you for the reference and the explanation, it was very helpful 😃

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