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

• Mar 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)
• Mar 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 $f$.
• Mar 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?
• Mar 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.
• Mar 22nd 2012, 01:48 PM
emakarov
Re: How do you prove that [3,4)~[3,4)\{pi} ?
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 $|A\times B|=|B|$.