I want to find an explicit bijection between $\Bbb R$ and $\Bbb R$\ $\Bbb N$. However, similar problems I have found solutions to relied on establishing equal cardinalities by means other than bijections. I don't have much in the way of a start, and I would appreciate any hints.
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Do you know of a bijection between $[0,1)$ and $(0,1)$? Can you see then how to use this to construct a bijection between $[0,2)$ and $(0,1)\cup (1,2)$? Can you see then how to use this to construct a bijection between $[0,n)$ and $\bigcup\limits_{i=0}^n(i,i+1)$? – JMoravitz Dec 13 '16 at 21:13
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HINT: It suffices to find a bijection between $[0,1)$ and $(0,1)$ and repeat it on each interval $[n,n+1)$ with $n\in\Bbb Z$. One way is to map $0$ to $\frac12$, $\frac12$ to $\frac13$, $\frac13$ to $\frac14$, and so on, leaving everything else in place.
Brian M. Scott
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3Just to emphasise that this is a very useful trick for finding a bijection between infinite $A$ and $A \cup {x}$: select a countable sequence from $A$ and then "put $x$ at the start of the sequence". – Patrick Stevens Dec 13 '16 at 21:16
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It's enough to find a bijection between $[0,1]$ and $[0,1)$, and iterate it. Choose to send each integer into the unit interval beneath it. This has almost surely been done on this site before, so give me a moment and I will find a link to that bijection.
Here is a thread with some examples of how such bijections could be structured. Adapt them to your situation as necessary.
A. Thomas Yerger
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Hint: Find a bijection between $\mathbb Z$ and $\mathbb Z \backslash \mathbb N$.
ThePortakal
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The Cantor-Schroeder-Bernstein theorem is constructive, and it's easy to inject each set into the other. For $\mathbb{R}$ into $\mathbb{R} \setminus \mathbb{N}$, just use your favourite bijection of $\mathbb{R}$ with $(0,1)$.
Patrick Stevens
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Hint: Fill in the gaps in $\mathbb R\setminus\mathbb N$ by shifting a sequence for each $n\in\mathbb N$.
Let $(a_{n,m})_{n,m\in\mathbb N}$ be distinct real numbers such that $a_{n,0}=n$ for all $n$. Let $a_{n,m}\mapsto a_{n,m-1}$ for all $m>0$ and fix all other points; then we get a bijection from $\mathbb R\setminus\mathbb N$ to $\mathbb R$.
It becomes more visible if we choose $a_{n,m}\to\infty$ for fixed $n$; e.g. $a_{n,m}=n+m\sqrt2$: the gaps at the natural numbers are filled by the numbers at a distance $\sqrt2$ to the right, whose places are filled by ... etc
Bart Michels
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