Prove that $|R-Z|=|R|$ where $R$ is reals and $Z$ is integers.
New approach Maybe the problem is reduced to finding a bijection between (0,1) and [0,1] if we find a bijection between these intervals, we can do the same thing for the other ones
What about this possible bijective function f(x)=1/(x-1)x
Thank you so much for your help, I have more problems like this to work on, so need more help.
$\endgroup$ 74 Answers
$\begingroup$Assume it is known that any open, nonempty subset of $\mathbb{R}$ is uncountable - this step is where your work may be. The rest is easy:
Let $S = \mathbb{R} \setminus \mathbb{Z}$.
First remark that $S \subset \mathbb{R}$, so we must have that $|S| \leq |\mathbb{R}|$.
The open interval $(0, 1)$ is a subset of $S \subset \mathbb{R}$, so $|(0, 1)| \leq |S| \leq |\mathbb{R}|$.
But $|\mathbb{R}| = |(0, 1)|$, and we're done.
NOTE: to see that $|\mathbb{R}| = |(0, 1)|$, we might construct a bijection $f:(0, 1) \longrightarrow (0, \infty)$ such that $f(x) = \frac{1}{1 - x} - 1$, and reason that this bijection can be easily extended to a bijection between $(0, 1)$ and $\mathbb{R}$.
$\endgroup$ 6 $\begingroup$We can construct a bijection between the sets using bijections from $\left(n,\,n+1\right)$ to $\left[n,\,n+1\right)$ for each $n\in\mathbb{Z}$, but we only need find such a bijection for $n=0$ because we can thereafter translate it, viz. $f\left( x+1\right) = f\left( x\right)+1$. So the exercise is verifying $\left(0,\,1\right)$ bijects with $\left[0,\,1\right)$. By Schröder-Bernstein, we only need verify each of these sets can be injected into a subset of the other. The identity function serves one way; the function $\left(x+1\right)/2$ serves the other way. (If you want an explicit choice for the function $f$, the theorem's proof provides a constructive procedure to obtain it.)
$\endgroup$ 7 $\begingroup$Here is one way to construct a bijection $\mathbb{R} \rightarrow \mathbb{R} - \mathbb{Z}$.
First, enumerate $\mathbb{Z} + \frac{1}{2}$ as: $a_1, a_2, a_3, \ldots$.
Now, define the bijection as follows:
Map $a_n$ to $a_{2n}$; in this way, all $a_n$ are bijected with $a_\text{even}$; but nothing has been sent to $a_\text{odd}$.
Now biject $\mathbb{Z}$ with $a_\text{odd}$.
For all other elements of $\mathbb{R}$, simply map them to themselves.
The result is a bijection between $\mathbb{R}$ and $\mathbb{R} - \mathbb{Z}$ as desired.
Thus, the two sets have the same cardinality. QED.
$\endgroup$ 6 $\begingroup$if you remove the countable set Z from uncountable set R, then R-Z still would be uncountable set which has the same cardinality as of the original uncountable set R
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