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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.

Martin Sleziak
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TheMathNoob
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  • You seem to have misunderstood the question. $\mathbb{R}-\mathbb{Z}$ is the set of all real numbers that are not integers. – Anon Jun 03 '16 at 06:27
  • Ummm right, but I have to find a bijection between the sets. Do I have to use a cantor's argument? – TheMathNoob Jun 03 '16 at 06:30
  • 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 – TheMathNoob Jun 03 '16 at 06:38
  • Your function is a map from R to R. Not R to R-Z. It won't work. – fleablood Jun 03 '16 at 06:39
  • I have a 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     – TheMathNoob Jun 03 '16 at 06:40
  • Notice if x is an integer then f (x) not in R-Z. So this isn't a map to R-Z. – fleablood Jun 03 '16 at 06:43
  • Cantor-Bernstein theorem is probably the easiest way to show this, but if you want to find an explicit bijection, you can look here. (Or here if you want to follow @TheMathNoob's suggestion.) – Martin Sleziak Jun 03 '16 at 09:51

4 Answers4

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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 $\mathbb{R}$.

Wes
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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.)

J.G.
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  • I don't get it. I got to the point in which we have to find a bijection from (0,1) to [0,1]. Can you elaborate on Schröder-Bernstein theorem – TheMathNoob Jun 03 '16 at 06:46
  • what about this one Y=1/(x-1)x – TheMathNoob Jun 03 '16 at 07:00
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    Two sets and have the same cardinality if there exists a bijection between them. According to the Schröder-Bernstein theorem, this is equivalent to the existence of injective maps in both directions. Now the inclusion gives an injection from $\mathbb{R}-\mathbb{Z}$ to $\mathbb{R}$. Since $R$ is the union of distinct sets $[z,z+1)$ ($z\in\mathbb{Z}$), and $(z,z+1)\subset\mathbb{Z}-\mathbb{R}$ as well as $(z,z+1)\subset[z,z+1)$ it now suffices to construct an injection from $[z,z+1)$ to $(z,z+1)$ for each $z$ (this yields an injective function from $\mathbb{R}$ to $\mathbb{R}-\mathbb{Z}$) – lattice Jun 03 '16 at 07:12
  • Finding a bijection from [0,1) to (0,1) is going to be practically impossible. (Where will 0 map to, then where will the point that zero mapped to be mapped to?) But finding a bijection from [0,1) to [1/2,1) $\subset $ (0,1) is easy and mapping (0,1) to (0,1) $\subset $ [0,1)$ is easy. So we have card [0,1)=card [1/2,1)<=card (0,1)<=card [0,1). Note: that means card [0,1)=card (0,1). – fleablood Jun 03 '16 at 07:15
  • @fleablood You mustn't confuse bijections with bijections that have nice properties such as continuity. If you follow the S-B theorem's proof, you'll see we can construct a bijection (but don't expect it to be pretty). – J.G. Jun 03 '16 at 12:35
  • As mentioned in my answer: Articulating an explicit bijection is not difficult (even without thinking about the SB Theorem)... – Benjamin Dickman Jun 12 '16 at 03:32
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    @BenjaminDickman For this problem, that approach is a nice one! – J.G. Jun 12 '16 at 07:09
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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.

Benjamin Dickman
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  • If you'd rather work with $[0, 1]$ and $(0, 1)$ then they can be put in bijection as follows: Pick any countable subset of $[0, 1]$ that doesn't contain $0$ or $1$ and enumerate it as $a_n$. Send $a_n$ to $a_{n+2}$; in this way, $a_1, a_2, a_3, \ldots$ are bijected with $a_3, a_4, a_5, \ldots$. Now send $0$ to $a_1$ and $1$ to $a_2$. The result is a bijection between $[0, 1]$ and $(0, 1)$ as desired. QED. – Benjamin Dickman Jun 03 '16 at 08:06
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    Hmm... I guess the difficulty of finding the bijection wasn't as great as I presumed. – fleablood Jun 03 '16 at 15:21
  • @fleablood Subtle the first time you see it (I think the idea goes back to Dedekind... cf. "Dedekind infinite") but pretty clear in retrospect, I hope! – Benjamin Dickman Jun 03 '16 at 15:54
  • Reminds me more of Hilbert's "Infinite Hotel". Should have been obvious that removing a countable infinite from N can be countable infinite but somehow the Real "background" threw me. Kicking myself. You don't nee to shove the reals aside to make one or countable holes. Just the integers. – fleablood Jun 03 '16 at 16:10
  • @fleablood (You can "upvote" the answer if it was useful!) Yes, if you know of the Hilbert Hotel then this should make good sense. The original argument of this nature seems to be from Cantor in a letter to Dedekind; search for the term "Hilbert Hotel" in the following wonderful paper and you will see the geometric representation that precedes HH: Gouvêa, Fernando Q. "Was Cantor Surprised?." American Mathematical Monthly 118, no. 3 (2011): 198-209. – Benjamin Dickman Jun 03 '16 at 16:21
  • @fleablood Here is a link to Gouvêa's AMM article (no paywall)! – Benjamin Dickman Jun 03 '16 at 23:48
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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

Sunny
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  • Your comment is the least helpful comment that I have every seen. And his answer is not that wrong: One can show that $\mathbb{R}$ is uncountable, and clearly $\mathbb{R}-\mathbb{Z}$ (i.e. $\mathbb{R}\setminus\mathbb{Z}$) is at least countably infinite (since it contains $\mathbb{Z}+\frac{1}{2}$), thus according to the continuum hypothesis must have a cardinality equal to either $\mathbb{Z}$ or $\mathbb{R}$. But in the first case, $\mathbb{R}$ would be a union of two countable sets which cannot be true. The only problem is that one cannot prove the continuum hypothesis from what I've heard. – lattice Jun 03 '16 at 06:45
  • Why is that unhelpful. It's correct. Finding a bijection between R and R-Z is going to be hard as hell. – fleablood Jun 03 '16 at 06:48
  • You don't need to construct a bijection, according to Schröder-Bernstein theorem injective maps in both directions are sufficient, which is not that hard to construct (compare J.G.'s answer). – lattice Jun 03 '16 at 06:52
  • My point was that we aren't going to find any bijection so we will have to some how do a cardinality of subset arguments. This post answer is incomplete in that we don't know if two uncountable sets have the same uncountable cardinality but it does show they are both uncountable – fleablood Jun 03 '16 at 07:06
  • Can you explain why my function above is not a bijection – TheMathNoob Jun 03 '16 at 07:07
  • @fleablood Right, so this answer is basically wrong because it assumes the continuum hypothesis (which is not or cannot be proven). – lattice Jun 03 '16 at 07:17
  • Your function is not defined for $x\in{0,1}$. – lattice Jun 03 '16 at 07:18
  • question how do we show $|(0,1)|= |R|$ – TheMathNoob Jun 03 '16 at 07:19
  • Firstly convince yourself, that if you "stretch" any open interval infinitely long, the result will be $\mathbb{R}$. Now describe this as a (bijective) function (this is a good exercise, so I want you to figure it out yourself). Hint: It is probably the easiest way to map the middle of the interval to $0$. – lattice Jun 03 '16 at 07:22
  • Yes, I am the noob here. I will do it anyways =) – TheMathNoob Jun 03 '16 at 07:23
  • I couldn't get it, so we map the middle of the interval from 0? – TheMathNoob Jun 03 '16 at 07:42
  • Found it y=Tan(pi(x-1/2)) and 0<x<1 – TheMathNoob Jun 03 '16 at 08:07
  • @lattice which is a far far cry from being the "most unhelpful post ever seen". – fleablood Jun 03 '16 at 15:11
  • @TheMathNoob f(x) = x/|x + 1| maps f:R ->(-1,1) – fleablood Jun 03 '16 at 15:27
  • @fleablood I agree! I found the answer interesting, and I think I caused some misunderstanding: When I wrote "your comment is the least helpful comment that I have ever seen" that was directed to someone who commented this answer to be completely unhelpful, but that person deleted that comment it seems. Anyway, indeed $f(x)=\tan(\pi(x-\frac{1}{2}))$ provides a solution for this problem. I was thinking of $f(x)=\frac{x-1/2}{1/2-|x-1/2|}$. This function is quite similarly used for some issues I think (for instance to show that the sphere $S^n$ without a point is homeomorphic to $\mathbb{R}^n$) – lattice Jun 04 '16 at 09:01
  • @lattice my comments were all also directed to the same deleted comment. I upvoted this answer simply to undo the downvote. I hadn't realized that finding the actual bijection was ... not hard. But one doesn't have to actually find the precise bijections. R - Z is uncountable because remove at most countably many terms from a higher than Aleph cardinality the cardinality is unaffected.... or because a countable union of Cardinal sets is Cardinal ... or a variety of other common sense ideas first. This post isn't complete but it's far from wrong. – fleablood Jun 04 '16 at 14:58