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This came up today when people showed that there is no linear transformation $\mathbb{R}^4\to \mathbb{R}^3$.

However, we know that these sets have the same cardinality. I was under the impression that if two sets have the same cardinality then there exists a bijection between them. Is this true? Or is it just that any two sets which have a bijection between them have the same cardinality.

Edit: the question I linked to is asking specifically about a linear transformation. My question still holds for arbitrary maps.

pancini
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4 Answers4

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"Same cardinality" is defined as meaning there is a bijection.

In your vector space example, you were requiring the bijection to be linear. If there is a linear bijection, the dimension is the same. There is a bijection between $\mathbb R^4$ and $\mathbb R^3$, but no such bijection is linear, or even continuous. (Space-filling curves, which are continuous functions from a space of lower dimension to a space of higher dimension, are not bijections since they are in no instance one-to-one.) If there is a bijection, then the cardinality is the same. And conversely.

Michael Hardy
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  • So, it is true that if cardinality of two sets are same then there must be a bijection between them . – A learner Apr 03 '21 at 15:58
  • In the above question, the given sets will have same cardinality if they are treated as $\mathbb{Q}$- vector spaces. Am I right? – A learner Apr 03 '21 at 16:02
  • Please respond! – A learner Apr 04 '21 at 10:10
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    @Alearner : If you mean $\mathbb R^n$ as a $\mathbb Q$-vector space, then it is an infinite-dimensional space, but the underlying set, and therefore the cardinality, is still the same. $${}$$ At this point I am wondering whether you meant they will have the same dimension, rather than the same cardinality, as vector spaces over $\mathbb Q. \qquad$ – Michael Hardy Apr 05 '21 at 00:06
  • I am taking mainly about cardinality,,you say same cardinality of sets implies bijection between them. But only as $\mathbb{Q}$- vector spaces there exists a linear bijection,, whether as $\mathbb{R}$- vector spaces, not possible. Right?! – A learner Apr 05 '21 at 13:20
  • @Alearner : Certainly there is no linear bijection between $\mathbb R^n$ and $\mathbb R^m$ as vector spaces over $\mathbb R$ if $m\ne n.$ As vector spaces over $\mathbb Q$ I think they are isomorphic, so there is a linear bijection between them. – Michael Hardy Apr 05 '21 at 17:03
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I asked myself something similar: We have two sets $A$ and $B$ and we have two injections $f,g$ i.e. $$ f:A\rightarrow B $$ and $$ g:B\rightarrow A $$ is this enough to conclude that both sets have the same cardinality? In fact I found this theorem (Cantor-Bernstein-Schröder) very illuminating and the answer is then, yes, it is enough.

If we have the two injections $f,g$ we can find a bijection $h$ between both sets which therefore means they have the same cardinality.

So if you find two injections in your example, then you got your answer. Unfortunately it is quite hard to find an injection $\mathbb{R}^4\to \mathbb{R}^3$. The example given below does not work.

[EDIT as pointed out in the comments,$g$ is actually not an injection, it is only an injection from $S^1\times S^1\to \mathbb{R}^3 $ check here, page 2 bottom (Wayback Machine] $$ g:\mathbb{R}^4\to \mathbb{R}^3, g(x_1,x_2,x_3,x_4)=((x_1(2 + x_3),x_2(2 +x_3), x_4) $$ and $$ t:\mathbb{R}^3\to \mathbb{R}^4,t(x_1,x_2,x_3)=(x_1,x_2,x_3,0) $$

Martin Sleziak
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user190080
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  • How did you conclude that $g:\mathbb R^4\to\mathbb R^3$ is injective? ${}\qquad{}$ – Michael Hardy Jul 13 '15 at 21:02
  • Right. Also if there is a function A onto B and B onto A – pancini Jul 13 '15 at 21:07
  • @MichaelHardy while I was googling around to look for nice examples myself I found these lecture notes (second page bottom). I didn't check it myself...but it would definitely be a good idea to do so. I don't find it very easy to come up with an injection... – user190080 Jul 13 '15 at 21:14
  • @ElliotG interesting, sounds just right, is this an easy equivalence to the stated theorem? – user190080 Jul 13 '15 at 21:19
  • Not sure about a rigorous proof, but for finite sets you could state that the number of elements is the same – pancini Jul 13 '15 at 21:24
  • @MichaelHardy since you pointed it out: is there any canonical way to construct such embeddings? – user190080 Jul 13 '15 at 21:27
  • @ElliotG well, in the finite case it is indeed easy to see the equivalence. But you are sure this holds for all sets? – user190080 Jul 13 '15 at 21:29
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    $g$ is not an injection, since $(0,0,0,0)$ and $(1,1,-2,0)$ have the same image. More generally, you should not expect it to be, since this is smooth, which means it is locally linear, and your linear algebra intuition that it is locally has a kernel of dimension at least. – Steven Gubkin Jul 13 '15 at 21:29
  • I'm not positive but I believe I saw this in a textbook. I think the only proof provided was an intuitive one, though. – pancini Jul 13 '15 at 21:31
  • @StevenGubkin good point...I'll think of a correct example. – user190080 Jul 13 '15 at 21:33
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    @user190080 For the same reason, you are unlikely to find an injection which is "given by a nice formula". The injections are all nasty, violent things which completely disregard the structure of the reals, and just pulverize it into a "mere set". – Steven Gubkin Jul 13 '15 at 21:40
  • @StevenGubkin this is quite poetic :) Do you have any reference where I could look into for further investigations? This reminds me a bit of the construction of the cantor set... – user190080 Jul 13 '15 at 21:43
  • @user190080 No references unfortunately. An idea for a bijection from $\mathbb{R}^2 \to \mathbb{R}$ is to first map $\mathbb{R}^2$ onto the unit square $(0,1) \times (0,1)$. Then map $(0.a_1a_2a_3...,0.b_1b_2b_3...)$ to $0.a_1b_1a_2b_2a_3c_3...$. Finally map $(0,1)$ back onto $\mathbb{R}$. This is probably not exactly correct because of nonuniqueness of decimal representation, but it gives an idea of the kind of thing which would work. – Steven Gubkin Jul 13 '15 at 22:09
  • @StevenGubkin: Yes it is not correct. It can be fixed by first proving that there are countably many reals with non-unique binary expansions and then fixing them by some ad-hoc method like in the bijection from $\mathbb{R}$ to $\mathbb{R} \smallsetminus \mathbb{Q}$. It would be rather messy to go directly from $\mathbb{R}^2$ to $\mathbb{R}$, so the easier method is to first establish a bijection from $\mathbb{R}$ to ${0,1}^\mathbb{N}$ using that idea and then interleaving would work perfectly. – user21820 Jul 14 '15 at 05:22
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Use the positional expansion–based mapping proposed by @StevenGubkin in this comment $$(0.a_1 a_2 a_3 \ldots,\, 0.b_1 b_2 b_3 \ldots) \mapsto 0.a_1 b_1 a_2 b_2 a_3 b_3 \ldots$$ with a quite simple fix:

  1. do not use expansions with infinitely repeating $9$ (if you use decimal system);
  2. do interleave single digits only if they are not $9$, but for nines take all consecutive nines as an interleave unit together with the following single non–$9$ digit.

This makes the mapping a bijection $\Bbb R^2 \supset [0,1)^2\leftrightarrow [0,1)\subset\Bbb R$. Extension to all reals is quite straightforward, for example by $\tan$ and $\arctan$.

Community
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CiaPan
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  • yeah, this works. I just gave it some thought, we could also work with the uniquely non-terminating decimal representation (actually, it is only important that we choose which unique representation we want), then interleave up to the next non-zero digit and change again, so $x=0.3407\ldots, y=0.995\ldots$ will map to $z=0.3949075\ldots$ – user190080 Jul 14 '15 at 17:18
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Actually, strictly speaking the statement has to be restricted even more:

There are no $\mathbb R$-linear bijections $\mathbb R^4\to\mathbb R^3$.

Now if speaking about $\mathbb R^n$, then one usually implicitly defines them as vector space over $\mathbb R$. But it is easily checked that they are also vector spaces over $\mathbb Q$. And when considering them as vector spaces over $\mathbb Q$, then also the notion of linear transformation changes.

We can make that explicit by talking of $\mathbb R$-linear functions when considering $\mathbb R^n$ as vector space over $\mathbb R$, and of $\mathbb Q$-linear functions when considering $\mathbb R^n$ as vector space over $\mathbb Q$.

And then we find:

There are, indeed, $\mathbb Q$-linear bijections $\mathbb R^4\to\mathbb R^3$.

That's because as vector spaces over $\mathbb Q$, both $\mathbb R^4$ and $\mathbb R^3$ are continuum-dimensional (that is, they both have a basis with the cardinality of the real numbers).

And obviously, if there are $\mathbb Q$-linear bijections, there are bijections. Note, however, that the reverse is not true: The $\mathbb Q$-vector spaces $\mathbb Q^4$ and $\mathbb Q^3$ have the same cardinality (there exist bijections between then), but there are no $\mathbb Q$-linear bijections between them (for the same reason why there are no $\mathbb R$-linear bijections between $\mathbb R^4$ and $\mathbb R^3$).

celtschk
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