Does same cardinality imply a bijection? 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.
 A: 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
$$
t:\mathbb{R}^3\to \mathbb{R}^4,t(x_1,x_2,x_3)=(x_1,x_2,x_3,0)
$$
A: "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.
A: 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:


*

*do not use expansions with infinitely repeating $9$ (if you use decimal system);

*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 $\arctan$.
A: 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 $\mathbb R^3$).
