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It is a problem I encountered when working on shape optimization(mech eng, not math one).

Consider two connected sets $A$ and $B$ in $\mathbb{R}^d$ (it would be nice if $d$ can be chosen arbitrarily, but I only need the result for $d\leq 3$), and each of them contain more than one point.

I am wondering if there always exists a bijective mapping from $A$ to $B$?

Updated assumption:

Further assume there is no degeneracy(I'm not sure if it's a correct word) between two sets.

An example: if $A$ is a 3-dim object, e.g. a cube, $B$ is also a 3-dim object, e.g. a ball. But $B$ cannot be an ojbect possible to be embedded in lower dimensional spaces, e.g. a flat plate is not allowed.

With this imposed assumption, would it further simplify the argument?

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    $\text{@}$newbie: Your question has already been answered in general, so you don't need further assumptions. Are you looking for something particular in a "simplified" argument based on further assumptions?2012-05-28
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    @JonasMeyer The ultimate goal is that I also want to receive a better property of such as differentiability, even smoothness, which seems unlikely to be achieved by the previous assumption.2012-05-28
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    Asking even for continuous bijections between "nondegenerate" connected subspaced of $\mathbb R^d$ is too much to ask in general. E.g., the surface of a sphere and the surface of a donut have no continuous bijection between them.2012-05-28
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    @JonasMeyer We can only consider compact set, even simply connected. What I really want to abtain is sufficient regularity and smoothness.2012-05-28
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    What about if $A$ and $B$ have the same topological properties?2012-05-28
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    [Homeomorphic](http://en.wikipedia.org/wiki/Homeomorphism) does not generally imply [diffeomorphic](http://en.wikipedia.org/wiki/Diffeomorphism). In the second article you'll find the statement, "In dimensions 1, 2, 3, any pair of homeomorphic smooth manifolds are diffeomorphic." But such diffeomorphisms might not extend differentiably to the ambient space.2012-05-28

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Yes, but the mapping may not be very nice. To see this, consider the projection mappings $\pi_{v}:\mathbb R^d\to \mathbb R$ which project an element of $\mathbb R^d$ onto the line with direction $v$ and map this to $\mathbb R$ in the usual manner. These are continuous, and since $A$ and $B$ are connected we have that $\pi_v(A)$ and $\pi_v(B)$ are connected for all $v$. Since there exist points $x_1,x_2\in A$ we can let $v=x_2-x_1$ and get that $\pi_v(A)$ has at least two points. But the only connected subsets of $\mathbb R$ are intervals, and any interval which is not a point contains $\mathfrak c$ (continuum) many points. Thus $A$ contains at least $\mathfrak c$ many points, and since $\mathbb R^d$ contains only $\mathfrak c$ many points it contains exactly $\mathfrak c$ many points. The same proof shows $B$ contains $\mathfrak c$ many points, so $A$ and $B$ have the same cardinality. Thus there exists a bijection between them.

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    Thanks for your quick reply. I have updated the question a little bit. Would you mind read it again? Cheers.2012-05-28
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    The answer is yes to your more general question, so it is of course yes to your new one. Are you looking for the bijection to have certain special properties, e.g. continuity?2012-05-28
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    Yes, exactly. Even smoothness, as mentioned in another comment.2012-05-28
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    @newbie Then the answer is certainly not. Consider $A=B_1(0)$, the open ball around the origin, and $B=\bar A$, the closure of $A$. Then $B$ is compact but $A$ is not, so there is no continuous map between them, and certainly no differentiable or smooth map.2012-05-28