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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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    [Homeomorphic](http://en.wi$k$ipedia.org/wi$k$i/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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    @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