# existence of a connected set with given connected projections.

Suppose A and B are compact connected sets in the XY plane and XZ plane respectively in R^3. Suppose further that the the range of x-values taken by A and B are the same (i.e, projections of A and B onto the x-axis are the same closed interval). Is there always a connected set in R^3 whose projections onto XY and XZ planes are A and B respectively?

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## 2 Answers

The answer is yes.

Let $\alpha,\beta:[0,1]\to[0,1]\times \mathbb R$ be two paths; $\alpha(t)=\left(\alpha_1(t),\alpha_2(t)\right)$ and $\beta(t)=\left(\beta_1(t),\beta_2(t)\right)$. Assume that $\alpha_1(0)=\beta_1(0)=0$, $\alpha_1(1)=\beta_1(1)=1$.

Claim. The points $a=\left(0,\alpha_2(0),\beta_2(0)\right)$ and $b=\left(1,\alpha_2(1),\beta_2(1)\right)$ lie in the same connected coponent of the set $\Sigma\subset \mathbb R^3$ formed by points of the following type $\left(\alpha_1(t),\alpha_2(t),\beta_2(\tau)\right)$ such that $\alpha_1(t)=\beta_1(\tau)$.

Proof. Note that for generic smooth choice of $\alpha$ and $\beta$ the set $\Sigma$ is a smooth 1-dimensional manifold which might be not connected, but it has only two boundary points in $a$ and $b$. Thus, in this case one can connect these points by a curve. The general case can be done by approximation. $\square$

The rest is easy: one can approximate $A$ and $B$ by path-connected sets $A'$ and $B'$ with the same property. Thus one can present $A'$ and $B'$ as a union of curves $\alpha$ and $\beta$ with the above property. Moreover we can assume that the ends of $\alpha$ and $\beta$ are fixed (i.e. $a$ and $b$ are fixed). For each pair $(\alpha,\beta)$, choose the connected component of $a$ in $\Sigma$ and take the union of all of them.

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Nice answer!. Could you point me to a reference to the fact: "a connected set in the plane can be approximated by path connected set". – arun s Mar 8 '10 at 7:53
A connected open set in the plane has to be path-connected (that is standard fact and it is easy to prove). So you may pass to $\epsilon$-neighborhood of your set (and take its closure if you want to keep it compact). – Anton Petrunin Mar 8 '10 at 15:44

Edit: This answer is wrong. But, since Anton based his answer on this idea, I am leaving it up.

No. Let $A$, resp. $B$, be the graph $x=f(y)$, resp. $x=g(z)$, for some continuous functions $f$ and $g$. Pick numbers $0<d<b<a<c<1$, and arrange things so that $f(0)=g(0)=0$, $f(1)=g(1)=1$, and $f(y)$ grows from 0 to $a$, then decreases to $b$, then finally increases to 1, whereas $g(z)$ grows from $0$ to $c$, decreases to $d$, and increases to 1. (Here, $a$, $b$, $c$, $d$ are function values, i.e., values of $x$, not of the arguments $y$ or $z$.)

Now if $C$ projects to $A$ and $B$, respectively, it is not hard to see that $(0,0,0)\in C$ and $(1,1,1)\in C$. Trying to move $(x,y,z)$ continuously from the former to the latter, a contemplation of how $x$ must grow and shrink in order to maintain $x=f(y)$ and $x=g(z)$ leads to a contradiction. (Hard to put in words, but a pair of graphs reveals it I think.)

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No, it is connected. – Anton Petrunin Mar 7 '10 at 18:27
Rats. You're right. I had mislabeled one of my pictures. – Harald Hanche-Olsen Mar 7 '10 at 19:07
It was nice mistake, I made my answer based on it :) – Anton Petrunin Mar 7 '10 at 19:12
To the "edit": No, you can not make an example this way; it follows directly from my answer. (It also contains an answers to the original question, but you need some extra technical steps.) – Anton Petrunin Mar 7 '10 at 20:09
Right; I see it now. I edited the “edit”. – Harald Hanche-Olsen Mar 7 '10 at 22:54