Examine the proof of the imbedding theorem in the case \(m = 1\) and show that the map \(g\) of part (2) actually maps \(X\) onto a linear graph in \(\mathbb{R}^3\).
The image \(g(X)\) itself need not be a linear graph. For \(X\) a Cantor set, which has dimension \(0\le1\), let \(U_1,U_2\) be overlapping open sets covering \(X\) and let \(\phi_1\) be a partition-of-unity function that takes infinitely many values in \((0,1)\) on \(X\cap U_1\cap U_2\), such as the restriction of a piecewise linear function on \([0,1]\). Then \(g(X)\cap(z_1,z_2)\) is an infinite compact totally disconnected set, which is not a finite union of points and segments. The proof gives \(g(X)\) inside the linear graph \(\bigcup_{i<j}[z_i,z_j]\).
Solution
In the proof of Theorem 50.5, when \(m=1\) we have \(N=2m+1=3\). The finite open cover \(\{U_i\}_{i=1}^r\) is chosen to have order at most \(m+1=2\), and a partition of unity \(\{\phi_i\}\) dominated by this cover is chosen. The approximating map is \[ g(x)=\sum_{i=1}^r\phi_i(x)z_i, \] where the points \(z_i\in\mathbb R^3\) are in general position.
At each point \(x\in X\), at most two of the numbers \(\phi_i(x)\) are nonzero. Since they are nonnegative and sum to \(1\), either \[ g(x)=z_i \] for some \(i\), or \[ g(x)=t z_i+(1-t)z_j\qquad(0<t<1) \] for some pair \(i\ne j\). Thus \[ g(X)\subset \bigcup_{1\le i<j\le r}[z_i,z_j], \] a finite union of straight line segments.
This finite union of segments is a linear graph with vertices \(z_1,\ldots,z_r\). Points in general position in \(\mathbb R^3\) have every four of them affinely independent. If \(\{i,j\}\cap\{k,l\}=\varnothing\), the segments \([z_i,z_j]\) and \([z_k,z_l]\) are disjoint, since a common point would put \(z_i,z_j,z_k,z_l\) in one plane. If the segments share exactly one endpoint \(z_i\), they meet only at \(z_i\), since \(z_i,z_j,z_l\) are not collinear. So two edges meet at most in a common vertex.
Therefore, in the case \(m=1\), the map constructed in part (2) of the embedding theorem maps \(X\) into a linear graph in \(\mathbb R^3\).