z sin -> z^2 sin (make it C^1 at least)
authorKevin Walker <kevin@canyon23.net>
Tue, 27 Jul 2010 21:20:32 -0400
changeset 488 efcc71e5489f
parent 487 c576b7868f05
child 489 bdbd890086eb
z sin -> z^2 sin (make it C^1 at least)
text/blobdef.tex
--- a/text/blobdef.tex	Tue Jul 27 15:29:45 2010 -0700
+++ b/text/blobdef.tex	Tue Jul 27 21:20:32 2010 -0400
@@ -159,8 +159,8 @@
 \begin{align*}
 A & = [0,1] \times [0,1] \times [-1,1] \\
 B & = [0,1] \times [-1,0] \times [-1,1] \\
-C & = [-1,0] \times \setc{(y,z)}{z \sin(1/z) \leq y \leq 1, z \in [-1,1]} \\
-D & = [-1,0] \times \setc{(y,z)}{-1 \leq y \leq z \sin(1/z), z \in [-1,1]}.
+C & = [-1,0] \times \setc{(y,z)}{z^2 \sin(1/z) \leq y \leq 1, z \in [-1,1]} \\
+D & = [-1,0] \times \setc{(y,z)}{-1 \leq y \leq z^2 \sin(1/z), z \in [-1,1]}.
 \end{align*}
 Here $A \cup B = [0,1] \times [-1,1] \times [-1,1]$ and $C \cup D = [-1,0] \times [-1,1] \times [-1,1]$. Now, $\{A\}$ is a valid configuration of blobs in $A \cup B$, and $\{C\}$ is a valid configuration of blobs in $C \cup D$, so we must allow $\{A, C\}$ as a configuration of blobs in $[-1,1]^3$. Note however that the complement is not a manifold.
 \end{example}