I am trying to work out the details of following example from page 101 of Tu's An Introduction to Manifolds:

Example 9.3. Let $\Gamma$ be the graph of the function $f(x) = \sin(1/x)$ on the interval $]0, 1[$, and let $S$ be the union of $\Gamma$ and the open interval $I=\{(0,y)\in \mathbb{R^2} |−1<y<1\}$.

The subset $S$ of $\mathbb{R^2}$ is not a regular submanifold for the following reason: if $p$ is in the interval $I$, then there is no adapted chart containing $p$, since any sufficiently small neighborhood $U$ of $p$ in $\mathbb{R^2}$ intersects $S$ in infinitely many components.

Using Tu's definitions, I need to show that given a neighborhood $U$ of $p$, there exists no homeomorphism $\phi$ from $U$ onto an open set of $V \subset\mathbb{R^2}$ with the property that $U \cap S$ is the pre-image with respect to $\phi$ of the $x$ or $y$ axes intersected with $V$.

I am not sure where the fact that $U \cap S$ has infinitely many components comes into play. I tried to derive a contradiction using this fact; but even if restrict my attention to connected neighborhoods $U$ of $p$, the intersection of the connected set $\phi(U)$ with the $x$ or $y$ axes might have infinitely many components (I think), so there's no contradiction with the fact that homeomorphism preserves components.

I would appreciate any help!


Let $X$ denote the $x$-axis. Assume that $U\ni p$ is open and $\phi(U)$ is an open subset of $\mathbb R^2$ with $\phi:U\to \phi(U)$ a homeomorphism. If $\phi(U)$ intersects $X$ in several components, let $C$ denote the connected component containing $\phi(p)$. Since $\phi(U)$ is open, $\phi(U)\cap X$ is open in $X$. But $X\approx\mathbb R$ is locally connected, therefore components of open subsets are open. So $C$ is open and $X-C$ is closed in $X$. Since $X$ is closed in $\mathbb R^2$, $X-C$ is closed in $\mathbb R^2$ and $\phi(U)-(X-C)$ is open.
Applying $\phi$ we get a homeomorphism $\phi^{-1}(\phi(U)-(X-C))\approx \phi(U)-(X-C)$, the later of which intersects the $x$-axis in just a single component, which is a contradiction.

  • $\begingroup$ Very helpful answer, thank you. $\endgroup$ – yuval Aug 4 '13 at 0:30
  • 1
    $\begingroup$ @yuval: I realized that a much simpler and more general solution exists: Whenever $(U,\phi)$ is a chart about a point $p$, then by the openness of $\phi(U)$ there is a ball $B_\epsilon(\phi(p))$ contained in $\phi(U)$. If $\phi(p)\in X$, then clearly this ball intersects $X$ in a connected set, and this ball can be used as a coordinate chart. $\endgroup$ – Stefan Hamcke Aug 4 '13 at 14:40

Let $U\supset U'$ where $U'$ is open in $\mathbb R^2$ and $p\in U'\cap I.$ Then $U'\cap S= (\{0\}\times J)\cup (\cup F)$ where $J$ is a non-empty open subset of $]-1,1[$ and $F$ is a non-empty family of subsets of $\{(x,\in 1/x):x\in ]0,1[\},$ and each member of $F$ is homeomorphic to a non-empty open interval of $\mathbb R.$

Suppose $\phi:U\to V$ is a homeomorphism with $\phi( U\cap S) = V\cap (\mathbb R \times\{0\}).$

Now $\phi$ maps any subspace of $U\cap S$ homeomorphically onto its image. So $\phi(\{0\}\times J) = K\times \{0\}$ where $K$ is a non-empty open subset of $\mathbb R .$ And for each $f\in F,$ we have $\phi(f)=G_f\times \{0\}$ where $G_f$ is a non-empty subset of $\mathbb R,$ with $G_f$ disjoint from $K.$

But then $\phi (\{0\}\times J)$ is disjoint from the closure of $\phi(\cup F),$ and since $\phi|_{U'\cap S}\to \phi(U'\cap S)$ is a homeomorphism, that implies that $\{0\}\times J$ is disjoint from the closure of $\cup F,$ and hence that $\{0\}\times J$ is open in the space $\Gamma \cup I$, which is FALSE.


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