Existence of a cone neighbourhood in an open disc with a point in the border

Question

I started reading C. P. Rourke and B. J. Sanderson’s “Introduction to Piecewise-Linear Topology” which is recommended by D. Rolfsen in “Knots and Links”, they provide the following definition:

1.1 A subset $P \subset \mathbb{R}^n$ is a polyhedron if each point $a\in P$ has a cone neighbourhood $N=aL$ in $P$, where $L$ is compact.

Before this we only get a definition of cone as follows:

For $a\in \mathbb{R}^n, B\subset\mathbb{R}^n$ and $a\notin B$, their join $aB=\{\lambda a+(1-\lambda)b:b\in B, \lambda \in[0,1]\}$ is called a cone with vertex $a$ and base $B$ if each point in $aB$ is expressed uniquely as $\lambda a+(1-\lambda)b$ with $\lambda \in [0,1]$ and $b\in B$ as before, except for $a$ of course.

I have a couple issues with this definition, for instance:

• By cone neighbourhood in P do they mean a cone $aL$ such that $aL\subset P$ is true and also there is an open subset $O\subset P$ —open in the subspace topology (?)— such that $a\in O\subset aL$?

My confusion arose after this example (the second one), to be found in the next page of the book:

I get that the given space is not by any means a polyhedron, but the author says that it is not because there is no cone neighbourhood, but because the one that exists has a non-compact base, but I’ve come to feel pretty sure that there isn’t one at all, since the only two reasonable candidates for the base $L$ would be an open chord or an interior circumference tangent to the disk’s boundary precisely at $a$ (excluding $a$ so that it is not in $L$); but both of these attempts leave “gaps” which make it impossible to fit a basic open of $P$ containing $a$ inside of them.

My question is: am I right about this little correction? Knowing that would really help me to move on confidently, since this is already one of the very first definitions of the book.

P.s. I omitted the rest of the tail in my previous reasoning because it is not relevant, and the true pathology of the example is a consequence of the point in the boundary.

Answer 1

If the authors meant anything else by "cone neighborhood" than a neighborhood of $a$ that is also a cone, they surely would have given a definition for the phrase. You will have to check whether they did, as I do not have access to a copy. You might try looking up the phrase in the index and checking the early references. If they didn't give such a definition, then the interpretation you gave must be correct.

In that case, you are correct that by the definitions given - particularly the "uniqueness" requirement - $a$ does not have any neighborhood in $X$ that is a cone. To see that this is true in general, let $D$ be the open disk in $X$. Any neighborhood of $a$ (to be clear, I mean any set containing an open set containing $a$) must have the intersection of an open disk $U$ centered on $a$ with $D$ as a subset. Note that a portion of the boundary of $D$ in the plane lies inside $U$, and $U \cap \overline D$ is convex.

Any ray eminating from $a$ and passing through a point of $\overline D$ must also contain points of $U$, So some initial segment of that ray must lie in $U\cap D$. If the neighborhood is a cone $aL$, then there has to be a unique point $l \in L$ lying on the ray. If $d \in \partial D\cap U, d\ne a$, then $l$ lies on the segment $ad$ but cannot be $d$ since $d$ itself is not in $X$. Thus there are interior points of $dl$ lying inside $U\cap D \subset U\cap X$ but are not in the cone $aL$, contradicting that $aL$ is a neighborhood.

---

It is possible the authors were thinking of a closed disk when they said $a$ has a cone neighborhood that is not compact. In that case $X$ itself would be a cone neighborhood of $a$ with base $L$ consisting of the end of the tail and the boundary of the disk with $a$ removed, because $a$ cannot be in $L$. Since $a$ is removed, $L$ is not closed, and therefore not compact.

Answer 2

By cone neighbourhood in P do they mean a cone $aL$ such that $aL\subset P$ is true and also there is an open subset $O\subset P$ such that $a\in O\subset aL$?

Yes, a cone neighborhood of $a \in P$ is a neighborhood of $a$ in $P$ which has the form $aL \subset P$ for some $L \subset P \setminus \{a\}$. A cone neighborhood in general is not an open neighborhood. If $L$ is required to be compact, then $aL$ is a compact neighborhood of $a$ (which may nevertheless be open in $P$).

Therefore polyhedra are locally compact subspaces of $\mathbb R^n$.

Thus the space $X$ in the second example is not a polyhedron because $a$ does not have any compact neighborhood in $X$. The point $a$ has a neighborhood which is a join: Let $D$ denote the open disk and $e$ the endpoint of the tail. Taking $L = D \cup \{e\}$ we get $aL = X$.

However, $a$ does not have a cone neighbourhood as shown in Paul Sinclair's answer, and this contradicts the claim of the authors. Taking the closed disk instead of the open disk would solve this issue, but in that case $X' = X \setminus \{a\}$ is not a polyhedron. In fact, no boundary point of $X'$ has a cone neighborhood with a compact base.

I think the best variation of the second example is to replace the open disk $D$ by an open triangle $T$ with a tail at one of its vertices $a$. Then one can take $L = G \cap T$, where $G$ is a line between $a$ and the edge of $T$ not containing $a$.