What is a "cone neighborhood"?

Question

I am trying to understand the following definition from the book "Introduction to Piecewise-Linear Topology":

In particular, I did not find a definition of a "cone neighborhood" - what is it?

There are some examples of non-polyhedra:

But I did not understand them: why is there no "cone neighborhood" in the first example, and no "cone neighborhood" with a compact $L$ in the second example?

Answer 1

By context, it looks like the definition of cone neighborhood is built in to the definition of polyhedron. The wording is rather terse, so let me rewrite that passage by adding words, at the risk of adding redundancy:

A subset $P \subset \mathbb R^n$ is a polyhedron if for each point $a \in P$ there exists a compact subset $L \subset \mathbb R^n$ such that the subset $aL$ is a neighborhood of $a$ in $P$, called a cone neighborhood of $a$ in $P$.

I presume that the notation $aL$ has been defined elsewhere, but to be safe I do believe that the meaning is $$aL = \{(1-t)a + t b \mid b \in L, 0 \le t \le 1\} $$ that is to say, the union over all $b \in L$ of the straight line segments $$ab = \{(1-t)a + tb \mid 0 \le t \le 1\} $$ There may also be some hidden requirements. For example, it may be that notation like $aL$ is only allowed under the precondition that $a \not\in L$. On top of that, it may also be required that if $b \ne b' \in L$ then $ab \cap ab' = \{a\}$.

In the second example of figure 5, the point $a$ has a neighborhood of the form $aL$ where $L$ is the disjoint union of a point and an open semicircle, but that $L$ is not compact. In fact $a$ has no compact neighborhood at all, and therefore it certainly has no neighborhood of the form $aL$ where $L$ is compact.

The first example of figure 5 is a little more mysterious to me. Perhaps the reproduction is flawed, but the picture appears to be either a closed disc or a circle; the words "open disc" do not seem to match the picture; and the words "with a tail" do not seem to be matched by anything in the picture. Because of this mismatch I think I will demur on attempting to interpret that picture, except to suggest an exercise: prove that the circle is not a polyhedron.