Why is $C \cap X^0$ discrete where $C$ is compact and $X^0$ is the set of vertices of a linear graph?

Munkres Topology

I understand:

• why $$C \cap X^0$$ is closed in $$C$$ ($$X^0$$ is closed in $$X$$ by coherence)
• why $$C \cap X^0$$ has no limit points (no limit points if and only if all isolated points if and only if discrete, which is concluded)
• why $$C \cap X^0$$ is finite (compactness implies limit point compactness)

I don't understand why $$C \cap X^0$$ is discrete.

I have deduced $$C \cap X^0$$ is a union of vertices of edges (arcs). To show each of these vertices is open in $$C \cap X^0$$, I must find a open set of a superset of $$C \cap X^0$$, such as an open set of $$X$$, to show that the vertex is equal to the intersection of such open set and $$C \cap X^0$$. Without loss of generality, assume the vertices correspond to $$\{0\}$$ in $$[0,1]$$, the interval to which each of the edges (arcs) is homeomorphic. Denote such vertices $$\{p_{\beta}\}_{\beta \in K \subseteq J}$$. We must find an open set $$B$$ in $$X$$ to have $$\{p_{\beta}\} = B \cap C \cap X^0$$. I tried to select $$B = A_{\beta} \setminus \{q_{\beta}\}$$, but I am not sure if this is open in $$X$$. Under coherence, $$A_{\beta} \setminus \{q_{\beta}\}$$ is open in $$X$$ if $$\forall \alpha \in J$$, $$A_{\alpha} \cap [A_{\beta} \setminus \{q_{\beta}\}]$$ is open in $$A_{\alpha}$$.

$$A_{\alpha} \cap [A_{\beta} \setminus \{q_{\beta}\}]$$ is either:

• $$\emptyset$$ - Clopen
• $$A_{\beta} \setminus \{q_{\beta}\}$$ - Open because $$[0,1)$$ is open in $$[0,1]$$
• $$\{p_{\beta}\}$$ - We don't know yet if open!

What other open set can you suggest?

On intuition, I this like picking $$0$$ from a $$K$$ copies of $$[0,1]$$. Instead of $$[0,1]$$, we can choose different closed intervals of $$\mathbb R$$.

Here are some facts that might be related:

• $$C \cap X^0$$ is compact because it is closed in a compact space $$C$$
• This lemma, Lemma 83.2 is kind of an analog of a previous lemma, Lemma 71.2.
• What is a "linear graph?" I suspect however you define that will imply that $X^0$ is discrete, and therefore that every subset of it is discrete too. – hmakholm left over Monica Nov 15 '18 at 12:48
• @HenningMakholm See my answer. – user198044 Nov 16 '18 at 6:15