Munkres Topology

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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. enter image description here enter image description here
  • 1
    $\begingroup$ 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. $\endgroup$ – hmakholm left over Monica Nov 15 '18 at 12:48
  • $\begingroup$ @HenningMakholm See my answer. $\endgroup$ – user198044 Nov 16 '18 at 6:15

Thank you Henning Makholm!

Here is the definition of a linear graph.

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I missed the last sentence of the paragraph because I assumed the whole paragraph after the first sentence is proving the first sentence.

However, after I realized (earlier) that an equivalent definition of discrete is every subset is closed, I should have recognized the relevance of any union of vertices or edges is closed.


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