$Y$ relatively compact in non-cpct riemann surface $X$. Choose domain $Y_1\subset X$ s.t. $Y_1$ relatively cpct in $X$ and $Y$ relatively cpct in $Y$

Question

Let $X$ be a non-compact riemann surface and $Y$ relatively compact open subset of $X$. The book says "Choose domain $Y_1$ s.t. $Y\subset Y_1\subset X$ and a point $a\in Y_1-Y$.(Since $X$ is non-compact and connected, $Y_1-Y$ is non-empty.)"

$\textbf{Q:}$ The following is my thought process. Since $Y$ is relatively compact, there can only be finite number of path components. Since $Y\subset X$ and $X$ is connected, $X$ is path connected. So connect path components of $\bar{Y}$ by path and add tubular neighborhood of those path s.t. it locally looks like product of intervals. Use those tubular neighborhood and original $\bar{Y}$ to form $Y_1$. Clearly $Y_1$ is relatively compact in $X$ and $Y$ is relatively compact in $Y_1$. Furthermore $Y_1-Y$ is non-empty as I need to add extra points away from $\bar{Y}$. How should I interpret the statement? How to choose domain $Y_1$ and show there is a point $a\in Y_1-Y$? It seems the book says this is supposed to be obvious.

Ref. Forster Lectures on Riemann Surface Chpt 14, Cor 14.14

Answer 1

I would not say it is obvious.

You have correctly shown that $Y$ is contained in a compact path connected set $C$. Now choose a relatively compact open neighborhhod $U$ of $C$. Riemann surfaces are locally path connected (i.e. have a base consisting of path connected open sets). In locally connected path spaces the path components of open sets are open. See the answers to Definition of locally pathwise connected.

Now let $Y_1$ be the path component of $U$ containing the path connected $C$. Clearly $Y_1 \subset \overline{U} \ne X$.

Edited:

I was wrong when I said that you have correctly shown that $Y$ is contained in a compact path connected set $C$. The statement ist true, but your argument was that a relatively compact set can only have a finite number of path components. Here is a counterexample: Let $Y \subset \mathbb{C}$ be the set of all $z$ such that $\text{Im} z \in (0,1)$ and $\text{Re} z \in (0,1) \setminus \{ 1/n \mid n \in \mathbb{N} \}$.

So let us give a complete proof. We assume that $X$ is a path connected topological space such that each point has a path connected open neighborhood $U$ such that $\overline{U}$ is compact and path connected. Each Riemann surfaces has this property (take suitable disks around the points).

(1) Each relatively compact $Y \subset X$ is contained in a path connected compact $C \subset X$.

$\overline{Y}$ is covered by path connected open sets $U$ such that $\overline{U}$ is compact. Choose a finite subcover $U_0,\dots,U_n$. Choose paths $u_i$ connecting $U_0$ and $U_i$ for $i =1,\dots,n$. Then $C = \bigcup_{i=0}^n \overline{U_i} \cup \bigcup_{i=1}^n u_i([0,1])$ will do.

(2) Each compact path connected $C \subset X$ has a relatively compact path connected open neighborhood $Y'$.

$C$ is covered by finitely many path connected open sets $V_j$ such that $V_j \cap C \ne \emptyset$ and $\overline{V_j}$ is compact. Then $Y' = \bigcup_j V_j = C \cup \bigcup_j V_j$ will do.