Compactification by ends

Question

I am trying to solve Exercise 1.19 in Spivak's Volume 1 of "A comprehensive introduction to Differential Geometry".

After the definition of an 'end' of $X$, a topology on the new space $X\cup \epsilon(X)$ is defined "by choosing as neighborhoods of an end $\epsilon_0$ the sets $N_C(\epsilon_0)= \epsilon_0(C)\cup \{\text{ends }\epsilon : \epsilon(C)=\epsilon_0(C)\}$, for all compact C."

Now, with this, I assume that he defines $V$ to be a neighborhood of $x$ in $X$ if there exists an open subset $U$ of $X$ with $x\in U\subset V$ and now with this system of neighborhoods at any point in $X$ or end in $\epsilon(X)$, there is only one topology on $X\cup \epsilon(X)$ generated by it. Namely, $V$ is open in $X\cup \epsilon(X)$ iff $V$ is a neighborhood of each one of its points.

I assume this is the topology we are talking about. Am I right?

I am trying to prove that $X\cup \epsilon(X)$ is a compact space for $X$ a connected, locally connected, locally compact Hausdorff space, but for one case this seems to be untrue.

Take the subspace of the plane defined by $X=\{(x,0) \mid x\in \mathbb{R}\}\cup \bigcup_{n\in\mathbb{Z}}\{(n,y)\mid y\geq 0\}$, this space satisfies all the properties listed before. Moreover, it has and end $\epsilon_n$ for each "ramification" $\{(n,y)\mid y\geq 0,n\in\mathbb{Z}\}$. However, we could cover $X$ with any collection of open subsets of $X$ (don't include any of the newly introduced neighborhoods of ends), and join it to the cover $\epsilon(X)$ with the cover ${N_{\overline{B(n,1/2)\cap X}}(\epsilon_n)}_{n\in \mathbb{Z}}$ of $\epsilon(X)$, and each $N_{\overline{B(n,1/2)\cap X}}(\epsilon_n)$ contains exactly one end, $\epsilon_n$.

Furthermore, $N_{\overline{B(n,1/2)\cap X}}(\epsilon_n)$ is obviously a neighborhood of each of its points, so it is open in $X\cup \epsilon(X)$. Hence, from the resulting open cover of $X\cup \epsilon(X)$ cannot be extracted a finite subcover, because it would have just a finite amount of $N_{\overline{B(n,1/2)\cap X}}(\epsilon_n)$, and hence it would include just a finite amount of ends, while $\epsilon(X)$ is infinite.

So, $X\cup \epsilon(X)$ is not compact.

This leads me to think I'm not working with the right topology. Any light on this?

Thanks in advance.

Answer 1

Your definition is correct. What you are missing is that $X$ has two more ends besides the ends $\epsilon_n$: it has an end $\epsilon_{\infty}$ on which $x\to\infty$ and an end $\epsilon_{-\infty}$ on which $x\to-\infty$. That is, if $C\subset X$ is compact, $\epsilon_\infty(C)$ is you the (unique) component of $X\setminus C$ in which the $x$ coordinate is unbounded above and $\epsilon_{-\infty}(C)$ is the (unique) component of $X\setminus C$ in which the $x$ coordinate is unbounded below. A neighborhood of $\epsilon_{\infty}$ must contain all but finitely many of the $\epsilon_n$. So if you were to extend your open cover by adding a set containing $\epsilon_\infty$, you would only have finitely many $\epsilon_n$ left uncovered, breaking your counterexample.