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.
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.