Show the cone on the integers is not locally compact Let $C\mathbb{Z}$ (the cone on $\mathbb{Z}$) denote the quotient space obtained from $\mathbb{Z} \times [0,1]$ by identifying all points in $\{(z,1)|z\in \mathbb{Z}\}$. How can I show that $C\mathbb{Z}$ is not locally compact? Obviously the "problematic point" in $C\mathbb{Z}$ which will not have a compact neighborhood will be the image of $\{(z,1)|z\in \mathbb{Z}\}$ under the quotient map from $\mathbb{Z} \times [0,1]$to $C\mathbb{Z}$.
Starting idea: any open set in $C\mathbb{Z}$ containing the problematic point must contain the image, under the quotient map, of $\mathbb{Z}\times(a,1)$ for some $a\in[0,1]$

HINT: Let $p$ be the point at the tip of the cone. Show that every nbhd of $p$ contains a set of the form 
$$q\left[\bigcup_{n\in\Bbb Z}\big(\{n\}\times[a_n,1]\big)\right]\;,\tag{1}$$
where $q$ is the quotient map, and $a_n\in[0,1)$ for each $n\in\Bbb Z$. Such sets are closed, so if the cone were locally compact, one of them would have to be compact. But each of these sets is homeomorphic to the whole cone, so the cone would have to be compact. But it isn’t even countably compact: find a countable open cover of the cone that has no proper subcover at all. (For that matter, you can do this almost as easily directly for a set of the form $(1)$.)
