TITLE: Is the cone locally compact
QUESTION [3 upvotes]: Let $X$ denote the cone on the real line $\mathbb{R}$. Decide whether $X$ is locally
compact. [The cone on a space $Y$ is the quotient of $Y \times I$ obtained by
identifying $Y \times \{0\}$ to a point.]
I am having a hard time showing that there exists a locally compact neighborhood around $Y \times \{0\}$.  Some help would be nice.

REPLY [2 votes]: Here is a way of showing that no neighborhood of $r=\Bbb R\times\{0\}\in X$ is compact. The idea is to find in any neighborhood $V$ of $r$ a closed subspace homeomorphic to $\Bbb R$. Since the subspace is not compact, $V$ cannot be compact.
So let $V$ be a neighborhood of $r$ in $X$. Then $V$ contains the image of an open set $U$ around $\Bbb R\times\{0\}$. Since the interval $[n,n+1]$ for any $n\in\Bbb Z$ is compact, there is an $\epsilon_n>0$ such that $[n,n+1]\times[0,ϵ_n]$ is contained in $U$. Let $b_n=\min\{ϵ_n,ϵ_{n-1}\}$. Define
$$
f(x) = (x-n)b_{n+1}+(n+1-x)b_n,\quad n\in\Bbb Z,\quad x\in[n,n+1]
$$
This map has a graph $\Gamma$ homeomorphic to $\Bbb R$ and contained in $U$. The quotient map $q:\Bbb R\times I\to X$ embeds $\Gamma$ as a closed subspace of $V$, so $q(\Gamma)$ had to be compact if $V$ were compact.
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