Is the cone locally compact 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.
 A: 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.
A: As $\mathbb{R}$ is homeomorphic to $(0,1)$, consider the cone on $(0,1)$. Now consider the open triangle with vertices $\{(0,0),(1,0),(1/2,1/2)\}$ union $(0,1)\times \{0\}$ in $\mathbb{R}\times [0,1]$. Its image in the cone is open which contains $(0,0)$ and whose closure is compact.
A: Hint: $X$ is homeomorphic to $C= \{(x,y,z) \mid z \geq 0, x^2+y^2=z^2 \}$ and $C_n= \{ (x,y,z) \mid 0 \leq z \leq 1/n,  x^2+y^2=z^2 \}$ is a compact neighborhood of $(0,0,0)$.
