When is the cone $C(X)$ on a space locally compact? There are a few questions in this forum dealing with special cases of this question, for example this for $X = \mathbb Z$ and this for $X = \mathbb R$. My question is

What are necessary and sufficient conditions on $X$ which assure the local compactness of the cone $C(X) = (X \times I)/(X \times \{1\})$?

The answer may depend on the precise definition of local compactness. Here are two variants:

*

*$X$ is locally compact if each $x \in X$ has a compact neigborhood.


*$X$ is locally compact if each $x \in X$ has a neigborhood basis consting of compact sets.
Clearly 2. is stronger than 1. If "compact" is assumed to include "Hausdorff", then 1. and 2. are equivalent. The same is true for Hausdorff $X$ independent of the interpretation of "compact".
The reader is encouraged to use his favorite interpretation.
An obvious sufficient condition is this:

If $X$ is compact, then $C(X)$ is compact and thus locally compact in the sense of 1. If $X$ is compact Hausdorff, then $C(X)$ is compact Hausdorff and thus locally compact in the sense of 2.

Similarly an obvious necessary condition is this:

If $C(X)$ is locally compact, then $X$ is locally compact.

In fact, $X$ is homeomorphic to the base $X \times \{0\}$ of $C(X)$ which is closed in $C(X)$, thus locally compact.
If $X$ is locally compact, then clearly the open subspace $C(X) \setminus \{*\} \approx X \times [0,1)$ is locally compact, where $*$ is the tip of $C(X)$, i.e the common equivalence class of the points in $X \times \{1\}$.
It seems to me that a non-compact $X$ cannot have a locally compact cone. The reason is that if $C(X)$ is locally compact, then $*$ must have a compact neigborhood. I can prove a partial result (see my answer to my own question). But I am interested whether there is a more general theorem.
 A: Here is a partial answer.
Let $X$ be a normal (including Hausdorff) countably paracompact space. Then the following are eqiuvalent:

*

*$X$ is compact.


*$C(X)$ is compact.


*$C(X)$ is locally compact.
This applies to all paracompact Hausdorff spaces $X$, in particular to all metrizable $X$.
The equivalence of 1. and 2. is obvious, and 2. implies 3. It remains to show that 3. implies 1. Our strategy is to embed $X$ as a closed subset of a compact neigborhood of the tip $*$ of $C(X)$. This will be done by shifting the base $X \times \{0\}$ of $C(X)$ towards $*$.
Let $U$ be an open neigborhood of $*$ in $C(X)$ with compact closure $K \subset C(X)$. If $p : X \times I \to C(X)$ denotes the quotient map, then $V = p^{-1}(U)$ is an open neigborhood of $X \times \{1\}$ in  $X \times I$. For each $x \in X$ let $f(x) = \inf\{ t \in I \mid \{x \} \times [t,1] \subset V \}$. Clearly $0 \le f(x) < 1$ because $V$ is open. Moreover $\{x \} \times (f(x),1] \subset V$. The function $f$ is upper semicontinuous: Let $f(x) < r$. Pick $t$ such that $f(x) < t < r$. Then  $\{x \} \times [t,1] \subset V$ and thus there exists an open neigborhood $W_x$ of $x$ in $X$ such that $W_x \times [t,1] \subset V$. Then $f(y) \le t < r$ for $y \in W_x$. Since $f(x) < 1$ for all $x$ and the constant function $1$ is lower semicontinuous, a theorem which was proved independently by Dowker (see "On countably paracompact spaces." Canadian Journal of Mathematics 3 (1951): 219-224 / Theorem 4) and by Katetov (see "On real-valued functions in topological spaces." Fund. Math. 38 (1951): 85-91 / Theorem 2) says that there exists a continuous $h : X \to \mathbb R$ such $f(x) < h(x) < 1$ for all $x$. Define $H : X \to C(X), H(x) = p(x,h(x))$. This is an embedding: In fact, the restriction $\bar p : X \times [0,1) \stackrel{p}{\to} C(X)$ is an embedding and $\bar h : X \to  X \times [0,1), \bar h(x) = (x,h(x))$, is an embedding. Moreover, $H(X)$ is closed in $C(X)$ and $\bar h(X) \subset V$, thus $H(X) \subset U \subset K$. We conclude that $H(X)$ is compact. Therefore $X$ is compact.
Update:
The above theorem says that a normal (including Hausdorff) countably paracompact space $X$ which is not compact cannot have a locally compact cone.
In the special case of a $\sigma$-compact locally compact Hausdorff $X$ we can give an alternative proof which does not use the above "sandwich theorem" for upper and lower semicontinuous functions.
So let $C(X)$ be locally compact,  $U$ be an open neigborhood of $*$ in $C(X)$ with compact closure $K \subset C(X)$ and $V = p^{-1}(U)$ which is an open neigborhood of $X \times \{1\}$ in $X \times I$.
We have $X = \bigcup_{n=1}^\infty K_n$ with compact $K_n \subset X$ such that $K_n \subset \operatorname{int}K_{n+1}$. There exists open $W_n \subset X$ and $t_n \in (0,1)$ such that $K_n \times \{1\} \subset W_n \times (t_n,1] \subset V$. W.l.o.g. we may assume that the sequence $(t_n)$ is non-decreasing. Note that $s_n = (1+t_n)/2$ is contained in $(t_n,1)$. Let $B_n = \operatorname{bd} K_n$ which is compact (but possibly empty; in that case $K_n$ is clopen). The sets $C_n = K_n \setminus \operatorname{int}K_{n-1}$ are compact and contain the disjoint set $B_n$ and $B_{n-1}$ (formally we set $K_0 = \emptyset$). We inductively construct continuous $f_n : C_n \to I$ as follows: For $n=1$ let $f_1(x) = s_2$. Given $f_1,\ldots, f_n$ such that $f_i(x) = s_i$ for $x \in B_{i-1}$, $f_i(x) = s_{i+1}$ for $x \in B_i$ and $f_i(x) \in [s_i,s_{i+1}]$ for all $x \in C_i$ we use the Urysohn theorem to find $f_{n+1} : C_{n+1} \to I$ such that $f_{n+1}(x) = s_{n+1}$ for $x \in B_n$, $f_{n+1}(x) = s_{n+2}$ for $x \in B_{n+1}$ and $f_{n+1}(x) \in [s_{n+1},s_{n+2}]$ for all $x \in C_{n+1}$. The collection of all these $f_n$, $n \in \mathbb N$, can be pasted to a continuous $f : X \to I$ having the property that $(x,f(x)) \in V \setminus X \times \{1\}$. In fact, for $x \in C_n$ we have $f(x) = f_n(x) \in [s_n,s_{n+1}] \subset (t_n,1)$ and thus $(x,f(x)) \in C_n \times (t_n,1) \subset W_n \setminus X \times \{1\} \subset V \setminus X \times \{1\}$. By construction $X' = \{(x,f(x)) \mid x \in X \}$ is a closed subset of $C(X)$ which is homeomorphic to $X$ and, being a closed subset of $K$, compact.
