Metric on the cone of a compact metric space It is known that if $X$ is a compact metric space (with bounded metric), then its cone $CX$ (the quotient $X \times [0,1] / X \times \{1\}$) is metrizable.
Is there a way to provide an explicit metric on $CX$ that induces its usual topology? I've been thinking about it but the vertex of the cone seems to be very problematic when dealing with the triangle inequality in a potential metric.
 A: Here is an alternative approach to finding an explicit metric on the cone $CX$.
Let us first assume that $X$ is a subset of a normed linear space $(E, \lVert - \rVert)$ and the metric $d$ on $X$ is given by $d(x,y) = \lVert x - y \rVert$. Define the geometric cone over $X$ as the subset $C'X \subset E'  = E \times \mathbb R$ obtained by taking the union all lines segments $L(x) = \{((1-t)x,t) \mid t \in I \} \subset E'$, $x \in X$, connecting $(x,0)$ with $(0,1)$. In other words
$$C'X = \{(1-t)x,t) \mid x \in x, t \in I \} .$$
Then $C'X$ is a subset of the normed linear space $(E', \lVert - \rVert')$ where $\lVert (x,t) \rVert' = \lVert x \rVert + \lvert t \rvert$ and therefore inherits a metric $d'$ given by $d'((x,t),(y,s)) = \lVert ((x,t) - (y,s)) \rVert' = \lVert x - y \rVert + \lvert t - s \rvert$.
Now define
$$\varphi : X \times I \to C'X, \varphi(x,t) = ((1-t)x,t).$$
This is a continuous map such that $\varphi(X \times \{ 1\}) = \{ (0,1) \}$, hence it induces a continuous $\phi : CX \to C'X$ which is obviously a bijection. If $X$ is compact, then $\phi$ is a homeomorphism. Hence $CX$ can be endowed with the metric
$$D([x,t],[(y,s]) = d'(\phi([x,t]),\phi([y,s])) = d'(\varphi(x,t),\varphi(y,s))  \\ = \lVert (1-t)x -(1-s)y \rVert + \lvert t - s \rvert$$
that induces its usual topology.
Now let $(X,d)$ be a metric space with a bounded metric $d$. If $X$ is compact, then $d$ is automatically bounded. Let $C_b(X)$ be the vector space of all continuous bounded maps $f :X  \to \mathbb R$ which is endowed with the $\sup$-norm $\lVert f \rVert = \sup_{x\in X} \lvert f(x) \rvert$. It is well-known that $(X,d)$ embeds isometrically into $(C_b(X),\lVert - \rVert)$ via $x \mapsto e(x) = d(x,-) : X \to \mathbb R$. The proof is straightforward. Identifying $X$ with $e(X) \subset C_b(X)$, our above construction yields a geometric cone $C'X$ with metric $d'$. If $X$ is compact, this yields the following metric on $CX$:
$$D([x,t],[y,s]) = \sup_{z \in X} \lvert (1-t)d(x,z) - (1-s)d(y,z) \rvert + \lvert t -s \rvert .$$
This is admittedly less transparent than Eric Wofsey's solution.
Let us close with a remark concerning the geometric cone. If $X$ is not compact, then $C'X$ is not homeomorphic to $CX$. In fact, $CX$ does not have a countable neighborhoood base at its tip whereas $C'X$ is first countable. See my answer to Equivalent definition $\text{Cone}(K)$ which can be generalized to the present case.
Nevertheless, $C'X$ has some essential features attributed to a cone: It is contractible and the inclusion $X \to C'X, x \mapsto (x,0)$, is a cofibration.
A: Just follow the geometric idea of a cone, in which the copies $X\times\{t\}$ of $X$ shrink as $t$ approaches $1$.  So concretely, if $d$ is a metric on $X$ such that $d(x,y)\leq 2$ for all $x,y\in X$, let's define a metric $d'$ on $CX$ by $$d'((x,s),(y,t))=(1-\max(s,t))d(x,y)+|t-s|.$$  Note that this is well-defined since if $s$ or $t$ is $1$ then the values of $x$ and $y$ don't matter, and it is easy to check it satisfies the triangle inequality.  The intuition here is that for $t\geq s$, we find the distance from $(x,s)$ to $(y,t)$ by first travelling from $(x,s)$ up to $(x,t)$ (a distance of $|t-s|$) and then travelling from $(x,t)$ to $(y,t)$ which is a distance $(1-t)d(x,y)$ since the metric on $X\times\{t\}$ is scaled by a factor of $1-t$.  The assumption that $d\leq 2$ guarantees that this process is the "shortest distance" between $(x,s)$ and $(y,t)$ when you're allowed to make any sequence of horizontal and vertical moves inside the cone, so that the triangle inequality holds.  (As pointed out by Paul Frost, if $d$ can be greater than $2$, then in some cases the shortest distance would instead be to just go straight vertically to the tip of the cone and then back down.)
Now let $C'X$ denote $CX$ with this metric; we wish to show the identity map $CX\to C'X$ is a homeomorphism (where $CX$ has the quotient topology).  First, the identity map is continuous since the quotient map $X\times[0,1]\to C'X$ is easily seen to be continuous.  But then $CX$ is compact and $C'X$ is Hausdorff, so the identity map $CX\to C'X$ is automatically a homeomorphism, being a continuous bijection.

Note that much more generally, any Hausdorff quotient of a compact metric space is automatically metrizable.  See Quotient of compact metric space is metrizable (when Hausdorff)? for proofs, although these proofs don't give a particularly explicit metric.
