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.


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.


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.

  • $\begingroup$ Very nice. On $X = X \times \{ 0 \}$ the metric $d'$ even agrees with $d$. $\endgroup$ – Paul Frost Sep 9 '19 at 22:45
  • $\begingroup$ Wow, it truly was very intuitive! $\endgroup$ – Rlos Sep 10 '19 at 3:01
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    $\begingroup$ @EricWofsey The triangle inquality is not satisfied for any metric $d$.The minimal requirement is that $d$ is bounded by $2$ ($d'((x_1,0),(x_2,0) )=d(x_1,x_2)$, $d'((x_i,0),*) = 1$). I didn't completely check whether this is sufficient to obtain the triangle inquality, but it looks okay. $\endgroup$ – Paul Frost Sep 12 '19 at 8:34
  • $\begingroup$ @PaulFrost Oh, you're right, I missed some cases when checking it. Really, the right definition of $d'$ is "the shortest distance when you're allowed to freely move either vertically or horizontally". I was assuming that was the formula I wrote down but it's not in general. The issue is that it might be faster to go up, then move horizontally, and then move back down, than it is to just go straight horizontally. $\endgroup$ – Eric Wofsey Sep 12 '19 at 14:27

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