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Let $(X,d)$ be a metric space and $\mathcal{F}$ a family of functions $f : X \times X \rightarrow [0,\infty)$ s.t.

(i) $f(x,y) \geq d(x,y) \ \forall f \in \mathcal{F}$

(ii) for each $x \in X$ there exists $f \in \mathcal{F}$ s.t. $f(x,x) = 0$

(iii) for each $f \in \mathcal{F}$ there exists $g \in \mathcal{F}$ s.t. $g(x,y) = f(y,x)$ for all $x,y \in X$

(iv) for any point $a \in X$ and $f,g \in \mathcal{F}$ we have that the function $h : X\times X \rightarrow \mathbb{R}$ def. by $h(x,y) = f(x,a) + g(y,a)$ is in $\mathcal{F}$

Show that $\rho(x,y) := \inf\limits_{f\in \mathcal{F}}\{f(x,y)\}$ defines a metric on $X$

I'm stuck on trying to prove triangle inequality $\rho(x,y) \leq \rho(x,a) + \rho(a,y)$. Haven't really got anywhere useful. Any ideas?

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Fix $f, g \in \mathcal{F}$ and $a \in X$. Let $h$ be as in condition (iv). Then $$f(x, a) + g(y, a) = h(x, y) \ge \rho(x, y).$$ Since this holds for any such $f$, (leaving $g, a, x, y$ fixed), then taking the infimum of $f$ over $\mathcal{F}$ implies $$\rho(x, a) + g(y, a) \ge \rho(x, y).$$ Similarly, leaving $a, x, y$ fixed, and taking the infimum over $g$, $$\rho(x, a) + \rho(y, a) \ge \rho(x, y).$$ This holds over all $a, x, y \in X$, i.e. we've proven the triangle inequality.

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Let $\varepsilon >0$ and $f,g$ s.t. for all $x,y$, $$ f(x,a)+g(a,y)\leq \rho(x,a)+\rho(a,y)+\varepsilon $$

Since $$h(x,y):=f(x,a)+g(a,y)\in \mathcal F,$$ we get $$\rho(x,y)\leq h(x,y)\leq \rho(x,a)+\rho(a,y)+\varepsilon ,$$ what prove the claim.

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