# Minimum of two metrics is metric

Let X be a set and d, e two metrics on it. How can I show that the function f defined by f(x,y) = min(d(x,y), e(x,y)) is again a metric? This has been bothering me for the past few days as it seems intuitively true but always get stuck trying to prove the triangle inequality for f. Perhaps there is something obvious I'm missing.

If the minimums are the same metric the problem is easy, but showing d(x,y) + e(y,z) >= d(x,z) and the similar cases have left me stuck. My best attempt so far has been to attempt to find a contradiction by assuming the inequality is false:

f(x,y) + f(y,z) < f(x,z)

d(x,y) + e(y,z) < e(x,z) (a specific case in the contradiction, specifics not important)

d(x,y) + d(y,z) + e(y,z) < e(x,z) + d(y,z)

d(x,z) + e(y,z) < e(x,z) + d(y,z) (Using fact d is a metric)

d(x,z) + e(y,z) - d(y,z) < e(x,z), but e(y,z) < d(y,z) so d(x,z) should be smaller than e(x,z) which contradicts f(x,z) = e(x,z).

Now clearly this is an incorrect argument, but it's the best I have at this point. Any ideas?

• Are you sure it is true? Have you looked for a counterexample? Commented Nov 28, 2013 at 3:28
• I believe you need to consider $f(x,y)=d(x,y)$, $f(y,z)=d(y,z)$, $f(x,z)=e(x,z)$. I don't consider obvious that this can't happen Commented Nov 28, 2013 at 3:56
• I think that if we replace ${\bf maximum}$ it holds. Commented Nov 28, 2013 at 4:05
• @user111995 : alternate counterexample based on answer below: on $\mathbb{R}^2$, $d(\mathbf{x},\mathbf{y})=\frac{1}{10}|x_1-y_1|+|x_2-y_2|$, $e(\mathbf{x},\mathbf{y})=|x_1-y_1|+\frac{1}{10}|x_2-y_2|$. Commented Nov 28, 2013 at 16:10

Counterexample : $$d_1(x,y) = \sqrt{(x_1-y_1)^2 + \frac{1}{100} (x_2-y_2)^2}$$

$$d_2(x,y) = \sqrt{\frac{1}{100}(x_1-y_1)^2 + (x_2-y_2)^2}$$

So let $x=(1,0),\ y=(0,1),\ z=(0,0)$ so that $$d(x,y)>1 > d(x,z) + d(y,z)=\frac{1}{5}$$ where $d$ is a minimum.

• Thank you everyone for your insightful and intelligent responses. I should say that after checking the numbers in Hee Kwon Lee's answer d(x,z) + d(y,z) appears to be 1/50 rather than 1/5, but that is a minor point and the counterexample still holds. Thanks again! Commented Nov 28, 2013 at 16:24
• Thank you for your appreciation. Commented Nov 29, 2013 at 13:39
• @user111995 I find that 1/5 is indeed (approx) correct Commented Sep 8, 2016 at 23:37

As people have pointed out, minimum of two metrics need not be a metric. The best way to construct examples and counterexamples of metric spaces is using the following method.

Let $$X$$ be any set, $$(Y,d)$$ be a metric space, and $$f:X\to Y$$ be an injective function. Define $$d_f:X\times X\to \mathbb{R}$$ as $$d_f(x_1,x_2):=d(f(x_1),f(x_2))$$. It is easy to verify that $$d_f$$ is a metric on $$X$$, let us call this as the metric induced by $$f$$.

Consider $$X=\{a,b,c\}$$, a three point set. Define $$f,g:X\to \mathbb{R}$$ as $$f(a)=0,f(b)=1,f(c)=3$$, and $$g(a)=0,g(b)=2,g(c)=3$$. Let $$d_f$$ and $$d_g$$ be the metrics on $$X$$ induced by the functions $$f$$ and $$g$$ respectively, where $$\mathbb{R}$$ is given the usual metric. We have \begin{align*} d_f(a,b)=1;\quad d_f(b,c)=2;\quad d_f(a,c)=3\\ d_g(a,b)=2;\quad d_g(b,c)=1;\quad d_g(a,c)=3. \end{align*} Hence, if we define $$d:X\times X\to \mathbb{R}$$ as $$d(x,y):=\min\{d_f(x,y),d_g(x,y)\}$$, then we have $$d(a,b)=1;\quad d(b,c)=1;\quad d(a,c)=3.$$ Clearly $$3=d(a,c)\nleq d(a,b)+d(b,c)=2,$$ i.e. the triangular inequality is not satisfied by $$d$$, and hence it can not be a metric.