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How can I prove that if $(X,d)$ and $(X,k)$ are metric spaces then $(X, \max(d,k))$ and $(X, \min(d,k))$ are metric spaces?

I try to replace j as $j=\min(d,k)$ or $j=\max(d,k)$ and prove that $j$ is positive definiteness, symmetric and satisfies the triangle inequality.

K.defaoite
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Begginner Julia
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – Kavi Rama Murthy Feb 20 '21 at 04:48
  • Which parts were you able to prove, and where did you get stuck? – saulspatz Feb 20 '21 at 05:47
  • I'm stuck in triangule inequality. if d,k,l ∈ X then, j(d,k)=max(d,k)=max(d-l,k-l) ... – Begginner Julia Feb 20 '21 at 06:08
  • I have already started to solve the inequality, I have to do the 6 possible cases for the values of d, k, l right? For example, d<k<l and and so on. My question now is how do I prove that the other two conditions are met? – Begginner Julia Feb 20 '21 at 06:28
  • $d, k l$ are points in a space, not real numbers. Statements like "$d < k < l$" make no sense as most metric spaces do not come with an order operator "$<$". – Paul Sinclair Feb 20 '21 at 14:41
  • So how could you approach the problem more generally? – Begginner Julia Feb 20 '21 at 16:07

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