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September 9[edit]

Triangle inequality in ultrametric spaces[edit]

Is the inequality $d(x,z)\leq max(d(x,y),d(y,z))$ in an ultrametric space always an equality when $d(x,y)\neq d(y,z)$ ? GeoffreyT2000 (talk) 02:44, 9 September 2016 (UTC)[reply]

Yes. According to the article (and it's straightforward to prove), for any three points x, y, z, at least one of d(x,y)=d(x,z), d(x,y)=d(y,z), d(x,z)=d(y,z) is true. Equivalently, if d(x,y)≠d(y,z) then d(x,z) must be either d(x,y) or d(y,z). Suppose the first case, d(x,z)=d(x,y). Then d(y,z)≤max(d(x,z), d(x,y)) = d(x,y), max(d(x,y), d(y,z))=d(x,y)=d(x,z) as required. The other case is similar. --RDBury (talk) 08:10, 9 September 2016 (UTC)[reply]

Description of an inequality[edit]

In graph theory we have the following inequality: connectivity <= edge connectivity <= minimum vertex degree of the graph. You can't improve on this inequality: you can construct a graph with any set of three integers that satisfy it. How would you describe this type of inequality that can't be improved on? 24.255.17.182 (talk) 22:50, 9 September 2016 (UTC)[reply]

Sharp inequality Sławomir Biały (talk) 22:58, 9 September 2016 (UTC)[reply]

Is that the answer? The property being discussed, described in Connectivity (graph theory)#Bounds on connectivity, is that κ(G) ≤ λ(G) ≤ δ(G) for any graph G. The fact that there exists some graph G with κ(G) = λ(G) = δ(G) is sufficient for the inequality to be sharp. But isn't our questioner going beyond this and stating that for any positive integers k, l, d where k ≤ l ≤ d, a graph G can be constructed such that κ(G) = k, λ(G) = l, δ(G) = d, and asking for a name for such an inequality? -- ToE 12:51, 10 September 2016 (UTC)[reply]

Sharp inequality sounds good to me. Dmcq (talk) 22:45, 10 September 2016 (UTC)[reply]