I want to prove this fact that if X is a topological space which is covered by a family of open subsets {U_i} than dimX=supdimU_i One direction I can see that the RHS is less than or equal to the that of LHS...but the other direction is I have not been able to prove..any help is appreciated

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  • See Keenan's answer here http://math.stackexchange.com/questions/140066/krull-dimension-of-a-scheme/140078#140078 – Seth Sep 11 '14 at 17:34
  • @Seth Is it clear that the question is about the topic of Keenan's answer, Krull dimension? Maybe it is covering dimension (tandra should clarify). – Stephen Sep 11 '14 at 17:44
  • here dim refers to the sup of the length of chain irreducible cloed proper subsets of X which corresponds to the krull dimension of its affine coordinate ring – tandra Sep 11 '14 at 18:05
  • So, the topological space you are considering is definitely a scheme then (and in fact an affine scheme?) – Siddharth Venkatesh Sep 11 '14 at 20:53
  • This is an exercise in Hartshorne's book. The question Seth linked to contains a proof. – Ayman Hourieh Sep 11 '14 at 22:24

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