I'm reading this book (Hartshorne), and it uses a funny definition of topological dimension, which I'm having a hard time convincing myself is the usual one. The definition is as follows:(adsbygoogle = window.adsbygoogle || []).push({});

dim X is the supremum of natural numbers such that there exists a chain [itex]Z_0\subset Z_1\subset \dotsb \subset Z_n[/itex] of distinct closed irreducible subsets of X.

And a subset Y of X is defined to be irreducible if it can expressed as the union of two proper subsets which are closed in Y.

Compare this with the normal definition that you learn in Munkres:

The dimension of a space X is one less than the infimum of the orders of refinements of open covers.

OK, so Hartshorne has some examples using his definitions, but only using the Zariski topology. That's fine, and the results agree with what you expect. But I wanted to prove that the Hartshorne definition will give the same result as the Munkres definition for, eg, the real line with its standard topology.

It seems to me that there simply are no closed irreducible sets in the real line with its standard topology (while there are closed irreducible sets in the Zariski topology). For take any closed interval, it can be written as the union of two closed intervals. Can I conclude then that the Hartshorne dimension of the standard real line is zero?

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# Funny definition of dimension of a topological space

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