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Let $X$ be a topological space. If $X = \bigcup U_\alpha$ is an open covering of $X$ then $$\dim X = \sup_\alpha \dim U_\alpha.$$ Now suppose that $X = \coprod U_\alpha$, i.e., $X$ is the disjoint union of the $U_\alpha$'s, but the $U_\alpha$'s are NOT open (the most we can assume is that they are locally closed). Can we infer anything about the dimension of $X$ from the dimensions of the disjoint components, other than the dimension of $X$ has to be greater or equal to the dimension of each component?

We can assume that the decomposition is finite, in fact, assume that there just two components.

By dimension I mean the supremum of lengths of chains of closed irreducible subsets, i.e., the notion of dimension used in algebraic geometry for example.

In fact I am interested in a spacial case of the situation above. For me $X = G/B$ is a flag variety. Then $G/B$ has a Schubert cell decomposition. Now suppose that I consider a subspace $Y$ composed of several Schubert cells. I conjecture that the dimension of $Y$ is equal to the dimension of highest-dimensional Schubert cell in $Y$. Is this true?

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    $\begingroup$ By $\textrm{dim}$, do you mean the covering dimension? $\endgroup$ – Unwisdom Mar 7 '14 at 21:12

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