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Let $\{{A_n}\}$ be the closed subsets of $X$, such that ${A_n} \subset \operatorname{Int}{A_{n + 1}}$ and $ \cup {A_n} = X$, if $A_1$ and all $\operatorname{cl}(A_n-A_{n-1})$ have the covering dimension at most $m$, does $X$ have the covering dimension at most $m$?

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"\rm{Int}" A doesn't look the same as "\operatorname{Int} A"; the latter provides proper spacing before and after "Int": $\operatorname{Int} A$. But curly braces around the first part wrecks that: {\operatorname{Int}} A: ${\operatorname{Int}} A$. When you figure out why that's how it ought to work, you'll understand something about typesetting. (This distinction between the effects of \rm and of \operatorname doesn't work on Wikipedia, but I think that may change when they adopt mathJax in a few months. –  Michael Hardy Apr 19 '12 at 15:27
    
Your example should be {\rm Int} A rather than \rm{Int} A, since \rm doesn't take a parameter — it is simply a switch that changes the current font. –  Davide Cervone Apr 20 '12 at 12:15

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