If $(X,T_d)$ is a complete metric space and we have a sequence of closed subsets ${K_n}$ s.t. $\operatorname{int}(K_n)$ is empty, then


I searched about Baire's theorem but it said something different, with open sets and the intersection, is it the negation of Baire's theorem?

Thanks in advance

  • $\begingroup$ They are equivalent statements. $\endgroup$ – HeMan Dec 24 '16 at 18:24

The Baire category theorem states the following:

If $(X, d)$ is a complete metric space and $U_n$ a sequence of dense open sets, the their intersection $\bigcap U_n$ is also dense.

Now set $U_n = K_n^c$. Then $U_n$ is a sequence of open sets and $\overline{U_n} = \overline{K_n^c} = ((K_n)^\circ)^c = \emptyset ^c = X$. Now from the Baire category theorem we know $$X = \overline{\bigcap U_n} = \overline{\big(\bigcup K_n\big)^c} = \Big(\big(\bigcup K_n\big)^\circ\Big)^c,$$ i.e. $\big(\bigcup K_n\big)^\circ = \emptyset$.


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