# DFCC (discrete finite chain condition )

A space $X$ has the DFCC (discrete finite chain condition ) provided every discrete family of nonempty open sets is finite. (see A survey on star covering propertiesMV Matveev - Topology Atlas, preprint, 1998)

Does the discrete family of nonempty open sets mean that every point $x \in X$ has a nbhd that intersects at most one set of the given family?

However, in the proof of the Theorem 14 in the charpter 2, why he said if the sequence $S=\{U_n:n\in \omega\}$ is not DFCC, then there exists $x \in X$ is an accumulation point for $S$, each nbhd of $x$ meets infinitely many $U_n$'s? Why is "infinitely"? This is puzzling me.

Thanks for any help:)

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Does the discrete family of nonempty open sets mean that every point $x\in X$ has a nbhd that intersects at most one set of the given family?
It seems that there is an error in the proof exposition. The contradiction is already implied from the fact that a neighborhood $V$ intersects two different $U_n$-s. Therefore no such neighborhood of $x$ can exist and the sequence $s$ should be discrete.