Proof.
It is clear that (a) implies (b). Assume (b). Let x \in X, x \not\in T. Let x \in C \subset X be the connected component of X containing x. By Lemma 5.12.10 we see that C = \bigcap V_\alpha is the intersection of all open and closed subsets V_\alpha of X which contain C. In particular, any pairwise intersection V_\alpha \cap V_\beta occurs as a V_\alpha . As T is a union of connected components of X we see that C \cap T = \emptyset . Hence T \cap \bigcap V_\alpha = \emptyset . Since T is quasi-compact as a closed subset of a quasi-compact space (see Lemma 5.12.3) we deduce that T \cap V_\alpha = \emptyset for some \alpha , see Lemma 5.12.6. For this \alpha we see that U_\alpha = X \setminus V_\alpha is an open and closed subset of X which contains T and not x. The lemma follows.
\square
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Comment #8514 by Elías Guisado on
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