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A topological space X is regular if :

  • A.Singleton sets are closed sets
  • B.For each pair \(x \in X\) and an open set B in X with \(x \notin B\) there exists an open set U such that \(x \in U\) and \(u \cap B = \phi\)
  • C.For each pair \(x \in X\) and a closed set B in X with \(x \notin B\) there exists an open set U such that \(x \in U\) and \(u \cap B = \phi\)
  • D.For each pair \(x \in X\) and a closed set B in X with \(x \notin B\) there exist an open sets U, V such that \(x \in U\), \(B \subset V\) and \(U \cap V = \phi\)

Answer: D

X is regular if each point and each closed set not containing it have disjoint open neighbourhoods.

That is the definition: for every \(x \in X\) and closed B with \(x \notin B\), there are open sets U and V with \(x \in U\), \(B \subset V\) and \(U \cap V = \phi\).

Closed singletons describe the T1 axiom, a different property. The two options asking only for an open U with \(x \in U\) disjoint from B are too weak: when B is closed, the complement of B is such a U in every space, so the condition holds trivially and separates nothing; when B is open, the condition is not regularity at all.

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