Discussion
Consider the space R with the topology induced by the open base :
\(\{u / u = (a, b) \text{ or } (a, b) \setminus F\}\)
where \(F = \{y_{n} / n \in \mathbf{N}\}\).
The space R with this topology is :
Answer: D
The space is \(T_{2}\), which is the one property that holds for every choice of the points \(y_{n}\).
Every usual open interval \((a, b)\) is a basic open set, so this topology is finer than the usual topology of R. The usual topology is Hausdorff, and a finer topology keeps any separating neighbourhoods, so the new space is \(T_{2}\) and in particular \(T_{1}\). That rules out 'not \(T_{1}\)'.
Regularity depends on F. This is the construction of the K-topology, where \(F = \{1/n\}\): there the point 0 and the closed set F cannot be separated by disjoint open sets, because every open set containing F meets every neighbourhood of 0, so the space is Hausdorff but not regular. Since the stem allows such an F, 'regular' and 'both \(T_{2}\) and regular' cannot be asserted, and only \(T_{2}\) is guaranteed.
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