Discussion
Let \(X = \{a, b, c\}\) and let \(\Im = \{\phi, \{a\}, \{b\}, \{a, b\}, X\}\) be a topology defined on \(X\). Then which of the following statements are TRUE?
P: \((X, \Im)\) is a Hausdorff space.
Q: \((X, \Im)\) is a regular space.
R: \((X, \Im)\) is a normal space.
S: \((X, \Im)\) is a connected space.
Answer: C
The space is normal and connected, so the answer is R and S.
The only open set containing \(c\) is \(X\) itself. So \(a\) and \(c\) cannot be separated by disjoint open sets, and the space is not Hausdorff (P fails).
The closed sets are \(X, \{b, c\}, \{a, c\}, \{c\}\) and \(\phi\). The point \(a\) and the closed set \(\{b, c\}\) cannot be separated, because any open set containing \(\{b, c\}\) is \(X\); so the space is not regular (Q fails).
Every non-empty closed set contains \(c\), so no two non-empty closed sets are disjoint and normality holds vacuously (R). In any split of \(X\) into two disjoint non-empty open sets, one of them would have to contain \(c\) and so be all of \(X\); hence the space is connected (S).
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