Mathematics:Teaching, Learning and Exploring.
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(Part C Question: NET June-2015)Consider the set \(\mathbb{Z} \) of integers, with the topology \(\tau \) in which a subset is closed if and only if it is empty, or \(\mathbb{Z} \) , or finite. Which of the following statements are true?
A. \(\tau \) is the subspace topology induced from the usual topology on \(\mathbb{R} \).
B. \(\mathbb{Z} \) is compact in the topology \(\tau \)
C. \(\mathbb{Z} \) is Hausdorff in the topology \(\tau \).
D. Every infinite subset of \(\mathbb{Z}\) is dense in the topology \(\tau \).
\(\textbf{Options B and D}\)
\(U \) is open in \(\tau \iff |\mathbb{Z}-U| < \infty \)
Option A: The set \((-1,1) \cap \mathbb{Z} =\{0\} \) is open in induced topology from the usual topology on \(\mathbb{R} \) but not in the topology \(\tau \).
Option B: Let \(\mathcal C\) be an open cover of \(\mathbb{Z}\).Let \(U \in \mathcal C\) be any set in \(C\). It covers all but a finite number of points of \(\mathbb{Z} \).So for each of those points we pick an element of \(\mathcal{C}\) which covers each of those points.
Hence we have a finite subcover of \(\mathbb{Z} \)
Option C: \(\mathbb{Z} \) is not Hausdorff in \(\tau \) as no two open sets in \(\tau \) are disjoint.
Option D: Since complement of any open set in \(\tau \) is finite, every infinite set in \(\mathbb{Z} \) intersects any open set in \(\tau \).
\(\textbf{Remark:} \) These results can be generalised to co-finite topology on any infinite set.
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