Exercise 5.1
Solution:
(a) Closed. As it has no limit points, is closed. Clearly, it is not open. is not compact since is not bounded (Heine-Borel Theorem).
(b) Closed and compact. It is closed as zero is only limit point. It is also bounded and hence compact by Heine-Borel Theorem.
(c) Open and Closed. Not compact because is not bounded.
(d) None of them. is a limit point but not in it. Hence it is not closed. is a limit point of its complement but not in the complement. Hence its complement is not closed. Therefore it is not open. Because it is not closed so it is not compact.
(e) None of them.
(f) Closed and compact (because it is bounded).
Exercise 5.2
Solution:
(a) Let be an arbitrary open cover of . Then is an open cover of . Since is compact, we have a finite open cover of , where is a finite set.
Similarly, because is an open cover of and is compact, we have a finite open cover of , where is a finite set.
Let , then is finite and Hence Therefore is a finite open cover of . Since is arbitrary, we conclude that is compact.
(b) See Exercise 5.10.
Exercise 5.3
Solution:
(a) Let , then is not open.
(b) Let , then is not closed.
Exercise 5.4
Solution:
(a) If is closed, then is open. By Proposition 5.3, is also open. Hence is closed. By De Morgan’s law (Fact 5.11), we haveHence is closed.
(b) If is closed, then is open. By Proposition 5.3, is also open. Hence is closed. By De Morgan’s law (Fact 5.11), we haveHence is closed.
Exercise 5.5
Solution: Suppose is a limit of . Then there exists a sequence converging to such that for all . For any , since , we can find an such that . Since for , we conclude that is a point in the intersection of and -neighborhood of intersecting other than .
Conversely, suppose for any , there is a point in the intersection of and -neighborhood of other than . For each , there exists a point in the intersection of and -neighborhood of other than . Hence and . It is clear that the sequence converges to and for all . Thus is a limit point of .
Exercise 5.6
Solution: Letthen we have Taking any finitely many sets . Let . Then for all . Then . HenceNote that is a proper subset of . Therefore do not cover .
It says that for this special open cover there are no finite open covers for . Hence is not compact.
Exercise 5.7
Solution: If is closed and bounded, then every sequence from is bounded. Hence by Bolzano-Weierstrass Theorem, it has a convergent subsequence. Let be the limit of this convergent subsequence. If , then we are done. If , then . Hence is a limit point of . But is closed, we must have . We get a contradiction. Hence we always have .
If every sequence from has a subsequence that converges to a point in , we would like to show that is closed and bounded. Let be a limit point, then there exists a sequence from converging to . By assumption, this sequence has a subsequence converging to a point in . But by Proposition 3.32, this subsequence must converge to as well. Hence . This shows that contains its limit points. Hence by Theorem 5.10, is closed.
Now we show that is bounded. Suppose is not bounded, e.g. is not bounded above. Then for any , there exists such that . Hence the sequence from diverges to , so do its any subsequences by Exercise 3.23. Hence we get a contradiction. This implies that must be bounded, completing the proof.
Exercise 5.8
Solution: Since is compact, by Heine-Borel Theorem, is bounded and closed. Because , is bounded as well. Since is also closed, we conclude that is both closed and bounded. It follows from Heine-Borel Theorem that is compact.
This statement also holds for general topological space. A closed subspace of a compact space is compact.
Proof: Let be a compact space. Let be a closed subspace of . Let be an open cover of . Since is closed, it follows by definition of closed that is open in . So if we add to , we see that is also an open cover of . As is compact, there is a finite subcover of , say . This covers by the fact that it covers . If is an element of , then it can be removed from and the rest of still covers . Thus we have a finite subcover of which covers , and hence is compact.
Exercise 5.9
Solution: Let , then it is clear that is closed, see Example 5.7. Since is also bounded, by Heine-Borel Theorem, is compact. Let , then is bounded. Note that is not closed. This is because is a limit point of as . We have which is not closed. Hence by Heine-Borel Theorem is not compact. This shows that is not compact.
Exercise 5.10
Solution: Since is compact for each , by Heine-Borel Theorem, each is closed and bounded. Recall from Exercise 5.4 (b), the intersection is closed as well. Note that is a subset of for some . As is bounded, so is . Therefore is both closed and bounded, by Heine-Borel Theorem, is compact.
Exercise 5.11
Solution: Suppose is nonempty, open and closed. Let . Finally suppose that ; then there is some ; WLOG we can assume that .
Then the set is bounded below (by ) and nonempty (). Therefore exists.
- Suppose . Then since is open, it contains an open neighborhood . Recall the definition of , then there would be a sequence , , . This is not possible, because .
- Suppose . Then since is closed, its complement is open, therefore there is an open neighborhood contained in . We claim that otherwise which contradicts that is contained in . Therefore and this contradicts the definition of .
It follows that .
Exercise 5.12
Solution: Let . We show that is neither open nor closed.
is not closed. This is because is a limit point of as , but . It follows from Theorem 5.10 that is not closed.
is not open. Consider . Since , we conclude that is a limit point of . But , it follows from Theorem 5.10 that is not closed. Hence is not open by definition 5.6.