Could someone please check my work and make sure I've correctly used the included theorems and definitions

Please state all definitions and theorems that you will need: Theorem 3.4.7 (a) A set S is open iff S = intS . Equivalently, S is open iff every point in S is an interior point of (b) A set S is closed iff its complement IR\\ S is open. -...m Theorem 3.5.5 (Heine-Borel) A subset S of IR is compact iff S is closed and bounded. Definition 2.4.3 A set S is said to be finite if S = 0 or if there exists n E N and a bijection f: {1, 2, . ., n} - S . If a set is not finite, it is said to be infinite. Definition 2.4.6 A set S is said to be denumerable if there exists a bijection f: N - S . If a set is finite or denumerable, it is called countable. If a set is not countable , it is uncountable. The cardinal number of a denumerable set is denoted by No . Theorem 2.4. 12 The set I of real numbers is uncountable. Let f: D - R be continuous. For each of the following, prove or give a counterexample. 1. If D is open, then f(D) is open. What happens to D under f? counterexample: f(a) = lac| D = R which is open by Theorem 3.4.7(a) since intR = R , specifically it is both open and closed by Theorem 3.4.7. It is closed because bdR = 0 and Q C R . f(D) exists by the definition of the image of D in R. f(D) = [0, co) which is not open by Theorem 3.4.7(a), specifically it is neither open or closed by Theorem 3.4.7. So, continuous functions don't necessarily map open sets to open sets. 2. If D is not compact, then f(D) is not compact. What happens to D under f ? counterexample: f(x) = sin x Er.P.8 matondl D = (-27, 27) which is not closed by Theorem 3.4.7(b) and thus not compact by The Heine-Borel Theorem 3.5.5. f(D) exists by the definition of the image of D in R . f(D) = [- 1, 1] which is closed by Theorem 3.4.7(b) and bounded by the definition of boundedness and thus compact by the Heine Borel Theorem 3.5.5 and thus not not compact. So, continuous functions don't necessarily map non-compact sets to non-compact sets. 3. If D is infinite, then f(D) is infinite. What happens to D under f? counterexample: f(x) = 2 D = R which is infinite by definition 2.4.3, since it is not finite. It is not finite by Theorem 2.4. 12 and definition 2.4.6 since uncountable sets such as IR are not finite sets. f(D) exists by the definition of the image of D in R . f(D) = {2} which is finite by definition 2.4.3 and the definition of equinumerous as it consists of a single point that is equinumerous with {1}. So, continuous functions don't necessarily map infinite sets to infinite sets