By L. Smith

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Clearly, B is closed under compositions; therefore, B is a subsemigroup of ZZ . Denote by C the closure of B in ZZ . Let us show that C is not closed under the product operation and, therefore, is not a subsemigroup of ZZ . For each k ∈ N, deﬁne fk : Z → Z as follows: fk (x) = x + k if x ∈ N, and f (α) = α. Let g be the constant mapping of Z to itself that brings each point of Z to the point α. Clearly, fk ∈ B, for each k ∈ N, and the sequence {fk (x) : k ∈ N} converges to g(x) in Z, for each x ∈ Z.

20. Fix an open neighbourhood U of the identity e in G such that U ∩ H = {e}. Since G is a topological group, there is a symmetric open neighbourhood V of e in G such that V 4 ⊂ U. 22 that the family η = {hV : h ∈ H} is discrete in G. Since η is a family of non-empty open subsets of the pseudocompact space G, it follows that the index set H in the deﬁnition of the family η is ﬁnite. 24. Every inﬁnite pseudocompact topological group G contains a non-closed countable subset. Proof. Take any inﬁnite countable subset A of G, and let H be the subgroup of G algebraically generated by A.

5. Therefore, H = K, that is, H is closed in G. 20. Every discrete subgroup H of a countably compact quasitopological group G is ﬁnite. Proof. 18, H is closed in G. Therefore, H is countably compact. Since H is discrete, it follows that H is ﬁnite. 21. Every discrete subgroup H of a Lindel¨of quasitopological group G is countable. 20 can be complemented in a non-trivial way. This requires a new concept. Suppose that U is a neighbourhood of the neutral element of a topological group G. A subset A of G is called U-disjoint if b ∈ / aU, for any distinct a, b ∈ A.