Math 140A Homework 5 (Solutions)
نویسنده
چکیده
for each n ∈ N. Note that each On is open since it is a union of open sets (cf. Theorem 2.24(a) in Rudin). We claim that A = ⋂∞n=1 On. First, if a ∈ A, then a ∈ (a− 1/n, a+ 1/n) for each n ∈ N, so a ∈ On for each n ∈ N, and hence a ∈ ⋂∞n=1 On. Thus, A ⊆ ⋂∞n=1 On. Conversely, assume x ∈ R and x / ∈ A. Since A is closed, it follows that x is not a limit point of A, so there exists an ε > 0 such that for all a ∈ A we have |x − a| ≥ ε. By the archimedean property, there exists an n ∈ N such that 1/n < ε. It follows that |x− a| ≥ ε > 1/n for each a ∈ A, so x / ∈ (a− 1/n, a+ 1/n) for any a ∈ A. Thus, x / ∈ On, and hence x / ∈ ⋂∞ n=1 On. Therefore, ⋂∞ n=1 On ⊆ A, and hence A = ⋂∞n=1 On.
منابع مشابه
Math 140A Homework 2 (Solutions)
where the coefficients are real numbers and bm 6= 0. (i) Define addition and multiplication of two elements of F to be the usual addition and multiplication of functions. Show that with this addition and multiplication, F is a field. (ii) We can define an order on F as follows. A rational function like (1.1) is positive if and only if an and bn have the same sign, i.e., anbm > 0. Now given two ...
متن کاملMath 140A Homework 7 (Solutions)
Solution. (a) We will prove the contrapositive. Suppose ⋃i∈I Ai is not connected. Then there exist nonempty separated sets E,F ⊆ X such that ⋃ i∈I Ai = E ∪ F . Since ⋂ i∈I Ai is nonempty, we may pick an x ∈ ⋂ i∈I Ai. Since ⋂i∈I Ai ⊆ ⋃i∈I Ai = E ∪ F , it follows that x ∈ E ∪ F . Without loss of generality, assume x ∈ E. Then x ∈ Ai ∩ E for all i ∈ I. Moreover, F is nonempty, so there exists a y ...
متن کامل