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2.12 Regard Q, the set of all rational numbers, as a metric space, with d(p, q) = |p − q|. Let E be the set of all p ∈ Q such that 2 < p2 < 3. Show that E is closed and bounded in Q, but that E is not compact. Is E open in Q? Rudin’s Ex. 16 Proof Since for any p ∈ E, we have 1 < p, since otherwise 1 ≥ p2 , which contradicts to the definition of E. Similarly, we have p < 2. Hence E is bounded. To show that E is closed in Q, we assume√x ∈ Q is a limit point of E. Apparently, x2 6= 2 and x2 6= 3. If x2 < 2, we let r1 = 2 − |x| > 0. If z ∈ Nr1 (x), then |z| ≤ |x − z| + |x| < r1 + |x| = √ 2, 2 which implies that z < 2, or z ∈ / E. This contradicts to the assumption that x is a limit point of E. Thus, necessarily 2 < x2 . Similarly, if x2 > 3, we let √ r2 = |x| − 3 > 0. For any z ∈ Nr2 (x), since √ |z| ≥ |x| − |x − z| > |x| − r2 = 3, we have z ∈ / E, which again contradicts to the assumption that x is a limit point of E. This implies x2 < 3. Hence, 2 < x2 < 3. Therefore, we conclude that if x ∈ Q is a limit point of E, then x ∈ E. This means that E is closed in Q. To show that E is not compact, we take the collection {Un }∞ n=3 , where Un = Q ∩ −(3 − n1 )1/2 , −(2 + n1 )1/2 ∪ (2 + n1 )1/2 , (3 − n1 )1/2 , n ≥ 3. x ∈ Q since Q is the “whole” space. We should first argue that if x2 = 3, then x∈ / Q. √ √ 2 and 3 are well-defined by Theorem 1.21. Another way to show E is closed is to prove Ec = Q − E = √ Q ∩√{(−∞, √ − √3) ∪ (− 2, 2) ∪ ( 3, ∞)}. E is bounded and closed in Q, but not in R. So, it is not necessarily compact. Each Un is open in Q by Theorem 2.30. For any x ∈ E ⊂ Q, since 2 < x2 < 3, there is a positive integer n ≥ 3, such that 2+ 1 n < x2 < 3 − n1 , by the Archimedean property. Hence x ∈ Un . It follows that the collection {Un }∞ n=3 forms an open cover of E. Let {Un }n∈I , |I| < ∞, be any finite subcollection. Put N = maxn∈I {n}. Then, since Q is dense in R by Theorem 1.20, there exists a rational number r satisfying 3 − N1 < r2 < 3. From this construction,we know that r ∈ E, but r ∈ / ∪n∈I Un . Hence, any finite subcollection of {Un }∞ n=3 does not cover E. We conclude that E is not compact. Finally, we show that E is open in Q.√ Suppose x ∈ √ E ⊂ Q, with 2 < x2 < 3. Take a number r such that 0 < r < min{ 3 − |x|, |x| − 2}. Consider a neighborhood Nr (x) in Q: Nr (x) = {z ∈ Q : |z − x| < r}. For any z ∈ Nr (x), we have √ 3 − |x| + |x| = 3, √ √ |z| ≥ |x| − |z − x| > |x| − (|x| − 2) = 2. |z| ≤ |z − x| + |x| < √ Hence Nr (x) ⊂ E. By the definition, E is open. 1 Or show E is open by proving √ √ E√= Q∩{(− 3, − 2)∪ √ ( 2, 3)}.