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# connect3 - 3 Example 12 In Example 8 we saw that C1 =[0 1...

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3 Example 12. In Example 8, we saw that C 1 = [0 , 1] in the metric space [0 , 1] (2 , 3) . Fol- lowing Lemma 9, this means that for any point x [0 , 1] , C x = [0 , 1] (indeed, this is easy to show directly). On the other hand, if y (2 , 3) , the reader can readily verify that C y = (2 , 3) . Hence, the metric space [0 , 1] (2 , 3) has two connected components: [0 , 1] and (2 , 3) . Example 13. Consider the metric space Q equipped with the usual metric d ( q, p ) = | q - p | . Let q 0 be any non-zero rational number. By the Archimedean property of Q , there exists a natural number n so that n · q 2 0 > 1 . Since n 2 + 1 > n , we therefore have ( n 2 + 1) · q 2 0 > 1 . Now, define A = { p Q ; p 2 > 1 1+ n 2 } and B = { p Q ; p 2 < 1 1+ n 2 } . By definition, q 0 A . Note, 1 + n 2 is not a perfect square (since n > 0 here), and therefore (following standard proofs like the one in the first lecture) 1 1+ n 2 is not the square of any rational number. It follows that Q = A B . What’s more, if p A , then there is a small rational distance ω so that the interval ( p - ω , p + ω ) Q is contained in A = B c
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