By McInnes L.

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The catch now is that we need to show not just that we are incapable of finding a one-to-one correspondence between these points on the continuum and the natural numbers, but that no such correspondence can exist. We do this by the somewhat backwards approach of assuming there is such a correspondence, and then showing that a logical contradiction would result. From that we can conclude that any such correspondence would be contradictory, and thus can’t actually exist (at least not in any system that doesn’t have contradictions).

More formally, if we have an infinite sequence S1 , S2 , S3 , . . then we require that for any > 0 there exists an integer N ≥ 1 such that, for all m, n ≥ N , |Sn − Sm | < (recalling that |x − y| gives the distance between numbers x and y). Such a sequence is called a Cauchy sequence. Now, since any Cauchy sequence converges to something, we can identify (consider equivalent) the sequence and the point at its limit. Furthermore, since we know that using fractions we can get arbitrarily close to any point on the continuum, there must be some sequence of fractions that converges to that point, and so if we consider all the possible infinite Cauchy sequences of fractions, we can cover all the points on the continuum – we are assured that no holes or gaps can slip in this time.

A 3. b 4. ab 5. ba 6. aba 45 Permutations and Applications and anything with four or more as and bs will be reducible. Why is that? Since aa = · and bb = · any sequence will have to alternate as and bs, otherwise we can just cancel down consecutive pairs. On the other hand, if we have a sequence of more than three alternating as and bs then we’ll have a sequence aba or bab that we can convert using the fact that aba = bab, and end up with a pair of consecutive as or bs that we can then cancel down.

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The narrow road to the interior: A mathematical journey by McInnes L.


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