Siz.1 Cantor's Zig-Zag Method

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siz.1 Cantor's Zig-Zag Method sfr:siz:zigzag: We've already considered some \easy" enumerations. Now we will consider explanation sec something a bit harder. Consider the set of pairs of natural numbersdefined by: N × N = fhn; mi : n; m 2 Ng We can organize these ordered pairs into an array, like so: 0 1 2 3 ::: 0 h0; 0i h0; 1i h0; 2i h0; 3i ::: 1 h1; 0i h1; 1i h1; 2i h1; 3i ::: 2 h2; 0i h2; 1i h2; 2i h2; 3i ::: 3 h3; 0i h3; 1i h3; 2i h3; 3i ::: : : : : : : : : : : : :: Clearly, every ordered pair in N × N will appear exactly once in the array. In particular, hn; mi will appear in the nth row and mth column. But how do we organize the elements of such an array into a \one-dimensional" list? The pattern in the array below demonstrates one way to do this (although of course there are many other options): 0 1 2 3 4 ::: 0 0 1 3 6 10 ::: 1 2 4 7 11 ::: ::: 2 5 8 12 ::: ::: ::: 3 9 13 ::: ::: ::: ::: 4 14 ::: ::: ::: ::: ::: : : : : : : : : : : : ::: :: This pattern is called Cantor's zig-zag method. It enumerates N × N as follows: h0; 0i; h0; 1i; h1; 0i; h0; 2i; h1; 1i; h2; 0i; h0; 3i; h1; 2i; h2; 1i; h3; 0i;::: And this establishes the following: sfr:siz:zigzag: Proposition siz.1. N × N is enumerable. natsquaredenumerable Proof. Let f : N ! N × N take each k 2 N to the tuple hn; mi 2 N × N such that k is the value of the nth row and mth column in Cantor's zig-zag array. This technique also generalises rather nicely. For example, we can use it to explanation enumerate the set of ordered triples of natural numbers, i.e.: N × N × N = fhn; m; ki : n; m; k 2 Ng We think of N × N × N as the Cartesian product of N × N with N, that is, 3 N = (N × N) × N = fhhn; mi; ki : n; m; k 2 Ng zig-zag rev: c8c9782 (2021-09-28) by OLP/ CC{BY 1 and thus we can enumerate N3 with an array by labelling one axis with the enumeration of N, and the other axis with the enumeration of N2: 0 1 2 3 ::: h0; 0i h0; 0; 0i h0; 0; 1i h0; 0; 2i h0; 0; 3i ::: h0; 1i h0; 1; 0i h0; 1; 1i h0; 1; 2i h0; 1; 3i ::: h1; 0i h1; 0; 0i h1; 0; 1i h1; 0; 2i h1; 0; 3i ::: h0; 2i h0; 2; 0i h0; 2; 1i h0; 2; 2i h0; 2; 3i ::: : : : : : : : : : : : :: Thus, by using a method like Cantor's zig-zag method, we may similarly obtain an enumeration of N3. And we can keep going, obtaining enumerations of Nn for any natural number n. So, we have: Proposition siz.2. Nn is enumerable, for every n 2 N. Problem siz.1. Show that (Z+)n is enumerable, for every n 2 N. Problem siz.2. Show that (Z+)∗ is enumerable. You may assume Prob- lem siz.1. Photo Credits Bibliography 2.

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