Cantor's first set theory article: Difference between revisions

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| style="padding-left: 1em; padding-right: 1em" |The denseness of sequence ''P'' is used to [[recursively define]] a nested sequence of intervals that excludes all of the numbers in ''P''. The [[base case (recursion)|base case]] starts with the interval (''a'',&nbsp;''b''). Since ''P'' is dense in [''a'',&nbsp;''b''], there are infinitely many numbers of ''P'' in (''a'',&nbsp;''b''). Let ''x''_''k''₁ be the number with the least index and ''x''_''k''₂ be the number with the next larger index, and let ''a''₁ be the smaller and ''b''₁ be the larger of these two numbers. Then, ''k''₁&nbsp;<&nbsp;''k''₂, {{nowrap|''a'' < ''a''₁ < ''b''₁ < ''b''}}, and (''a''₁,&nbsp;''b''₁) is a [[proper subinterval]] of (''a'',&nbsp;''b''). Also, {{nowrap|''x''_''m'' ∉ (''a''₁,&nbsp;''b''₁)}} for ''m''&nbsp;≤&nbsp;''k''₂ since these ''x''_''m'' are the endpoints of (''a''₁,&nbsp;''b''₁). Repeating the above proof with the interval (''a''₁,&nbsp;''b''₁) produces ''k''₃, ''k''₄, ''a''₂, ''b''₂ such that {{nowrap|''k''₁ < ''k''₂ < ''k''₃ < ''k''₄}} and {{nowrap|''a'' < ''a''₁ < ''a''₂ < ''b''₂ < ''b''₁ < ''b''}} and {{nowrap|''x''_''m'' ∉ (''a''₂, ''b''₂)}} for ''m''&nbsp;≤&nbsp;''k''₄.<ref group=proof name=p/>

The [[recursive step]] starts with the interval {{nowrap|(''a''_''n''–1, ''b''_''n''–1)}}, the inequalities {{nowrap|''k''₁ < ''k''₂ < . . . < ''k''_{2''n''–2–3} < ''k''_{2''n''–1–2}}} and {{nowrap|''a'' < ''a''₁ < . . . < ''a''_''n''–1 < ''b''_''n''–1 . . . < ''b''₁ < ''b''}}, and the fact that the interval {{nowrap|(''a''_''n''–1, ''b''_''n''–1)}} excludes the first 2''n''{{space|hair|2}}–2 members of the sequence {{nowrap|''P''{{space|hair|2}}—{{space|hair|2}}that}} is, {{nowrap|''x''_''m'' ∉ (''a''_''n''–1, ''b''_''n''–1)}} for ''m''&nbsp;≤&nbsp;''k''_2''n''–2. Since ''P'' is dense in [''a'',&nbsp;''b''], there are infinitely many numbers of ''P'' in {{nowrap|(''a''_''n''–1, ''b''_''n''–1)}}. Let ''x''_{''k''_{2''n''{{space|hair|2}}–1}} be the number with the least index and ''x''_{''k''_2''n''} be the number with the next larger index, and let ''a''_''n'' be the smaller and ''b''_''n'' be the larger of these two numbers. Then, ''k''_{2''n''{{space|hair|2}}–1}&nbsp;<&nbsp;''k''_2''n'', {{nowrap|''a''_''n''–1 < ''a''_''n'' < ''b''_''n'' < ''b''_''n''–1}}, and (''a''_''n'',&nbsp;''b''_''n'') is a [[proper subinterval]] of {{nowrap|(''a''_''n''–1, ''b''_''n''–1)}}. Combining these inequalities with the ones for step ''n''{{space|hair|2}}–1 of the recursion produces {{nowrap|''k''₁ < ''k''₂ < . . . < ''k''_2''n''–1 < ''k''_2''n''}} and {{nowrap|''a'' < ''a''₁ < . . . < ''a''_''n'' < ''b''_''n'' . . . < ''b''₁ < ''b''}}. Also, {{nowrap|''x''_''m'' ∉ (''a''_''n'',&nbsp;''b''_''n'')}} for ''m''&nbsp;=&nbsp;''k''_2''n''–1 and ''m''&nbsp;=&nbsp;''k''_2''n'' since these ''x''_''m'' are the endpoints of (''a''_''n'',&nbsp;''b''_''n''). This together with {{nowrap|(''a''_''n''–1, ''b''_''n''–1)}} excluding the first 2''n''{{space|hair|2}}–2 members of sequence ''P'' implies that the interval (''a''_''n'',&nbsp;''b''_''n'') excludes the first 2''n'' members of {{nowrap|''P''{{space|hair|2}}—{{space|hair|2}}that}} is, {{nowrap|''x''_''m'' ∉ (''a''_''n'', ''b''_''n'')}} for ''m''&nbsp;≤&nbsp;''k''_2''n''. Therefore, for all ''n'', {{nowrap|''x''_''n'' ∉ (''a''_''n'', ''b''_''n'')}} since ''n''&nbsp;≤&nbsp;''k''_2''n''.<ref group=proof name=p/>

The sequence ''a''_''n'' is increasing and [[Upper and lower bounds|bounded above]] by ''b'', so the limit ''A''&nbsp;=&nbsp;lim_{''n''&nbsp;→&nbsp;∞}&nbsp;''a''_''n'' exists. Similarly, the limit ''B''&nbsp;=&nbsp;lim_{''n''&nbsp;→&nbsp;∞}&nbsp;''b''_''n'' exists since the sequence ''b''_''n'' is decreasing and [[Upper and lower bounds|bounded below]] by ''a''. Also, ''a''_''n''&nbsp;<&nbsp;''b''_''n'' implies ''A''&nbsp;≤&nbsp;''B''. If ''A''&nbsp;<&nbsp;''B'', then for every ''n'': ''x''_''n''&nbsp;∉&nbsp;(''A'',&nbsp;''B'') because ''x''_''n'' is not in the larger interval (''a''_''n'',&nbsp;''b''_''n''). This contradicts ''P'' being dense in [''a'',&nbsp;''b'']. Hence, ''A''&nbsp;=&nbsp;''B''. For all ''n'', {{nowrap|''A'' ∈ (''a''_''n'',&nbsp;''b''_''n'')}} but {{nowrap|''x''_''n'' ∉ (''a''_''n'',&nbsp;''b''_''n'')}}. Therefore, ''A'' is a number in [''a'',&nbsp;''b''] that is not contained in ''P''.<ref group=proof name=p/>