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Undid revision 1096011298 by 2603:6010:B002:3E4:57:9234:1D65:EA1C (talk) Cantor's proof starts by assuming the negation of the thing to be proved, and proceeds as in any proof by contradiction. |
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Either the number of intervals generated is finite or infinite. If finite, let (''a''<sub>''L''</sub>, ''b''<sub>''L''</sub>) be the last interval. If infinite, take the [[Limit of a sequence|limit]]s ''a''<sub>∞</sub> = lim<sub>''n'' → ∞</sub> ''a''<sub>''n''</sub> and ''b''<sub>∞</sub> = lim<sub>''n'' → ∞</sub> ''b''<sub>''n''</sub>. Since ''a''<sub>''n''</sub> < ''b''<sub>''n''</sub> for all ''n'', either ''a''<sub>∞</sub> = ''b''<sub>∞</sub> or ''a''<sub>∞</sub> < ''b''<sub>∞</sub>. Thus, there are three cases to consider:
{{Anchor|Case1}}[[File:Cantor's first uncountability proof Case 1 svg.svg|thumb|350px|alt=Illustration of case 1. [[Real line]] containing closed interval [''a'', ''b''] that contains nested open intervals (''a''<sub>''n''</sub>, ''b''<sub>''n''</sub>) for ''n'' = 1 to ''L''. Two distinct numbers ''y'' and one ''x''<sub>''n''</sub> are in (''a''<sub>''L''</sub>, ''b''<sub>''L''</sub>).|Case 1: Last interval (''a''<sub>''L''</sub>, ''b''<sub>''L''</sub>)]]
:Case 1: There is a last interval (''a''<sub>''L''</sub>, ''b''<sub>''L''</sub>). Since at most one ''x''<sub>''n''</sub> can be in this interval, every ''y'' in this interval except ''x''<sub>''n''</sub> (if it exists) is not
{{Anchor|Case2}}[[File:Cantor's first uncountability proof Case 2 svg.svg|thumb|350px|alt=Illustration of case 2. Real line containing interval [''a'', ''b''] that contains nested intervals (''a''<sub>''n''</sub>, ''b''<sub>''n''</sub>) for ''n'' = 1 to ∞. These intervals converge to ''a''<sub>∞</sub>.|Case 2: ''a''<sub>∞</sub> = ''b''<sub>∞</sub>]]
:Case 2: ''a''<sub>∞</sub> = ''b''<sub>∞</sub>. Then ''a''<sub>∞</sub> is not
{{Anchor|x_n not in}}
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