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Some notes on diagonal method Tags: Reverted possible vandalism Visual edit |
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Then all its elements can be written in an enumeration ''s''<sub>1</sub>, ''s''<sub>2</sub>, ... , ''s''<sub>''n''</sub>, ... .
Applying the previous lemma to this enumeration produces a sequence ''s'' that is a member of ''T'', but is not in the enumeration. However, if ''T'' is enumerated, then every member of ''T'', including this ''s'', is in the enumeration. This contradiction implies that the original assumption is false. Therefore, ''T'' is uncountable.<ref name="Cantor.1891"/>
'''Review please:'''
'''Notes''' The diagonal method is not applicable for the issue above. Consider the set ''T'' of all sequences of binary digit ''1s''. (Sequences might be extended with trailing ''0s'' for readability and/or to be infinite.) Let the following enumeration of elements from ''T'':
''s''<sub>1</sub> = ()
''s''<sub>2</sub> = (1)
''s''<sub>3</sub> = (1, 1)
''s''<sub>4</sub> = (1, 1, 1)
''s''<sub>5</sub> = (1, 1, 1, 1)
''s''<sub>6</sub> = (1, 1, 1, 1, 1)
''s''<sub>7</sub> = (1, 1, 1, 1, 1, 1)
...
The sequence ''s'' = (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...), by construction, is a member of ''T'' that differs from each ''s<sub>n</sub>'', since their ''n''<sup>th</sup> digits/or number of digit ''1s'' differ. Hence, ''s'' cannot occur in the enumeration.
Since there is a clear bijection between ''T'' and ''<math>{\mathbb N}</math>'', it could be also shown that ''<math>{\mathbb N}</math>'' were uncountable. As no original assumptions have been set, therefore the method of proof has to be inaccurate: the creation of diagonal element assumes that such a comparison of the ''n''<sup>th</sup> digits/or the counting of digit ''1s'' are valid for such infinite sequences, however they are not.
=== Real numbers ===
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