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Topic:
Lists
Replies:
7
Last Post:
Jan 24, 2013 5:06 PM




Re: Lists
Posted:
Jan 24, 2013 1:04 PM


"Don Deluise" <thedon@tomato.red> wrote in message news:1xh71bla4hvjd$.wf5fd9paee0w$.dlg@40tude.net... > This list enumerates all 2 bit binary sequences: > > 00 > 01 > 10 > 11 > > This one enumerates all 3 bit binary sequences: > > 000 > 001 > 010 > 011 > 100 > 101 > 110 > 111 > > Using diagonalization on the first to produce a sequence that is not in the > list fails. It produces '10' which *is* in the list. > > Using diagonalization on the second list also fails. It produces '111' > which *is* in the list. > > Clearly, increasing the length of the sequences does not produce lists in > which diagonalization will achieve its purpose, i.e. to produce binary > sequences of a given length which are not already in the list.
..at least, not while we are restricted to finite binary strings lengths, and there is no limit placed on the allowed list length.
We can have a sort of equivalent theorem for finite binary strings of length n:
Theorem: If (s_1, s_2, ...., s_n) is a sequence of nbit binary sequences, there will always be an nbit binary sequence not present in the sequence. Proof: usual diagonalisation argument
> > So if we were to compile an enumeration of infinite length binary > sequences, how do we know that diagonalization produces a sequence not > already in the list?
The difference is that with infinite length binary sequences, there is an obvious 11 correspondence that can be exploited between "entry number in the list" and "binary digit position". As you've observed above, the proof does not work without some such correspondence.
In Cantor's proof, the list length is deliberately restricted to being "countable"  in fact this is implicit in the definition of the term "list" used in the proof, a list being a map from N to the target set in question. If Cantor had allowed arbitrarily large "lists" (i.e. uncountable "lists") the proof would fail in the same way as your finite examples.
So, Cantor restricts consideration to "countable lists" (generally just called lists!) and the proof works. Similarly if we suitably restrict the list lengths in the finite case as I did above, the proof also works in the same way. (No surprise really I suppose...)
Regards, Mike.
> > P.S. This is not an attempt to prove that the Real numbers are either > countable or not. It is simply to raise a question about one of the proofs > that they are not



