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Continum of numbers and Turing Machines
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Perry E. Metzger writes:
[Countability proofs deleted...]
For simplicity, lets just try to map the reals between zero and one to
the integers, and lets consider them expressed as binary numbers.
Imagine that I had built a mapping between this subset of the reals
and the positive integers. Any such mapping implies a list, that is,
that I could build a table like
1 .1010101101010010010010010101001.....
2 .0100001010100010100101001001010010...
3. .11000101001010110100010100010101001....
etc.
I can now construct a number that is not in the table. Take the first
binary digit from the first number in the table, and complement it. That
is the first digit in my constructed number. Take the second digit
from the second number and complement it -- that is the second digit
of the constructed number. Add in the complement of the third digit of
the third, the fourth digit of the fourth, etc. The number I have just
constructed can't be the first number in the imaginary table because
the first digit didn't match. It can't be the second because the
second didn't match. It can't be the third because the third doesn't
match. Indeed, it can't be any of them. Thus, you can't map the reals
to the integers.
The reals are thus in some sense a "bigger" infinite set than the
integers.
Small but important correction: the number that you contructed may in
fact be a binary equivalent to one already in the list.
Example:
.0111111...
.1000000...
Claim: For a given real x, there exist at most a finite number of
equivalent binary representations. (In fact, just 2.)
Proof: Left as an excercise.
I think everyone can see how to splice this little lemma into the
proof. Of course, the proof isn't nearly as clean as before, so it may
take more than 5 minutes for a 12 year old (or 12 minutes for a 5 year
old :-).
Dietrich Kappe
[email protected]
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