Re: An infinite debate


Tim Peters wrote:
[Ajeet]
Hi All,

Consider the following set :

S = U S_i (0<=i<infinity and U denotes the set operation of union)

where S_i is the set of all numbers, whose decimal representations are
of length i. For example the number 3.4 belongs to S_2. (assume the
decimal does not add to the length, so 34 would also belong to S_2).

Then this consequence follows:

If r is an element of S, r's decimal representation is finite.

Simply because there must be an i such that r is an element of S_i, in which
case r's decimal representation has i digits.

Argument : S = R (set of all real numbers)
========

Good luck ;-)

Proof:
=====
Suppose a real number "r" does not exist in S. let the decimal
representation of "r" be d_0.d_1d_2d_3......(to infinity) where d_i is
the ith digit.
+Since r does not belong to S, there will be some digit k which will be
off.

Why? This effectively assumes what you're trying to prove. What /does/
follow is that r's decimal representation is not finite.

More precisely

the prefix d_0.d_1d_2d_3....d_k-1 belongs to S but,
the prefix d_0.d_1d_2d_3....d_k-1d_k does not belong to S

But this is not possible because that prefix would have been added in
all sets S_j where j >= k+1.

Whereas what actually obtains is that every finite prefix of r is contained
in S. The idea that there must be some k such that r's first k digits
appear in S while r's first k+1 digits don't is simply false.


Yes ... I have been given this argument. I understand it but dont get
the premise. For example, you will probably argue that the set of all
finite strings will not contain an infinite string. But then what is
the length of the largest finite string in the set? If it is finite,
then shouldnt there be such a maximum finite length string? There is no
such maximum finite number .. i.e. it is infinitely large. The only way
out is that the set should contain strings of infinite length. Is this
right or wrong?

If the above is right, then the set of all finite prefixes will contain
infinite length prefixes and therefore the set will contain all reals.

If not, whyyyy not ??

Therefore no such k exists.

True.

Therefore r belongs to S.

False. If you /assume/ r's decimal representation is finite, then you can
claim that such a k must exist, so this is really a proof of that S contains
all reals whose decimal representations are finite.

Counter - Proof:
=============
Consider the real sqrt(2). The decimal representation of this real
would be infinite.

Now all the sets in the union are disjoint. Therefore there must exist
a set S_m which contains sqrt(2) and no other set can contain this
number.

However, no such m can exist, because then m would need to be infinite.
(there is no natural number "infinity")

Hence S cannot be R.

True.

Paradox anyone?

Infinity is really, really, /really/ big :-)

.

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