12 ms·
Sigh. Here we go, a definition of the real numbers. I'm only doing this once. Let's look at the set of all cauchy sequences of rational numbers. A sequence
by jibiki 17y ago
Sigh. Here we go, a definition of the real numbers. I'm only doing this once.
Let's look at the set of all cauchy sequences of rational numbers. A sequence of rational numbers is just a map N->Q. We write a_0, a_1, a_2, ... for the sequence which maps 0|->a_0, 1|->a_1, 2|->a_2, ... Such a sequence is called "cauchy" if it has the property that:
For all epsilon > 0 there exists an N such that n,m > N => |a_n-a_m|<epsilon.
Basically, this means that the tail of the sequence bunches together.
Now, we define an equivalence relation ~ on the space of cauchy sequences. We say:
a_0, a_1, a_2... ~ b_0, b_1, b_2... if:
a_0-b_0, a_1-b_1 ,a_2-b_2... converges to zero
(A sequence a_0,a_1,... converges to zero if, for every epsilon > 0, there exists N such that n>N => a_n < epsilon.)
Now, the real numbers are the set of equivalence classes of rational cauchy sequences under ~, as defined above.
Clearly, .999... is (the equivalence class of):
0.9, 0.99, 0.999, ...
And 1 is the equivalence class of:
1, 1, 1, ...
So are these the same? Well, we have to check if our representatives are equivalent. Does this sequence converge to 0?
1-0.9, 1-0.99, 1-0.999, ...
Yes, it does. So 1 = .999...
- RiderOfGiraffes 17y agoSweet, I've not seen this approach before. It still doesn't convince my colleague, because he says that the sequence never gets to zero, but now I know where I stand. He can't deal with the formalisms of convergence and limits. It's also interesting to try to prove this via the Dedekind Cut construction, rather than the equivalence classes of Cauchy Sequences. I'll have a go at that later tonight.