The Unapologetic Mathematician

Mathematics for the interested outsider

Cuts and Sequences are Equivalent

Sorry to not get this posted until so late, but the end of the semester has been a bit hectic.

We’ve used Dedekind cuts to “complete” the order on the rational numbers — to make sure that every nonempty set of numbers with an upper bound has a least upper bound. We’ve also used Cauchy sequences to “complete” the uniform structure on the rational numbers — to make sure that every Cauchy sequence converges. But do we actually get the same thing in each case?

If we take a real number x represented by a Cauchy sequence x_n it’s easy to come up with a cut. Given a rational number q we use the constant sequence q_n=q and compare it to x_n. If x_n-q_n is eventually nonnegative then q is less than x$, and should go into the left set X_L. On the other hand, if it’s eventually nonpositive then q is greater than x and should go into the right set X_R. It’s straightforward to show that this function from \mathbb{R} to the set of cuts preserves the order.

Now let’s start with the cut (X_L,X_R) and write down a Cauchy sequence. Pick some x_L\in X_L and x_R\in X_R, and construct the sequence as follows. First write down x_0=x_L and x_1=x_R. Now set x_2=\frac{x_L+x_R}{2}. This value will either be in X_L or it won’t. If it is, replace x_L by x_2, and otherwise replace x_R by x_2. Then define x_3 as the midpoint between our two left and right points, and again replace either the left or the right point. Keep going, and we see that all future numbers in the sequence are closer to each other than the current x_L and x_R are to each other. And these two always keep moving closer and closer to each other, halving their distance at each step. So the sequence has to be Cauchy. If we picked a different x_L and x_R to start with, we’d get an equivalent sequence. I’ll leave this to you to show.

Notice here that the points in the sequence that lie in X_L are moving steadily upwards towards the cut, and those in X_R are moving steadily downwards towards it. Eventually, the sequence will rise above any point in X_L and fall below any point in X_R, and so if we take this sequence and build a cut from it we will get back the exact same cut we started with. Also, if we build a cut from a Cauchy sequence, and then a sequence from it, we get back an equivalent sequence. Thus we have set up a bijection between the set of cuts and the set of equivalence classes of Cauchy sequences, and we’ve already seen that it preserves the order structure.

Now let’s look at the map from sequences to cuts and verify that it preserves addition and multiplication of positive numbers. This will make the map into an isomorphism of ordered fields, and so both constructions are describing essentially the same thing. So if we have Cauchy sequences x_n and y_n, which give right sets X_R and Y_R of rational numbers, then what’s the right set of the sequence x_n+y_n? It’s the set of rational numbers q so that x_n+y_n-q is eventually nonnegative. But any such q can be broken up as q=q_x+q_y, where x_n-q_x and y_n-q_y are both eventually nonnegative. That is, (X+Y)_R is the set of sums of elements of X_R and Y_R, and so addition is preserved. The proof for multiplication is essentially the same.

So both methods of extending the real numbers give us essentially the same ordered field, which is thus both complete as a uniform space and Dedekind complete.

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December 7, 2007 - Posted by | Fundamentals, Numbers

1 Comment »

  1. [...] a sequence of real numbers, then we’re taking a bunch of infima and suprema, all of which are guaranteed to exist. Thus the limits superior and inferior of any sequence must always [...]

    Pingback by Limits Superior and Inferior « The Unapologetic Mathematician | May 2, 2008 | Reply


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