# The Unapologetic Mathematician

## 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.

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 […]

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