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Revision of Creating real numbers using limiting arguments from Sat, 18/04/2009 - 20:36

Quick description

If you find yourself wanting to prove the existence of a real number with certain properties, then almost always you will need to do so by means of a limiting argument of some kind, though sometimes this can be disguised if you use other results in your proof that are themselves proved using limiting arguments. This article gives examples of the main general techniques that are used.


Basic concepts and theorems of analysis. In particular, the notion of a real number, sequences, limits, the supremum of a set, the intermediate-value theorem, the nested-intervals property, the Bolzano-Weierstrass theorem. Some of these are briefly explained in the article but advance familiarity with them would definitely be helpful.

Example 1

The nested-intervals property of the real numbers is the truth of the following theorem.

Theorem 1 Let [u_1,v_1],[u_2,v_2],\dots be a collection of non-empty closed intervals, and suppose that each one contains the next as a subset. Then the intersection of all the intervals is non-empty.

Just before we see how to prove this result, let us see why it is not a trivial statement by observing that the corresponding statement for open intervals is false: for example, each of the intervals (0,1),(0,1/2),(0,1/3),\dots contains the next one, each is non-empty, and yet no real number is contained in every single one of them.

General discussion

Note that we are trying to prove that a real number x exists with a certain property: the property of belonging to all the intervals [u_n,v_n]. Thus, the property we want to prove is naturally regarded as the simultaneous occurrence of a countable sequence of properties P_1,P_2,\dots, where P_n is the property of belonging to the interval [u_n,v_n]. In such a situation, we cannot normally hope to create x all in one go. Instead, we create some auxiliary and more complicated object such as a convergent sequence or a non-empty set that is bounded above, and define x in terms of that. (Given the sequence, one would take x to be the limit, and given the set one would take the supremum.)

This instantly makes the task easier. For instance, if we use a sequence, then normally we have to ensure that each term in the sequence satisfies finitely many properties rather than infinitely many, though in this example it is even easier than that since the sequence suggests itself instantly.

Example 1, continued

Here, then, is a proof of Theorem 1.

Proof. The sequence u_1,u_2,\dots is increasing (though not necessarily strictly increasing) and bounded above (by v_1). Therefore, it converges. Let x be the limit. Then x\geq u_n for every n, since u_m\geq u_n for every m\geq n. (This follows very easily from a principle that is proved as Example 1 in I have a problem to solve in real analysis and I do not believe that a fundamental idea is needed.) We can also show that x\leq v_n for every n, since u_m\leq v_m\leq v_n for every m\geq n. Therefore, x belongs to every interval [u_n,v_n], as required.

General discussion

Note that in order to create our real number we used a basic result of real analysis: that bounded monotone sequences converge. This is typical of the way one proves that real numbers exist, and the next few examples will do the same, but using different basic results.

From now on, we shall think of the nested-intervals property not as a theorem but as one of the basic results about real numbers that we are free to use. The next example illustrates not just that different basic results can be used, but that different basic results can be used for the same problem. In fact, we shall give four proofs of the intermediate value theorem. (A fifth proof can be found as Example 1 of Discretization followed by compactness arguments.)