Nonoverlapping IntervalsDate: 04/08/99 at 22:47:22 From: Kai - Yuan Cheng Subject: Uncountable number Show that there can not be an uncountable (nondenumerable) number of nonoverlapping intervals on the real line. Thanks. Date: 04/09/99 at 11:23:22 From: Doctor Floor Subject: Re: Uncountable number Hi, Kai-Yuan Cheng, Thanks for your question! First we have to be sure that the number of rational numbers is countable. You might like to read why that is true in the Dr. Math archives: http://mathforum.org/dr.math/problems/stephanie9.9.98.html Then we note that any (non-empty) interval on the real line contains a rational number. Or, in other words, that between any two unequal real numbers there is a rational number. To see that, think of the two real numbers in decimal progression, e.g.: 0.12491287491827409182734091827342341235123519... 0.12491287491827409182734091827342341235123520... There must be a point where the two progressions become different, which is the last decimal shown in the example. It is clear that we can construct a rational number between these two. There seems only to be a problem when the first progression would go on with 9's only, and the second with 0's only. But then the two numbers are equal. So we have that in any interval there is a rational number. Since the intervals are nonoverlapping, it follows that there cannot be more of those intervals than there are rational numbers. As a consequence of the countability of the rational numbers, the number of nonoverlapping intervals on the real line cannot be uncountable. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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