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The set contains a subset that is isomorphic to the ordered field of rational numbers, and hence subsets that are isomorphic to and
REMARK. The proof of [link] is immediate from part (b) of Exercise 1.7. In view of this theorem, we will simply think of thenatural numbers, the integers, and the rational numbers as subsets of the real numbers.
Having made a definition of the set of real numbers, it is incumbent upon us now to verify that this set satisfies our intuitive notions about the reals. Indeed, we will show that is an element of and hence is a real number (as plane geometry indicates it should be),and we will show in later chapters that there are elements of that agree with our intuition about and Before we can proceed to these tasks, we must establish some special properties of the field The first, the next theorem, is simply an analog for lower bounds of the least upper bound condition that comes from the completeness property.
If is a nonempty subset of that is bounded below, then there exists a greatest lower bound for
Define to be the set of all real numbers for which That is, is the set We claim first that is bounded above. Thus, let be a lower bound for the set and let us show that the number is an upper bound for If then So, implying that Since this is true for all the number is an upper bound for
Now, by the completeness assumption, has a least upper bound We claim that the number is the greatest lower bound for To prove this, we must check two things. First, we must show that is a lower bound for Thus, let be an element of Then and therefore Hence, showing that is a lower bound for
Finally, we must show that is the greatest lower bound for Thus, let be a lower bound for We saw above that this implies that is an upper bound for Hence, because is the least upper bound for we have that implying that and this proves that is the infimum of the set
The following is the most basic and frequently used property of least upper bounds. It is our first glimpse of “ limits.”Though the argument is remarkably short and sweet, it will provide the mechanism for many of our later proofs, so master this one.
Let be a nonempty subset of that is bounded above, and Let denote the least upper bound of i.e., Then, for any positive real number there exists an element of such that
Let be given. Since it must be that is not an upper bound for ( is necessarily less than or equal to any other upper bound of ) Therefore, there exists an element for which This is exactly what the theorem asserts.
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