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The set $R$ contains a subset that is isomorphic to the ordered field $Q$ of rational numbers, and hence subsets that are isomorphic to $N$ and $Z.$
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 $R$ satisfies our intuitive notions about the reals. Indeed, we will show that $\sqrt{2}$ is an element of $R$ 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 $R$ that agree with our intuition about $e$ and $\pi .$ Before we can proceed to these tasks, we must establish some special properties of the field $R.$ 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 $S$ is a nonempty subset of $R$ that is bounded below, then there exists a greatest lower bound for $S.$
Define $T$ to be the set of all real numbers $x$ for which $-x\in S.$ That is, $T$ is the set $-S.$ We claim first that $T$ is bounded above. Thus, let $m$ be a lower bound for the set $S,$ and let us show that the number $-m$ is an upper bound for $T.$ If $x\in T,$ then $-x\in S.$ So, $m\le -x,$ implying that $-m\ge x.$ Since this is true for all $x\in T,$ the number $-m$ is an upper bound for $T.$
Now, by the completeness assumption, $T$ has a least upper bound ${M}_{0}.$ We claim that the number $-{M}_{0}$ is the greatest lower bound for $S.$ To prove this, we must check two things. First, we must show that $-{M}_{0}$ is a lower bound for $S.$ Thus, let $y$ be an element of $S.$ Then $-y\in T,$ and therefore $-y\le {M}_{0}.$ Hence, $-{M}_{0}\le y,$ showing that $-{M}_{0}$ is a lower bound for $S.$
Finally, we must show that $-{M}_{0}$ is the greatest lower bound for $S.$ Thus, let $m$ be a lower bound for $S.$ We saw above that this implies that $-m$ is an upper bound for $T.$ Hence, because ${M}_{0}$ is the least upper bound for $T,$ we have that $-m\ge {M}_{0},$ implying that $m\le -{M}_{0},$ and this proves that $-{M}_{0}$ is the infimum of the set $S.$
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 $S$ be a nonempty subset of $R$ that is bounded above, and Let ${M}_{0}$ denote the least upper bound of $S;$ i.e., ${M}_{0}=supS.$ Then, for any positive real number $\u03f5$ there exists an element $t$ of $S$ such that $t>{M}_{0}-\u03f5.$
Let $\u03f5>0$ be given. Since ${M}_{0}-\u03f5<{M}_{0},$ it must be that ${M}_{0}-\u03f5$ is not an upper bound for $S.$ ( ${M}_{0}$ is necessarily less than or equal to any other upper bound of $S.$ ) Therefore, there exists an element $t\in S$ for which $t>{M}_{0}-\u03f5.$ This is exactly what the theorem asserts.
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