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I wish to prove that for every real number, $x$, there is a subset of the rationals, $S$, such that $x$ is the least upper bound of $S$.

At first glance this appeared axiomatic to me, however upon further inspection I supposed it was a trivial fact, once I started working on a proof I found it quite difficult. I may be approaching this in the wrong manner but here is my work so far:

If $x$ is rational, this is the case because $x$ is the least upper bound of $\{x\}$.

If $x$ is irrational things get a little murkier, and much hand wavier.

Let us have some set $A$ and its least upper bound $a$. If we take some element of $A$, lets call it $b$ and some set $B$ that has a least upper bound of $b$ we can create a new set $\left(A \setminus \{b\}\right) \cup B$ that is also has a least upper bound of $a$. (I haven't proven this is true but it makes intuitive sense)

Via this process we know that if there is no subset of the rationals with a least upper bound of $x$ every set with a least upper bound of $x$ must also contain at least one element with this property otherwise we would be able to expand our set into a set with a least upper bound of $x$.

And that's all I have. The final conclusion seems impossible but I can't seem to prove that it is not the case.

In addition this pre-supposes that every real number is the least upper bound of a set that does not contain itself, a fact I cannot myself prove.

I would appreciate any tips or suggestions on finishing this proof.

Community
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Sriotchilism O'Zaic
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1 Answers1

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Just take $S$ to be the set of rationals $r$ such that $r \le x$.

Robert Israel
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  • Thanks for the answer! Being that it is technically a correct answer I have marked it as accepted. However I will note that this assumes that there are rational numbers between any two real numbers, which was, unfortunately, what I had set out to prove in the first place. This is obviously my fault as I did not mention this fact anywhere in the body of my question so I thank you for your correct answer. – Sriotchilism O'Zaic Sep 18 '17 at 16:09
  • How are you defining real numbers? – Robert Israel Sep 18 '17 at 17:16
  • We were defining the real numbers to be the Dedkind completion of the field of rationals with an ordering such that $a>0 \land b>0 \implies a+b > 0 \land ab > 0$, $a > b \iff a-b > 0$ and for all $a$ exactly one of the following holds $a > 0$, $a = 0$, $-a > 0$. I don't know if this is a standardized axiom system, it seems to not be one of the ones listed on wikipedia for what its worth. – Sriotchilism O'Zaic Sep 18 '17 at 17:29