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Conclusion : This question makes no sense unless we consider it in the set of real numbers


How to prove the greatest lower bound for all positive rational numbers is $0$?

I can only figure out the following right now:

(1). $0$ is a lower bound for all positive rational numbers;

(2). Any positive rational number is not the greatest lower bound for all positive rational numbers;

Density property of rational numbers may be helpful to prove the conclusion, but it is not easy for me to give a complete rigorous proof right now.


I think

1). There is no guarantee that the greatest lower bound in the question must be a rational number .

2).The question doesn't want to prove that zero is the greatest lower bound in Q, I don't need the proof that 0 is the greatest rational lower bound.

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  • $\begingroup$ How do you define positive? $\endgroup$ – Dando18 Jul 24 '17 at 13:13
  • $\begingroup$ @Dando18 greater than $0$ $\endgroup$ – iMath Jul 24 '17 at 13:14
  • $\begingroup$ Then let $L$ be any lower bound of $\mathbb Q$. You just need to show that $L<0$ for all $L$. $\endgroup$ – Dando18 Jul 24 '17 at 13:16
  • $\begingroup$ What sort of proof are you looking for? Would it suffice to exhibit a sequence of positive rationals that goes to $0$? $\endgroup$ – lulu Jul 24 '17 at 13:19
  • $\begingroup$ @lulu Better to prove it without using limit $\endgroup$ – iMath Jul 24 '17 at 13:33
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Argue by contradiction. Suppose $c\in\Bbb Q$ satisfies $c>0$ and $c$ is a lower bound for all rational numbers. Then $c/2\in\Bbb Q$ and $c/2<c$ since if $c/2\geq c$ then $1/2\geq1$. A contradiction.

Thus the greatest lower bound must be less or equal to zero. Now zero is the largest number in that range and zero works so it must be the greatest lower bound.

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    $\begingroup$ Why is it obvious that the lower bound is rational? $\endgroup$ – lulu Jul 24 '17 at 13:19
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    $\begingroup$ I thought we were looking for the greatest lower bound in $\Bbb Q$. Is that not the question? $\endgroup$ – Gregory Grant Jul 24 '17 at 13:20
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    $\begingroup$ The "greatest lower bound" for a set is not necessarily in that set- just as "0" is not a positive rational number. For example, the "greatest lower bound" of the set of all rational numbers whose square is larger than 2 is $\sqrt{2}$, not a rational number. $\endgroup$ – user247327 Jul 24 '17 at 13:24
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    $\begingroup$ @lulu It is clear that $0$ is a lower bound. Moreover, this answer proves that $0$ is the greatest rational lower bound. What more do you want? $\endgroup$ – Omnomnomnom Jul 24 '17 at 13:26
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    $\begingroup$ @lulu The question is "prove that $0$ is the greatest lower bound". To start, we prove that $0$ is a lower bound. Then, we prove (by contradiction) that any other lower bound is at most $0$. How does this proof fail? $\endgroup$ – Omnomnomnom Jul 24 '17 at 13:38
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Since there has been a discussion below the other answer, I will elaborate a bit more.

The hint still is:
Use contradiction. And don't read further if you don't want to get spoilered.


First I want to note that having a "lower/upper bound" of a set $M$ does not make any sense if it is not refered to a superset $N$ of $M$.
(Note: $M$ is also a superset of itself, so $N=M$ isn't a problem at all.)

So the definition for (greatest) lower bound is:

  • A lower bound of the subset $M$ of $N$ is an element $n∈N$, such that $n≤m$ for all $m∈M$.
  • The greatest lower bound (or infimum) of $M$ is a lower bound $n_{inf}∈N$, if for all lower bounds $n'$ it holds $n'≤n_{inf}$.

Therefore the question of OP is clear:

Assuming all numbers you know are rational numbers, [...]

We only now rational numbers $ℚ$. Hence, we have to proof $0$ is the infimum of $ℚ^+(=M)$ with respect to $ℚ(=N)$.

Then the proof is:

  1. $0$ is a lower bound.
  2. Assume there is another lower bound $q∈ℚ$ that is larger than $0$. That would mean, that then $0$ would not be the infimum. But since $q>0$ the number $q':=\frac{q}{2}∈ℚ^+$ fulfils $0<q'<q$. Hence, $q$ is not a lower bound, and therefore $0$ must be the infimum of $ℚ^+$
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  • $\begingroup$ I have some doubt on your statement that having a "lower/upper bound" of a set $M$ does not make any sense if it is not refered to a superset $N$ of $M$., what about saying 1 is a lower bound in $\{x\in \mathbb Q \,|\,x<\sqrt 2\}$ for $\{x\in \mathbb Q \,|\,x>\sqrt 2\}$? here I have not mentioned the superset of $\{x\in \mathbb Q \,|\,x>\sqrt 2\}$ $\endgroup$ – iMath Jul 24 '17 at 15:26
  • $\begingroup$ @iMath Your comment is a good example, why my statement [...] does not make any sense[...] is "correct". You need a superset ($ℚ$) of both $M_1=\{x∈ℚ|x<√2\}$ and $M_2=\{x∈ℚ|x>√2\}$ with a relation "$≤$" to define that $1$ is a lower bound of $M_2$. Otherwise how is $$1≤x$$ for $x∈M_2$ well-defined? $\endgroup$ – P. Siehr Jul 24 '17 at 15:52
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The greatest lower bound (ie. infimum) $I$ is characterized by two things: 1) $\forall x \in \mathbb{Q}_+: I \leqslant x$, and 2) there exists a sequence of positive rationals that converges to $I$. The last condition is satisfied by eg. $\lbrace 1/n \rbrace_{n=1}^\infty$.

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Assume that we are "living in a rational world", and consider the set $P$ of strictly positive rationals. Does this set have a greatest lower bound in ${\mathbb Q}$?

Denote the set of lower bounds of $P$ by $L$. Then $0\in L$, but $P\cap L$ is empty: If $c\in P$ then ${c\over2}\in P$ as well. Since ${c\over2}<c$ we may conclude that $c\notin L$.

It follows that $L\subset{\mathbb Q}_{\leq0}\,$, hence $0$ is obviously the maximal element of $L$.

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