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The definition of least upper bound I am working with states that

If $A\subset F$, where $F$ is an ordered field, then an element $a\in F$ is a least upper bound if

$1$. $a$ is an upper bound of $A$;

$2$. $a\leq b$ for every upper bound $b$ of $A$.

So say I wish to find the least upper bound of $A=\{a\in\mathbb{Q} \ | \ a<3\}$. Here $F=\mathbb{Q}$, and the way we define the least upper bound does not restrict it, if it exists, to being in $A$. I understand that. What I don't understand is how we know that in this case the least upper bound of $A$ is $3$. Does $A$ not have some "ending point" that is really close to $3$, but is not identical to $3$? Or is it because we could keep inching closer and closer to (but never actually reach) $3$ that a least upper bound is not found until we actually leave $A$ and choose $3$?

I guess I am having a hard time writing out my thoughts mathematically in this case why $3$ is the least upper bound. Perhaps someone can help me in this regard?

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  • $\begingroup$ You got your answer in your own question... "Or is it because we could keep inching closer and closer to (but never actually reach) $3$ that a $l.u.b$ is not found until we actually leave $A$ and choose $3$. $\endgroup$ – Anik Bhowmick Aug 21 '18 at 6:32
  • $\begingroup$ Ideas like inching closer etc are more confusing than most people think. Just see that the number $3$ fits the definition of least upper bound (ie verify both the conditions, also the first condition is obvious). $\endgroup$ – Paramanand Singh Aug 21 '18 at 14:51
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Well, no matter how "close" to $3$ you get, you can always get closer. Say, you choose $3-\epsilon \in \mathbb{Q} \,(\epsilon>0)$, you can always choose $3-\epsilon/2 \in \mathbb{Q}$ which will be closer. But, when you choose $3$, it's trivially true that $x<3 \, \forall x \in A$ and notice that nothing else less than $3$ does the job. So, $3$ is the LUB. It does help that $A \subset \mathbb{R}$ since $\mathbb{R}$ has the LUB property.

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  • $\begingroup$ This does not sound convincing. If I claim that some $3-\varepsilon$ is the least upper bound, why should I care about your $3-\varepsilon/2$? That's larger than my number. $\endgroup$ – hmakholm left over Monica Aug 21 '18 at 17:51
  • $\begingroup$ @HenningMakholm Well, he wants to find the least upper bound. For that, it has to be an upper bound first and there are infinitely many rationals between those two numbers all of which are in $A$. P.S. I added this to the answer. $\endgroup$ – SinTan1729 Aug 21 '18 at 17:55
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    $\begingroup$ x @SinTan: Excellent. (But perhaps not necessary after all -- I had missed that he was looking for a l.u.b. in $\mathbb Q$ rather than in $\mathbb R$. And in that case $3-\varepsilon$ would be rational and so would $3-\varepsilon/2$, proving that $3-\varepsilon$ was not an upper bound). $\endgroup$ – hmakholm left over Monica Aug 21 '18 at 17:59
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" Or is it because we could keep inching closer and closer to (but never actually reach) 3 that a least upper bound is not found until we actually leave A and choose 3?"

This is informally correct: You have a motonone sequence of rationals increasing to $3$.

Usually to find the $\sup$ you do the $2$ following things:

$1)$ You show that something is a bound

$2)$ You show that this bound is the least

In your case, $3$ is obviously a bound. To show that it is the least one, notice that the sequence $x_n=3-\frac{1}{n}$ is a sequence of rationals in your interval that converges to $3$. If there was a bound $L<3$, you just use the definition of convergence to $3$ to show that there is some $x_i$ between $L$ and $3$ which contradicts the fact that $L$ is an upper bound. QED.

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$3$ is clearly an upper bound of $A$. Suppose $3\not\le b$ for every upper bound $b$ of $A$, i.e. $3>b$ for some upper bound $b$ of $A$. Since $F=\mathbb{Q}$, $b$ is a rational number. We use the fact that between any two rationals $c$ and $d$ is a rational $e$ s.t. $c<e<d$, so in particular there is a rational $e$ s.t. $b<e<3$. But then $e<3$ so $e\in A$, and $b<e$, i.e. $b$ is not an upper bound of $A$, contradiction.

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  • $\begingroup$ This answer does the best job pointing to the mathematical property which makes the argument work: the density of $\mathbb Q$. It becomes clear that if we changed the setting to, e.g., $\mathbb Z$, the same argument would not work. $\endgroup$ – BallBoy Aug 21 '18 at 18:36
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Expanding on my comment we see that $3$ is an upper bound for $A$. For the second condition in definition of least upper bound let us assume that $b$ is an upper bound for $A$ and show that $b\geq 3$. This is not difficult to show. Assume on the contrary that $b<3$ then $c=(b+3)/2$ is such that $c<3$ so that $c\in A$ and $c>b$ so that $c\notin A$. This contradiction shows that $b\geq 3$ and hence $3$ is least upper bound of $A$.


The chosen example with set $A$ is particularly easy and if you have assimilated the definition of least upper bound then it will be obvious that $3$ is the least upper bound of $A$.

On the other hand the set $B=\{x\mid x\in\mathbb {Q}, x>0,x^2<2\}$ is somewhat difficult to handle. One can easily see that $2$ is an upper bound for $B$ but it is not so easy to prove that there is no least upper bound for set $B$ in $\mathbb{Q} $.

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