Why is $3$ the least upper bound of $A=\{a\in\mathbb{Q} \ | \ a<3\}$?

Question

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?

Answer 1

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.

Answer 2

" 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.

Answer 3

$3$ 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.

Answer 4

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 $A$.

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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} $.