0
$\begingroup$

I'm learning set theory and I got to bounded sets. I got to the following definition and I have a question about it:

enter image description here

The problem that I have is the quantified variable $a$. If I would write the quantified statement down it would look like this: $\forall \epsilon \in \mathbb{N}, \exists a \in A, M - \epsilon < a$. My question is wouldn't the right quantified variable $a$ be a $\forall$ as well, like this: $\forall \epsilon \in \mathbb{N}, \forall a \in A, M - \epsilon < a$. Now, with the definition from the book, I'm not fully targetting all of the elements of $A$, as I actually should since a least lower bound is targetting the whole set. So the question is, which version is right, mine or the one from the book ?

$\endgroup$
2
  • $\begingroup$ "as I actually should" No. You should not. There are two aspects of least upper bound. The "upper bound aspect", that does target all $\alpha \in A$. ANd the least aspect which most certainly does NOT target all elements of $A$. In fact it doesn't target any elements of $A$. As lest upper bound its target are the REAL NUMBERS (nothing to do with $A$) that are all smaller than $M$. What is is saying is "$\forall w; w< M;w$ is not upper bound. Now its the target of $w$, not of $M$, that is the elments of $A$. Namely $w$ is not u.b. so $\exists a\in A; w< a$...... $\endgroup$
    – fleablood
    Sep 19, 2022 at 18:17
  • $\begingroup$ The actual statement is twofold i) $\forall a\in A: a \le M$. (that is "$M$ is an upper bound") and ii) $\forall w\in \mathbb R, w<M: \exists a\in A, w< a$. $\endgroup$
    – fleablood
    Sep 19, 2022 at 18:20

1 Answer 1

1
$\begingroup$

There are two parts to the definition:

  • Upper bound: $\forall a\in A, a \le M$
  • Least: $\forall \epsilon>0, \exists a \in A, M - \epsilon < a$

For the "least" part, it might help to think of an equivalent statement: $\forall \epsilon>0, \lnot(\forall a \in A, M - \epsilon \ge a)$, which says that $M - \epsilon$ is not an upper bound. That is, there is no upper bound smaller than $M$.

$\endgroup$
4
  • $\begingroup$ I understood that. The problem that I have is with how the variable $a$ has been quantified. My problem is understanding why the variable $a$ has been quantified the way it is. Shouldn't the variable $a$ be quantified with a $\forall$ as well ? So $\forall \epsilon \in \mathbb{N}, \forall a \in A, M - \epsilon < a$ ? $\endgroup$
    – David
    Sep 19, 2022 at 17:49
  • 3
    $\begingroup$ No, the $\forall a$ goes only with the upper bound part. Also, note that $\epsilon >0$ (any positive real number), not $\epsilon \in \mathbb{N}$. $\endgroup$
    – RobPratt
    Sep 19, 2022 at 17:51
  • $\begingroup$ I added a little bit of explanation that might help. $\endgroup$
    – RobPratt
    Sep 19, 2022 at 17:57
  • $\begingroup$ The least upper bound of $(0,1)$ is $1$. Why? Because for all $a\in A$ we have $a\le 1$ and for all $\epsilon >0$ the number $1-\epsilon$ is not an upper bound. Let's test that for $\epsilon =0.1$. Is $0.9$ an upper bound? No, because there exists $a\in A$ ($0.95$ is an example) such that $a \not\le 0.9$. It just isn't true (or relevant) that all elements $a \in A$ satisfy $a \not\le 0.9$. $\endgroup$ Sep 19, 2022 at 18:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .