0
$\begingroup$

Set 1 defined as $\left\{ {{m \over n}:m,n \in N\;{\rm{with}}\;m < n} \right\}$, and
Set 2 defined as $\left\{ {{m \over {m + n}}:m,n \in N} \right\}$.
So far my approach to problem
For set 1 since ${m < n}$, therefore, numerator is always less than denominator, suggesting the ${{m \over n}}$ is always less than 1. As values of $m$ and $n$ keeps increasing the ${{m \over n}}$ gets closer to $->$ 1. But as I know ${{m \over n}}$ is $<1$. Therefore, there is no upper bound, and no sprema. For infima, if I set $m=1$ then increasing value of $n$ for higher and higher value ${{m \over n}}$ will approach $0$. Therefore, infima of set is $0$. Am I correct? If yes, is there a better way to present answer, technically, a mathematical language, reducing story telling with words.
For set 2 ${{m \over {m + n}}}$ does not have upper bound, so no suprema. Right ? why? because ah snap! I just got intuitively, I might be wrong. Please explain what is suprema for set 2. Ironically, here I like to have some storytelling. For infima, if I set $m=1$ smallest one and keep on increasing $n$ then ${{m \over {m + n}}}$ will have infima $=0$ ? Please explain where I am going wrong?

  • 0
    I was hoping you'd get hint the second time, but I guess I have to make this request explicit: stop using the set theory tags. They are irrelevant to your question. – 2017-01-23
  • 0
    $m/(m+n)=1-(n/(m+n))$ is certainly bounded above by $1$. – 2017-01-23
  • 0
    They both are the same set. They both have an infinum and supremum. – 2017-01-23
  • 1
    Let m < n. Let a =n-m. Then m/n =m/(m+a). So set 1 is a subset of set 2. Let m,n be in N. Let k=m+n > m. m/(m+n) = m/k. So set 2 is a subset of set 1. So the sets are equal. 0 < m/n < 1 so 0 is a lower bound and 1 is an upper nound. You just need to shown 0 is the inf and 1 is the sup. – 2017-01-23
  • 1
    " But as I know m/n <1 . Therefore, there is no upper bound, and no supremum" ??? You know the *exact* opposite! m/n < 1 so 1 is an upper bound. So the set must have a supremum. Why on earth did you say there *wasn't* an upper bound? You gave the *exact* definition of what an upper bound is! – 2017-01-23
  • 0
    @fleablood yes I am wrong in stating there is no upper bound. It certainly has upper bound but is it least upper bound. I can always have m/n which is greater than previous m/n approaching 1. It is approaching 1 but does it have properly defined least upper bound ? I am bit confused here in the fact that it is approaching 1 but never equals 1. – 2017-01-23
  • 0
    @fleablood in both the cases I am confused with finding least upper bound. They both have upper bounds as you explained they are both same. – 2017-01-23
  • 1
    The least upper bound need not be in the set. The least upper bound is the smallest number that is as larger *or larger* than all elements. But no element needs to actually equal it. – 2017-01-23
  • 1
    No element in the set equals 1. But that is not in any way a requirement for a supremum. The two requirements for a supremum are i) sup >= to all elements (which is true for 1) ii) if x < sup there is an element, a, so that x < a <= sup (which might be true of 1). There is absolutely no requirement that there is an element so that a = sup. So it remains to show ii. – 2017-01-23
  • 1
    You are thinking of "maximum". That set has no maximum. But maximum and supremum are not the same thing. Max is largest element in set. Sup I smallest number at least as large as all elements. Sup need not be an element. The fact that it frequently is not is the whole point of real analysis. – 2017-01-23
  • 1
    The simplest example of sup is the sup of the set (0,1). 1 is sup (0,1) but 1 is *not* an element of the set. 1 is larger or equal than all elements so it is upper bound. Any x <1 is not an upper bound as there will be an element a so that x < a <1. So 1 is least upper bound aka sup. – 2017-01-23
  • 0
    Thanks a lot. I think I understood what you stated but I need to ponder upon it more. I guess it takes time to understand "real" analysis. – 2017-01-23
  • 0
    @fleablood, suppose I construct a set ${{{{{\left( { - 1} \right)}^m}} \over n}:m,n \in N}$, so the set is $\left( { - \infty ,\infty } \right)$. Hence, finding suprema and infima for this set is not possible right? – 2017-01-23
  • 0
    1) $-1 \le \frac {(-1)^m}n \le 1; \forall m,n$ so the set is is bounded above by 1 and bounded below by -1. 2) The reals have the "least upper bound" property is the sup and inf *must* exist although there is no reason to assume they will be easy to identify. BUT in this case they are easy to do. They are -1 (inf) and 1 (sup). – 2017-01-23
  • 0
    The set ($-\infty, \infty$) has no sup or inf because it is not bounded either above or below. But the set you describe is not $(-\infty, \infty)$. It is $(-1,1) \cap \mathbb Q$ and it is bounded above and below but 1 and -1. The set I think you wanted to say was $\{\frac{(-1)^m*m}{n}\} = (-\infty, \infty) \cap \mathbb Q = \mathbb Q$ which is not bounded so it has not sup or inf. – 2017-01-23
  • 0
    Yes, indeed I constructed it wrong but you were able to catch the idea. – 2017-01-23

1 Answers 1

0

Let $M$ be Set 1. i.e. $M = \{\frac mn| m,n \in \mathbb N; m < n\}$.

Definition: A set $S\subset X$ is bounded above (below) if there is an $x \in X$ so that for all $a \in S$ we have $x \ge a$ ($x \le a$). And we say that $x$ is an upper (lower) bound of $S$.

Let $q =\frac mn \in M$ then $0 < m < n$ so $0 < \frac mn = q < 1$. So for all any $x \ge 1$ we have that all possible $q \in M$ thatn $x > q$. So $M$ is bounded above by $1$ and $1$ is an upper bound of $M$.

Notice there are an infinite number of upper bounds. $27 > q$ for all $q \in M$ so $27$ is an upper bound. $5,432\sqrt{\pi} > q$ for all $q \in M$ so $5,432\sqrt{\pi}$ is an upper bound.

Definition: An element $x \in X$ is said to be a least upper (greatest lower) bound or synonymously a supremum (infimum) if 1)$x$ is an upper (lower) bound and 2) $x$ is the least (greatest) upper (lower) bound; that is; if $y < x$ ($y > x$) then $y$ is not an upper (lower) bound... or in other words for any $y < x$ ($y > x$) there will be an $a \in S$ so that $y < a \le x$ ($y > a \ge x$).

So although $27$ is an upper bound it is not the least upper bound because $26 < 27$ and $26$ is also a least upper bound.

Notice. 1) there is absolutely no condition whatsoever that $\sup A \in A$. A good example is $A = (0,1)$. The $1 > a; \forall a \in A$ so $1$ is an upper bound. If $x < 1$ then there is some $a \in A$ so that $x < a < 1$. So $x \not \ge a$. So $x$ is not an upper bound. So $1$ is the least upper bound. So $\sup A = 1 \not \in A$.

2) Depending on $X$ there may not be an least upper bound at all. Consider $S = \{q \in \mathbb Q| q^2 < 2\}\subset Q$. In other words $S = \{q \in \mathbb Q| -\sqrt{2} < q < \sqrt{2}\}$ there is no rational number that that is the smallest rational number that is larger or equal to $\sqrt{2}$ so $sup S$ simply does not exist.

BUT

Theorem/Fundamental Principal of Analysis. $\mathbb R$ has the "least upper bound property". That is every set that is bounded above (or below) does have a least upper (greatest lower) bound.

So $M$ is bounded above by $1$. So $\sup M$ must exist. But we have NO reason to believe $\sup M \in M$. None at all.

So $1$ is a upper bound. But is it the least upper bound?

If $x < 1$ then is there always a $\frac mn$ so that $x < \frac mn \le 1 ; m < n$?

Intuitively we should say yes. But to prove it may be irksome.

Assume $0 < x < 1$. Let $e = 1 -x > 0$ then there is some $n \in \mathbb N$ so that $0 < 1/n < e$. So $0 < x = 1- e < 1 - 1/n = \frac {n-1}n < 1$. So $ x$ is not an upper bound.

So $\sup M = 1$.

Likewise if $q = \frac mn \in M$ then $m > 0; n>0$ so $\frac mn > 0$. So $0$ is a lower bound of $M$. But is it the greatest lower bound?

Well, yes. If $ x > 0$ then there is a $1/n$ so that $x > 1/n > 0$ and $1/n \in M$. So $x$ is not an lower bound for any $x > 0$ so $0$ is the greatest lower bound.

Note: $-5,678,2984,2984$ is also a lower bound. But it isn't that greatest lower bound.

  • 0
    This is the best explanation one can get. – 2017-01-23