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Need help in vetting my answers for the questions here in CRM series book by MAA: Exploratory Examples for Real Analysis, By Joanne E. Snow, Kirk E. Weller.

Suppose there are two nonempty subsets of real numbers, $L$ and $R$, such that:
- if $x \in L, y \in R$, then $x \lt y$, and
- $L \cup R =\mathbb{R}$.
Answer the following questions :
(a) If the set $L$ has a supremum, but no maximum, can the set $R$ have an infimum? minimum?
If your answer is no, explain. If your answer is yes, give an example.
(b) If the set $L$ has a maximum, can the set $R$ have an infimum? minimum?
If your answer is no, explain. If your answer is yes, give an example.
(c) If the set $R$ has an infimum, but no minimum, can the set $L$ have a supremum? maximum?
If your answer is no, explain. If your answer is yes, give an example.
(d) If the set $R$ has an minimum, can the set $L$ have a supremum? maximum?
If your answer is no, explain. If your answer is yes, give an example.
(e) Is there a relationship between the infimum of $R$ and supremum of $L$?

The sets' first condition is taken to mean that any element $y$ of set $R$ is bigger than any element of set $L$.
The second condition divides the reals into two disjoint sets.
Together, they mean that the two ordered disjoint sets divide real numbers, with all the second set elements to the right of all first set elements on real number line.

(a) It is guaranteed for first set supremum to belong to second set. In fact, in that case the second set ($R$) both minimum as well as infimum 'must' be the same as supremum of first set ($L$).
An example $L= (-\infty, 3), R = [3, \infty)$

It applies irrespective of the fact that the supremum of $L$ is a rational or irrational, as between any two rationals or irrationals, there are infinitely many irrationals.
Also, as an irrational number cannot be attained/reached, and only an approximate value or actual expression as say $\sqrt {5}$ is stated.
So, if a value is stated as supremum of $L$ but not a maximum; then it must belong to the other set, due to the sets together defining the real number line in a non-overlapping manner.

(b) It is guaranteed for set $R$ to have infimum as it is the same as supremum of $L$, due to no gap between the upper bound of $L$ & lower bound of $R$. But, not possible to have minimum in $R$.
An example to show that the opposite is not possible:
$L= (-\infty, 3], R = [3, \infty)$
Here, both sets have an overlapping point: $3$.

An example to show that this is the only possibility:
$L= (-\infty, 3], R = (3.0001, \infty)$
Here, there is gap in the real number line, as there might be infinite irrationals apart from possibly few rationals in the open interval $(3,3.0001)$.

It applies irrespective of the fact that the infimum of $L$ is a rational or irrational, as between any two rationals or irrationals, there are infinitely many irrationals.
The reason is that an irrational number cannot be attained/reached, & can be only approximated.
This reasoning is the same as that applies for the corresponding open bound in reals, for finding minimum / maximum as say for $(3,5)$.

(c) The set $R$ has an infimum, but no minimum, then it is always possible for the set $L$ to have a supremum as well as maximum. This supremum will then be the same as the infimum of $R$.

Below is shown that out of two possibilities for $L$, given this property of $R$; only one is possible due to no gap between the two sets.

(i) supremum of $L$ lies in $L$:
$L = (-\infty, 3], R=(3, \infty)$
Here, $L$ has maximum.

(ii) supremum of $L$ does not lie in $L$:
$L= (-\infty, 3), R = (3, \infty)$
This is not possible as the value $3$ is in none of the two sets, hence $3\not \in \mathbb{R}$, which is wrong.
Although primitive, but the second example highlights the case.

(d) If the set $R$ has a minimum, the set $L$ will have a supremum, but never a maximum. The reason is that between any two rationals or irrationals there are infinitely many irrationals. If minimum of $R$ exists, then the supremum of $L$ can only be specified in terms of infimum/minimum of $R$.
Say, $R = [\sqrt{5}, +\infty)$, then supremum of $L$ is the same as infimum of $R$, but maximum of $L$ does not exist.

(e) There a relationship between the infimum of $R$ and supremum of $L$?
They can be the same only, as only one of the two is specifiable, given the property of the real number line that between any two rationals or irrationals there are infinitely many irrationals. Also, the two sets define reals in a non-overlapping manner.

jiten
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  • Request response, & vetting. There is no similar one being asked earlier, so will serve many in future. – jiten Apr 29 '19 at 14:40
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    This is Dedekind theorem which says that under those conditions either $L$ has a maximum (exclusive) or $R$ has a minimum. A similar thing when done with $\mathbb{Q} $ gives another possibility : neither $L$ has a maximum nor $R$ has a minimum. – Paramanand Singh Apr 29 '19 at 16:26
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    A proper answer to the question assumes knowledge of some form of completeness of real numbers. Perhaps you are aware of this form: any non-empty subset of $\mathbb {R} $ which is bounded above has a supremum. – Paramanand Singh Apr 29 '19 at 16:32

1 Answers1

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  • Be concise. Answer what is being asked. If it is a yes, give an example, if no, explain.

  • Being rational or not is irrelevant to this question. Yes, you know that those statements are irrelevant but you mentioned them which is puzzling to me.

  • You seems to be have the misconception that irrational number can't be attained (depends on your definition of attain). But yup, this is irrelevant to this question.

  • Part $(a)$, great example, perhaps add a few lines to explain what are the infimum, minimum, maximum of your example rather than random stuff.

  • Part $(c)$, the second example, why should it be there? Just show the first example and do similar stuff as part $(a)$.

  • Part $(b)$, for the part about infimum, I can't find your example. Mentioning of $(3,5)$ and being rational or not are just distractions.

  • Part $(e)$:

    $L$ is bounded above and $R$ is bounded below. $L$ has a supremum, $l$ and $R$ has an infimum, $r$. Suppose on the contrary that $l > r$, by definition of supremum, we can find a sequence in $L$ such that $l_n \to l$, where $l_n$ can get arbitrarily close to $l$. Hence for some $m$, we have $l_m > r$ where $l_m \in L$ but $l_m$ is a lower bound of $R$, which contradicts to the fact that $r$ is the greatest lower bound. Hence, we certainly have $l\le r$.

    Now, suppose on the contrary that $l < r$. Consider $x = \frac{l+r}2$. We have $x > l$, hence $x \notin L$. We have $x < r$, hence $x \notin R$ which contradicts to $L \cup R = \mathbb{R}$. Hence $l=r$.

Now, after answering part $(e)$, then we use part $(e)$ to explain $(b)$ and $(d)$.

Siong Thye Goh
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  • Thanks a lot. Please see my post at : https://math.stackexchange.com/q/3207045/424260. I am sorry for asking before solving here completely, but I hoped for no response for this post to come, & was busy in preparing the new post. Also, it may take me a lot of time in dwelling on your answer here, so better stated the new post now. – jiten Apr 29 '19 at 15:34
  • New to formal(abstract) way of stating for part (e), & it should be preferred approach, as not ambiguous based on the words used & their interpretation. Want to edit all answers to show this approach, but get confused when you say to 'only' explain part $(b), (d)$. Say, want to provide proof for part $(a)$, but unable to extend approach. My attempt fails to incorporate $L$ not having maximum. In first part ($l\gt r$), still $l_n$ is there, & $l_n\rightarrow l$, but unable to put in the concept of maximum lacked in $L$. Please state how to extend this approach to incorporate proof of part (a). – jiten Apr 29 '19 at 23:53
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    The answer to part $(a)$ is yes. Provide an example and explain why do they work. That's it. – Siong Thye Goh Apr 30 '19 at 01:19
  • It means that if I were to take a formal proof, then the implication's premise should have all positive terms (as in (b), there is both maximum & supremum of $L$), i.e. for premise $\implies$ conclusion. Still not sure that this method by contradiction works only for such cases. I am unclear that this strategy does not work where any component of premise is negative, as in part (a) where there is no maximum of $L$. I hope this proof is adaptable to both easily, but unable to fathom. I hope it is only comparatively easy to give an example there, but that does not help to enhance my knowledge. – jiten Apr 30 '19 at 01:35
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    The supremum of $L$ is in $L$ but we have shown that the infimum of $R$ is equal to the supremum of $L$ but we know that $L \cap R=\emptyset$. Hence, we can't have minimum in $R$. – Siong Thye Goh Apr 30 '19 at 01:47
  • Thanks a lot for this (hopefully, am correct in terming it) direct proof. But, still request a proof by contradiction for case (a) (even though I suspect the reason you are not presenting it is to have simpler approach, rather than convoluted one). – jiten Apr 30 '19 at 02:08
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    Just add a line at the beginning. Suppose on the contray that $R$ has a minimum. You can then find a contradiction at the very end when we fidn that $R$ can't have a minimum. Btw, please don't make the comments too long. – Siong Thye Goh Apr 30 '19 at 02:15
  • Thanks a lot for that. One last comment: am I correct in stating the proof in your second-last comment as direct. A direct proof is of the form If $p$, then $q$. Here, have $p =$ 'supremum of $L$ is in $L$'and 'shown that infimum of $R$ is equal to the supremum of $L$' and '$L\cap R = 0$'. $q = $ cannot have minimum in $R$. – jiten Apr 30 '19 at 02:21
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    yes, it's a direct proof. – Siong Thye Goh Apr 30 '19 at 02:24
  • Please (please) give a detailed answer to my post at: https://math.stackexchange.com/q/3207045/424260 – jiten Apr 30 '19 at 03:51
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    It's beyond my ability to give a meaningful answer. – Siong Thye Goh Apr 30 '19 at 03:52
  • please give some hint as directly integrating the below is not helping : $|\frac{x^n}{1-x}| \lt x^n(1-a)^{-1} \implies |\frac{x^n}{(1-x)(1-a)}| \lt x^n$

    Taking the lhs, get: $\implies |\int_a^b \frac{x^n}{(1-x)(1-a)}|$

     – jiten Apr 30 '19 at 03:58
  • Request to put a bounty on my question, as might be it attracts responses. Sorry, for requesting you, as my score is too small to place more than a few bounties. – jiten Apr 30 '19 at 05:07