# Supremum, bounds of subsets of real numbers.

Need help in vetting my answers for the questions 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.

• Request response, & vetting. There is no similar one being asked earlier, so will serve many in future. Commented Apr 29, 2019 at 14:40
• 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. Commented Apr 29, 2019 at 16:26
• 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. Commented Apr 29, 2019 at 16:32

• 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 $$(d)$$.

• The answer to part $(a)$ is yes. Provide an example and explain why do they work. That's it. Commented Apr 30, 2019 at 1:19
• 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 $R$. Commented Apr 30, 2019 at 1:47
• 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. Commented Apr 30, 2019 at 2:15
• yes, it's a direct proof. Commented Apr 30, 2019 at 2:24
• It's beyond my ability to give a meaningful answer. Commented Apr 30, 2019 at 3:52