Why Don't Minimal and Maximal Elements Imply Minimum and Maximum?

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In summary: The subsets of set $A$ are the following:$$\{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}$$The subsets of set $A$ are the following:$$\{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}$$In summary, the minimal subset of set $A$ is $\varnothing$.
  • #1
evinda
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Hello! (Smile)

Let $(A, \leq)$ be an ordered set.
We say that $a \in A$ is:
  • minimal, when it does not exist in $A$ an element that is previous of $a$ and different from it, i.e. $(\forall x \in A)(x \leq a \rightarrow x=a)$
    $$$$
  • maximal, when it does not exist in $A$ an element that is next of $a$ and different from it, i.e. $(\forall x \in A)(a \leq x \rightarrow a=x)$
    $$$$
  • minimum when $(\forall x \in A) a \leq x$
    $$$$
  • maximum when $(\forall x \in A) x \leq a$
Remark:

  • If $a$ is minimum in $A$ then it is also minimal.
  • If $a$ is maximum in $A$ then it is also maximal.

The converse of the above does not hold, in general.Could you explain me why the converse does not hold? (Thinking)
 
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  • #2
What you call minimum and maximum and usually called least and greatest.

Consider subsets of some set ordered by inclusion.
 
  • #3
Evgeny.Makarov said:
Consider subsets of some set ordered by inclusion.

Could you explain it further to me? (Thinking)
 
  • #4
I will if you confirm that you considered subsets of some set ordered by inclusion, for example, subsets of a 3-element set. By considering I mean determining which subsets are greatest and which are maximal.
 
  • #5
Evgeny.Makarov said:
I will if you confirm that you considered subsets of some set ordered by inclusion, for example, subsets of a 3-element set. By considering I mean determining which subsets are greatest and which are maximal.

If we take for example this set: $A=\{ 1,2,3\}$ the subsets will be the following:
$$\{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}$$

To check which subsets are greatest and which are maximal do we have to look at the number of elements that the subsets have? (Thinking)
 
  • #6
evinda said:
If we take for example this set: $A=\{ 1,2,3\}$ the subsets will be the following:
$$\{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}$$
Yes.

evinda said:
To check which subsets are greatest and which are maximal do we have to look at the number of elements that the subsets have?
No.

evinda said:
Let $(A, \leq)$ be an ordered set.
We say that $a \in A$ is:

minimal, when it does not exist in $A$ an element that is previous of $a$ and different from it, i.e. $(\forall x \in A)(x \leq a \rightarrow x=a)$
Here the word "previous" should be replaced by "less", and $a$ is less then $b$ means $a\le b$ w.r.t. the the order $\le$ we are considering. If our order is set inclusion, then we say $a\le b$ if $a\subseteq b$. So $a$ is minimal w.r.t. set inclusion if there does not exist a set $b$ such that $b\subseteq a$. If we consider all subsets of some set $X$, then the only minimal set is $\varnothing$. But what if we consider not all subsets, but only nonempty ones? When $A=\{1,2,3\}$ the nonempty subsets are $\{1\}$, $\{2\}$, $\{3\}$, $\{1,2\}$, $\{1,3\}$ and $\{2,3\}$. Which of them are minimal subsets?
 
  • #7
Evgeny.Makarov said:
Yes.

No.

Here the word "previous" should be replaced by "less", and $a$ is less then $b$ means $a\le b$ w.r.t. the the order $\le$ we are considering. If our order is set inclusion, then we say $a\le b$ if $a\subseteq b$. So $a$ is minimal w.r.t. set inclusion if there does not exist a set $b$ such that $b\subseteq a$. If we consider all subsets of some set $X$, then the only minimal set is $\varnothing$. But what if we consider not all subsets, but only nonempty ones? When $A=\{1,2,3\}$ the nonempty subsets are $\{1\}$, $\{2\}$, $\{3\}$, $\{1,2\}$, $\{1,3\}$ and $\{2,3\}$. Which of them are minimal subsets?

Each subset has itself as a subset, right? (Thinking)
If so, would it mean that there is no minimal element? (Thinking)
 
  • #8
evinda said:
Each subset has itself as a subset, right?
Yes, a non-strict subset.
evinda said:
If so, would it mean that there is no minimal element
Sorry, I'll have to rephrase what I said in post #6.

"Here the word 'previous' should be replaced by 'less', and $a$ is less then $b$ means $a< b$ w.r.t. the the order $<$ we are considering. If our order is set inclusion, then we say $a<b$ if $a\subsetneqq b$, i.e., $a$ is a strict subset of $b$. So $a$ is minimal w.r.t. set inclusion if there does not exist a set $b$ such that $b\subsetneqq a$. If we consider all subsets of some set $X$, then the only minimal set is $\varnothing$. But what if we consider not all subsets, but only nonempty ones? When $A=\{1,2,3\}$ the nonempty subsets are $\{1\}$, $\{2\}$, $\{3\}$, $\{1,2\}$, $\{1,3\}$ and $\{2,3\}$. Which of them are minimal subsets?"
 
  • #9
Evgeny.Makarov said:
"Here the word 'previous' should be replaced by 'less', and $a$ is less then $b$ means $a< b$ w.r.t. the the order $<$ we are considering. If our order is set inclusion, then we say $a<b$ if $a\subsetneqq b$, i.e., $a$ is a strict subset of $b$. So $a$ is minimal w.r.t. set inclusion if there does not exist a set $b$ such that $b\subsetneqq a$. If we consider all subsets of some set $X$, then the only minimal set is $\varnothing$. But what if we consider not all subsets, but only nonempty ones? When $A=\{1,2,3\}$ the nonempty subsets are $\{1\}$, $\{2\}$, $\{3\}$, $\{1,2\}$, $\{1,3\}$ and $\{2,3\}$. Which of them are minimal subsets?"

The minimal subsets are $\{1\}$, $\{2\}$ and $\{3\}$, right? (Thinking)
 
  • #10
evinda said:
The minimal subsets are $\{1\}$, $\{2\}$ and $\{3\}$, right?
Yes. So you see that minimal is not necessarily the least because $\{1\}$ is not strictly smaller that all other subsets. In particular, it is not the case that $\{1\}<\{2\}$, i.e., $\{1\}\subset\{2\}$ (by $\subset$ I mean strict inclusion).

Could you explain what you understand now what you did not understand in the beginning of the thread? That would help me answer questions more effectively.
 
  • #11
Evgeny.Makarov said:
Yes. So you see that minimal is not necessarily the least because $\{1\}$ is not strictly smaller that all other subsets. In particular, it is not the case that $\{1\}<\{2\}$, i.e., $\{1\}\subset\{2\}$ (by $\subset$ I mean strict inclusion).

I see... So, in our case we don't have a least element, right? (Thinking)

Can we say in the same way that $a$ is maximal w.r.t. set inclusion if there does not exist a set $b$ such that $b \supsetneq a$ ? (Thinking)

Evgeny.Makarov said:
Could you explain what you understand now what you did not understand in the beginning of the thread? That would help me answer questions more effectively.

I thought that we would consider the elements of a set. For example, I thought that if $A=\{1,2,3\}$ then since $(\forall x \in A)(x \leq 1 \rightarrow x=1)$ we would conclude that $1$ is the minimal element of $A$ and in this case it would also be the least element.
 
  • #12
evinda said:
So, in our case we don't have a least element, right?
Yes, the family of nonempty subsets of $\{1,2,3\}$ does not have the least element.

evinda said:
Can we say in the same way that $a$ is maximal w.r.t. set inclusion if there does not exist a set $b$ such that $b \supsetneq a$ ?
Yes, and it does not mean that maximal element is the greatest, though the greatest element is maximal.
 
  • #13
Evgeny.Makarov said:
Yes, the family of nonempty subsets of $\{1,2,3\}$ does not have the least element.

Nice... (Smile)

Evgeny.Makarov said:
Yes, and it does not mean that maximal element is the greatest, though the greatest element is maximal.

Could you give me an example of a set which has the greatest element that is also maximal? (Thinking)
 
  • #14
evinda said:
Could you give me an example of a set which has the greatest element that is also maximal?
As I said, greatest elements are always maximal. The family of all subsets of some set has the greatest element, which is the set itself. Also, in a total order the concepts of the maximal element and the greatest element coincide. The difference is meaningful only in partial orders.
 
  • #15
Evgeny.Makarov said:
As I said, greatest elements are always maximal. The family of all subsets of some set has the greatest element, which is the set itself. Also, in a total order the concepts of the maximal element and the greatest element coincide. The difference is meaningful only in partial orders.

$a$ is maximal w.r.t. set inclusion if there does not exist a set $b$ such that $b \supsetneq a$.

In our case, we cannot find a subset of $\{ 1,2,3 \}$ that is different from itself, therefore it is maximal, right? (Smile)

In order to check if a subset is the greatest do we check if $(\forall x \in A) x \subset a$ where $x$ is a subset of $A$? :confused:
 
  • #16
evinda said:
$a$ is maximal w.r.t. set inclusion if there does not exist a set $b$ such that $b \supsetneq a$.
Yes.

evinda said:
In our case
The concepts of maximal and greatest elements only make sense when the complete (partially) ordered set is specified. In this thread, several ordered sets have been mentioned, in particular, the family of all subsets of $\{1,2,3\}$ and the set of nonempty subsets of $\{1,2,3\}$. You need to specify which ordered set you are talking about.

evinda said:
we cannot find a subset of $\{ 1,2,3 \}$ that is different from itself
Your phrase is ambiguous. Do you mean we cannot find an $a\subseteq\{1,2,3\}$ such that $a\ne a$? Then of course we can't because $a=a$ always holds. Or do you mean we cannot find an $a\subseteq\{1,2,3\}$ such that $a\ne\{1,2,3\}$? Then we can: for example, $\varnothing\ne\{1,2,3\}$. Neither of these interpretations involves order, and you probably intended it to be involved. You need to express yourself more precisely.

evinda said:
therefore it is maximal, right?
$\{1,2,3\}$ is both maximal and greatest in the powerset of $\{1,2,3\}$.

evinda said:
In order to check if a subset is the greatest do we check if $(\forall x \in A) x \subset a$ where $x$ is a subset of $A$?
Your statements "$x\in A$" and "$x$ is a subset of $A$" contradict each other because an element in general is not a subset. Please write which subset are you checking for being the greatest element and in which ordered set. Also please remind if you use $\subset$ to denote arbitrary subset or strict subset.
 
  • #17
Evgeny.Makarov said:
Yes.

The concepts of maximal and greatest elements only make sense when the complete (partially) ordered set is specified. In this thread, several ordered sets have been mentioned, in particular, the family of all subsets of $\{1,2,3\}$ and the set of nonempty subsets of $\{1,2,3\}$. You need to specify which ordered set you are talking about.
I meant the set of nonempty subsets of $\{1,2,3\}$.
But thinking about it, both of the ordered sets, the family of all subsets of $\{1,2,3\}$ and the family of nonempty subsets of $\{1,2,3\}$, have as maximum and greatest element the set $\{ 1,2,3 \}$.
It is maximum since there does not exist a set $b \in \text{ family of all subsets of } \{1,2,3\} / \text{ family of nonempty subsets of } \{1,2,3\} $ such that $b \supsetneq \{ 1,2,3 \}$ and it is the greatest since $\forall x \in \text{ family of all subsets of } \{1,2,3\}/ \text{ family of nonempty subsets of } \{1,2,3\} $ we have that $x \subseteq \{ 1,2, 3\}$.

Or am I wrong? (Thinking)
 
  • #18
You are correct. I insisted on specifying the whole ordered set just in case, to avoid potential confusion. For example, $\mathcal{P}(\{1,2,3\})\setminus\{\{1,2,3\}\}$ has three maximal elements but no greatest element for the same reason that $\mathcal{P}(\{1,2,3\})\setminus\{\varnothing\}$ has three minimal elements but no least element.
 
  • #19
Evgeny.Makarov said:
You are correct. I insisted on specifying the whole ordered set just in case, to avoid potential confusion. For example, $\mathcal{P}(\{1,2,3\})\setminus\{\{1,2,3\}\}$ has three maximal elements but no greatest element for the same reason that $\mathcal{P}(\{1,2,3\})\setminus\{\varnothing\}$ has three minimal elements but no least element.

I understand... Thank you soo much! (Happy)
 

FAQ: Why Don't Minimal and Maximal Elements Imply Minimum and Maximum?

Why doesn't the converse hold?

The converse of a statement is not always true because it is a different statement with a different meaning. Just because a statement is true, it does not mean that its converse is automatically true.

What is the difference between a statement and its converse?

A statement is a proposition that is either true or false, while its converse is a statement that is formed by switching the hypothesis and conclusion of the original statement. They can have different meanings and truth values.

Can you provide an example of a statement and its converse?

An example of a statement and its converse is "If it is raining, then the ground is wet" and its converse "If the ground is wet, then it is raining." These two statements have different meanings and may not both be true.

Why is it important to understand that the converse of a statement may not be true?

Understanding that the converse of a statement may not be true is important because it helps us to avoid making incorrect assumptions. It allows us to critically analyze statements and not jump to conclusions based on the structure of a statement.

Can the converse of a statement ever be true?

Yes, the converse of a statement can be true in some cases. This usually happens when the original statement and its converse have the same meaning or when they are both logically equivalent. However, it is important to evaluate each statement and its converse separately to determine their truth values.

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