Why doesn't the converse hold?

[evinda]

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)

 

[Evgeny.Makarov]

What you call minimum and maximum and usually called least and greatest.

Consider subsets of some set ordered by inclusion.

 

[evinda]

Consider subsets of some set ordered by inclusion.

Could you explain it further to me? (Thinking)

 

[Evgeny.Makarov]

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.

 

[evinda]

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)

 

[Evgeny.Makarov]

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.

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.

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?

 

[evinda]

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)

 

[Evgeny.Makarov]

Each subset has itself as a subset, right?

Yes, a non-strict subset.

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?"

 

[evinda]

"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)

 

[Evgeny.Makarov]

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.

 

[evinda]

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)

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.

 

[Evgeny.Makarov]

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.

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.

 

[evinda]

Yes, the family of nonempty subsets of $\{1,2,3\}$ does not have the least element.

Nice... (Smile)

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)

 

[Evgeny.Makarov]

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.

 

[evinda]

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 $A$?

 

[Evgeny.Makarov]

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

Yes.

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.

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.

therefore it is maximal, right?

$\{1,2,3\}$ is both maximal and greatest in the powerset of $\{1,2,3\}$.

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.

 

[evinda]

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)

 

[Evgeny.Makarov]

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.

 

[evinda]

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)