Homework Help: Posets and minimal elements - Looking for an inductive proof

1. Dec 13, 2012

Kolmin

1. The problem statement, all variables and given/known data

Suppose $R$ is a partial order on a set $A$. Then every finite, nonempty set $B \subseteq A$ has an $R-minimal$ element.

2. Relevant equations

Partial orders are characterized by:

Reflexivity: $xRx$
Transitivity: $xRy \wedge yRz \rightarrow xRz$
Antisimmetry: $xRy \wedge yRx \rightarrow x=y$

Minimal elements can be defined in two equivalent ways:

$\neg \exists x \in X (xRb \wedge x \neq b)$
$\forall x \in X (xRb \rightarrow x=b)$

Problems:
First of all I am not sure if the following is a real proof of this statement. I have some problems with inductive proofs and I am particularly worried by the "Assume the subset $B$ of $A$ has cardinality n and it has a $R-minimal$ element" you are gonna find in the proof I wrote down. Can I really assume that?
Secondly, if the proof works, how is it? Too wordly and fuzzy? Efficient and perspicuous?
I have the feeling there is too much, but what can I cut?

Thanks a lot for any of your feedbacks. I am really looking forward to read them.

3. The attempt at a solution

Proof:
Let $B$ be an arbitrary subset of $A$. We prove the theorem by induction on the cardinality of $B$.
i) Base step:
Assume the subset $B$ of $A$ has cardinality 2. By assumption $R$ is a partial order on $A$, thus we have two cases. Either by antisimmetry the two elements are equal, or they are different. If they are equal, by definition, they are both $R-minimal$ elements of $B$. If they are different, one of the two has to be in the relation $R$ with the other element. In both cases, we are assured to have a $R-minimal$ element in $B$.
ii) Inductive step:
Assume the subset $B$ of $A$ has cardinality n and it has a $R-minimal$ element. Adding an element to the subset $B$ improves the cardinality of $B$ to n+1. We define this new set of cardinality n+1 as $B'$. The addition of a new element to $B$ to construct $B'$ gives us three cases.
Case 1. The new element is equal to the minimal element of $B$. Thus, by antisimmetry $B'$ has a two minimal elements.
Case 2. The new element is higher than the minimal element $B$. Thus $B'$ has a minimal element that is the same of $B$.
Case 3. The new element is lower than the minimal element $B$. Thus, $B'$ has a minimal element, that is the element added to $B$ to construct $B'$.
This exhausts all the possibilities. Henceforth the result is proven.

2. Dec 13, 2012

HallsofIvy

Your proof looks good to me but the word you want in the last line is "hence". "Henceforth" means "from now on".

3. Dec 13, 2012

Kolmin

Non native writer...

BTW, thanks a lot. I really felt it was too wordly.

4. Dec 13, 2012

Michael Redei

Your "base step" is flawed. You say "Either by antisimmetry the two elements are equal, or they are different." Who needs antisymmetry (with a Y in "symmetry" by the way) for that? Anyway, if there are two elements, they can never be equal, because then they'd just be one element.

So you begin with two elements, say, a and b. Who says that one needs to bear the relation R to the other? (Remember: "poset" = "PARTIALLY ordered set", i.e. there can be elements a,b that fulfil neither aRb nor bRa.)

If you have two elements a,b, then you can have aRb (which means ¬bRa, because a≠b), in which case a is minimal. Or bRa, which means that b is minimal. Or neither aRb nor bRa, and so both a and b are minimal.

In your "Inductive step" you say that the cardinality of B is "improved" to n+1, i.e. made better. I think you mean "increase" here. And you need to check your three cases:

Case 1 is impossible. If the "new" element is equal to any "older" one, it can't be new. (If you add the element "banana" to the set {apple,banana,carrot}, you're not increasing the cardinality of that set.)

You'r missing a Case 4: what happens if the new element is neither "higher" nor "lower" than the minimal element of B? Again, since R is only a partial ordering, this is entirely possible.

5. Dec 13, 2012

Kolmin

Huge mistake..I completely forgot that the notion of set implies that.

For a second, I thought that I was implicity assuming that it was a loset and not a poset. So, here we are: indeed I assumed completeness.

I don't see why it is the case.

This. I was looking for that word!

Lot of stuff to think about. Give me a sec.

Btw, I think that my problem is related to the fact that I don't see why you don't need completeness to get the result. In particular I think the main issue is that I don't see why, if neither aRb nor bRa, then both a and b are minimal.

Oh, btw...really thanks a lot.

6. Dec 13, 2012

Kolmin

Statement:
Suppose $R$ is a partial order on a set $A$. Then every finite, nonempty set $B \subseteq A$ has an $R-minimal$ element.

Proof:
Let $B$ be an arbitrary subset of $A$. We prove the statement by induction on the cardinality of $B$.
i) Base step:
Assume that $B$ is a singleton. Then the only element is by definition a $R-minimal$ element of $B$, by reflexivity of $R$.
ii) Inductive step:
Assume that $B$ has cardinality $n$ and that it has a $R-minimal$ element called $b$. Increase the cardinality of $B$ to $n+1$ by adding an element, say $b'$, and define this new set $B \cup \{b'\}$ of cardinality $n+1$ as $B'$. Thus we have three possible cases that define the relation $R$ between $b$ and $b'$.
Case 1. $bRb'$ : Thus $B'$ has a minimal element that is the same of $B$.
Case 2. $b'Rb$: Thus, $B'$ has a minimal element, that is the element added to $B$ to construct $B'$.
Case 3. $\neg (bRb' \vee b'Rb)$: Thus $b$ is still the $R-minimal$ element of $B'$.
Since this exhausts all the possibilities, the result is proven.

7. Dec 13, 2012

Kolmin

The server stop gave me enough time to get something hopefully decent that should work.

Still, I have some doubts.

Can I prove the result without going backward to the singleton cases?
Or, in other words, is there a way to prove it having as a base case the two elements one I used in my first attempt?

Is the "by reflexivity of R" redundant?

Is stylistically decent the way in which I introduce those new elements and sets?
Is it mathematically sound?
Too wordly?

Not enough explanations?
Are those lines a bad explanation?

Anyway, thanks a lot for any feedback.

8. Dec 13, 2012

Michael Redei

Perhaps we need more of those server pauses. I realised that I may have sounded a bit abrupt. I didn't mean to seem impolite, so I'm sorry if I did appear that way.

Yes, "reflexivity" is redundant here. For a singleton element there exists no smaller one, so it must be minimal. In fact, this "no smaller" definition of "minimal" is one that you could use more. Suppose you have two elements a and b such that no smaller elements exist and neither aRb nor bRa is true. Then both a and b are minimal. I'd keep the singleton as your base case though.

A bit repetitious, I'd say, since you're saying the same things more than once. You could shorten this as follows:

Assume that $B$ has cardinality $n$, and that $b$ is an $R$-minimal element of $B$. Now we consider a new set $B'=B\cup\{b'\}$ of cardinality $n+1$. Then we have three possible cases for the relation $R$ between $b$ and $b′$.

Case 3 is missing a part. Suppose neither $bRb'$ not $b'Rb$, as you have done. For $b'$ to be minimal there must be no other element $a$ that is smaller than $b'$. How can you be sure of that?

This is where your whole proof becomes complicated. Instead of looking at a minimal element $b$ of $B$ and constructing three cases, I'd begin with $b'$ and ask: what elements of $B$ stand in the Relation $R$ to $b'$? Suppose $S$ is the set of all these elements, i.e. $S = \{a\in B : aRb' \lor b'Ra\}$. This set has at most $n$ elements, so you can use it for your inductive step.

Now you'll need to fiddle around a bit, depending on whether $b'$ is minimal in $S$ or not, and you need to consider the elements outside $S$ (but still in $B$). You can probably use the fact that $R$ is transitive to show that the elements outside $S$ won't interfere with what happens inside $S$.

9. Dec 13, 2012

pasmith

You want your result to be true for all non-empty subsets, and singletons are non-empty subsets. I would say that it was obvious from the definition that a singleton subset has an R-minimal element.

The key point is that you're adding an element which isn't in $B$, so you'd better make that clear:

"Assume $B$ has cardinality $n$ and has an R-minimal element $b$. Let $B' = B \cup \{b'\}$ with $b' \in A \setminus B$. Then $B'$ has cardinality $n+1$."

I don't think you need to refer expressly to the cardinality of $B$. Actually much of the set-up could be abbreviated:

"Since a singleton subset has an R-minimal element, it is enough by induction on the cardinality of $B \subset A$ to show that if $B$ has an R-minimal element $b$ then the set $B' = B \cup \{b'\}$ where $b' \in A \setminus B$ has an R-minimal element."

And then you consider the possibilities.

For Case 3 you also have to show that there is no other element $a \in B$ such that $aRb'$.

10. Dec 13, 2012

Kolmin

Well, don't worry. I appreciate the fact that you had this thought, but it's really not a problem and I didn't feel it. Actually, to be honest, when I started to read your post and I saw "base case" with the inverted commas, at the beginning I thought you wanted to be sarchastic and I kinda liked it, cause the inverted commas with a sarchastic inflexion is the default way in which I would always present my - indeed! - "proofs"...
So, thanks a lot (probably I start to sound repetitive).

Not sure if this is a typo.
Actually with case 3, with the two elements that bear no relations, I wanted to show more or less exactly the opposite, which is that we are at least sure that $b$ (and not $b'$) is a minimal element of $B'$, and we can have at most two different minimal elements (indeed $b$ and $b'$).

But are we "at least sure" of it?

11. Dec 13, 2012

pasmith

I think you can argue that if $aRb'$ then $a$ must be an R-minimal element of $B$, because otherwise one has $bRa$ and $aRb'$ so that $bRb'$, which we are assuming not to be the case. But $aRb'$, so $a$ is an R-minimal element of $B'$.

12. Dec 13, 2012

Kolmin

Is it not a bit redundant?
My line of reasoning is more or less the same of that I wrote down in my last reply to Michael Redei.
I add an element to $B$ and I come out with $B'$. Now, I assume that $B$ has a minimal element. So, if the new element and the minimal element of $B$ bear no relation, still $b$ is a minimal element (of - let's say - the right side of the Hasse diagram). So whatever $a$ we prove bears certain relation with $b'$ (on - let's say - the left side of the Hasse diagram), at most it makes us find another minimal element, but we already have it. Thus, it should be redundant.

Is it right or not? This was my line of reasoning when I wrote down my "proof" (yes...inverted commas with sarchastic inflexion!, but not sure it is sound.

Btw, thanks a lot for your feedback.

13. Dec 13, 2012

Kolmin

14. Dec 13, 2012

pasmith

Not so much redundant as wrong: I should have said "For Case 3 you also have to show that if there is any other element $a \in B$ such that $aRb'$ then $a$ is an R-minimal element of $B$."

Anyway, I think between the various responses we now have a proof.

15. Dec 14, 2012

Kolmin

To me even that one looks redundant for the reasons I specified.

16. Dec 14, 2012

pasmith

On reflection, you are right. Clearly all minimal elements of B are the same size, so either b' is less than all of them, b' is greater than all of them, or b' is the same size as all of them. In all three cases B' has a minimal element.

17. Dec 14, 2012

Michael Redei

What do you mean by "the same size"? Since R is antisymmetric, if a and b are "the same size", they're actually equal. How would your argument work for the following relation R?

For any two rational numbers $x$ and $y$ we define the relation $R$ as follows: Let $x={n_x}/{d_x}$ and $y={n_y}/{d_y}$ where $(n_x,d_x) = (n_y,d_y) = 1$, i.e. the numerator and denominator of a fraction have no common factor. Then we set
$$xRy ~~ \mbox{iff} ~~ d_x=d_y ~~ \mbox{and} ~~ n_x\leq n_y.$$
Obviously, $xRy$ can only be true if $x\leq y$, but $x\leq y$ doesn't necessarily imply $xRy$.

Now let $B$ be the set of all positive fractions whose numerator is even and whose denominator is a single digit. Then $\frac21,\frac23,\frac25,\frac27,\frac29$ are all minimal elements of $B$, but are they all "the same size"? Suppose they are (since none can be said to be "smaller" or "larger" than an other), what happens if you set $b'=\frac13$? This is $R$-less than $\frac23$, but not comparable to the other elements of $B$.

18. Dec 14, 2012

pasmith

"a is the same size as b" is defined by the relation $aSb \Leftrightarrow (\neg(aRb \vee bRa)) \vee (a = b)$.

S is obviously reflexive and symmetric and is also transitive, so that for all a, b, and c, $aSb \wedge bSc \Rightarrow aSc$. This is obvious if
any of a,b, and c are equal, so let all three be distinct. Then
$$aSb \wedge bSc \Leftrightarrow (\neg(aRb \vee bRa)) \wedge (\neg(bRc \vee cRb)) \Leftrightarrow \neg(aRb \vee bRa \vee bRc \vee cRb) \\ \Leftrightarrow \neg((aRb \vee bRc) \vee (cRb \vee bRa)) \\ \Leftrightarrow \neg(aRb \vee bRc) \wedge \neg(cRb \vee bRa) \\ \Rightarrow \neg(aRb \wedge bRc) \wedge \neg(cRb \wedge bRa)$$
where the last line follows because for all statements P and Q, $\neg(P\vee Q) \Rightarrow \neg(P \wedge Q)$. We then have
$$\neg(aRb \wedge bRc) \wedge \neg(cRb \wedge bRa) \Leftrightarrow \neg(aRc) \wedge \neg(cRa) \Leftrightarrow \neg(aRc \vee cRa) \Leftrightarrow aSc$$
by transitivity of R. Thus S is an equivalence relation.

It should be obvious that for all distinct a and b there are three possibilities: either aRb, bRa, or aSb. It should also be obvious that if a and b are minimal elements of a subset then aSb.

EDIT: Alternatively one can let $b \in B$ be minimal, but otherwise arbitrary. Then if $b'Rb$ then $b'$ is minimal in $B'$, and if $bRb'$ then $b$ is minimal in $B'$, and if neither $bRb'$ nor $b'Rb$ for any minimal $b$ then minimal elements of $B$ are minimal elements of $B'$.

Last edited: Dec 14, 2012
19. Dec 14, 2012

Michael Redei

So "same size as" means "incomparable or equal". This is the first time I've seen that definition.

Not at all obvious, and, in fact, false. S need not be transitive.

The transitivity of R implies $\neg(xRy\land yRz) \Leftarrow \neg(xRz)$, but not the converse, $\neg(xRy\land yRz) \Rightarrow \neg(xRz)$. So the first "$\Leftrightarrow$" in that line should only be "$\Leftarrow$".

That follows trivially from your definition of S. Either a is less than b, greater than b, equal to b or incomparable to b.

I still don't see how you arrive at this result though, which seems contrary to the example I have given before:

20. Dec 14, 2012

Kolmin

I don't enter in the discussion about "same size" because my mathematical skills are not that advanced, even if basing on my limited knowledge I would agree with Michael Redei.

Btw, going back to my original Case 3 of the proof, I think that this example based on fractions supports exactly my line of reasoning. Indeed, if we set $b'=\frac13$, it does change the composition of the set of the minimal elements (it takes the place of $b'=\frac23$), but we don't care, as far as we know that we do have a minimal elements.

In other words, considering for example this random Hasse diagram (it is really not important how it is), let's imagine that $b'$ is something that bears no relation to $\{∅\}$, but it's related to $\{z\}$. Well, we did have before the addition of $b'$ a minimal element, namely $\{∅\}$, and we still have it.
Adding $b'$ doesn't alter the fact that we do have a minimal element: it simply alters the composition of the set of the minimal elements of a given poset, but that's beyond the result we have to prove (I would call it a corollary).

In my "proof", the minimal element that still stands, beyond any addition, is $b$, thus the result is proven.