Understanding of Proof of Local Duality Principle

Question

From https://proofwiki.org/wiki/Duality_Principle_(Order_Theory)/Local_Duality

Theorem: Let $\Sigma$ be a statement about ordered sets, and $\Sigma^*$ be the dual statement of $\Sigma$. Let $(S,\preceq)$ be an ordered set, and $(S,\succeq)$ be its dual. Then the following are equivalent:

$\qquad\qquad$(1): $\Sigma$ is true for $(S,\preceq)$.

$\qquad\qquad$(2): $\Sigma^*$ is true for $(S,\succeq)$.

To give an example, let the set $S$ be $\{0,1\}$ ordered by the standard ordering $\leq$.

So here is a true statement ($\Sigma$): $0\leq1$.

Here is the dual statement ($\Sigma^*$): $0\geq1$. But this is not true, right? Am I missing something?

For context, ProofWiki defines dual statements to be those obtained simply by replacing every reference to $\preceq$ in $\Sigma$ with a reference to its dual $\succeq$.

Answer 1

You seem to be "over-intepreting" (is that a thing?) the statements, and using the intuitive notions of "smaller" and "bigger".

Ok, so we start with an order $\leq$. Whenever we say that "$a$ is smaller than $b$", we just mean that $a\leq b$, where $\leq$ was our given preorder. More formally, we simply mean that "the statement written as '$a$', followed by the relation '$\leq$', followed by '$b$', is true".

Then we define a new relation $\geq$, in terms of $\leq$, as: $$a\geq b\iff b\leq a$$

But it just so happens that $\geq$ itself is also a preorder, in the formal meaning:

• $\geq$ is reflexive: $a\geq a$
• $\geq$ is anti-symmetric: $a\geq b$ and $b\geq a$ imply $a=b$
• $\geq$ is transitive: $a\geq b$ and $b\geq c$ imply $a\geq c$.

Cool, so now we can say that $a$ is smaller than $b$ using this new relation: $a$ is smaller than $b$ if "the statement written as '$a$', followed by the relation '$\geq$', followed by '$b$', is true". This means that $a\geq b$, which, by definition, means that $b\geq a$, so "$b$ is smaller than $a$"...

This is a problem with the notion of "smaller than" (or "greater than"): it depends on what order relation we have previously fixed! When we interpret a statement in a poset, we need to be clear about which order we are considering.

So the statement is "$0$ is smaller than $1$". If we interpret it in $S$, it means that '$0$ is smaller than $1$ in the preorder given by $\leq$'.

The dual statement is "$0$ is greater than $1$". If we interpret it in the dual $S^*$, it means that '$0$ is greater than $1$ in the preorder given by $\geq$', or equivalently '$1$ is smaller than $0$ in the preorder given by $\geq$', that is '$1\geq 0$', which is true enough.

The best way to avoid these problem is to avoid using ambiguous phrases such as "smaller" and "bigger", and really focus on how the statement are being interpreted.

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Formal details:

The problem is really with the proper formalization of the dualization process. Here's a way to formalize it using the same nomenclature as you (in the spirit of Universal Algebra). I'll gloss over some details, which hopefully won't be of much problem.

Definition: The theory of posets consists of two binary relation symbols "$\leq$" and "$\geq$" (in this order) and the following axioms:

• $\leq$ is reflexive, anti-symmetric and transitive;
• for all $x$ and all $y$, $x\leq y$ if and only if $y\geq x$.

The language of posets is simply the language that we need to talk about posets. We do not care about what posets are at this point. We do it in the next one, though:

Definition: A poset is a set $S$ with two given interpretations $\leq^S$ and $\geq^S$ (i.e., binary relations on $S$) which satisfy the axioms for posets.

But remember, the order of $\leq^S$ and $\geq^S$ matters. So to be more explicit whenever necessary, we will call the triple $(S,\leq^S,\geq^S)$ a poset instead.

Then we define terms, formulas and sentences as "phrases which make sense when you read them" (again, I'm glossing over details). If we have a given poset $(S,\leq^S,\geq^S)$, we may also form phrases involving the elements of $S$.

If $\sigma$ is a sentence, the dual sentence $\sigma^*$ is just what we obtain from $\sigma$ by swapping $\leq$'s and $\geq$'s.

Now we need to know what it means to say that a statement is true (again, glossing over details):

If $\sigma$ is a sentence in the language of posets and $(S,\leq^S,\geq^S)$ is a poset, then the value of $\sigma$ in $(S,\leq^S,\geq^S)$ is

• true, if the phrase we get by interpreting the symbols $\leq$ and $\geq$ as $\leq^S$ and $\geq^S$, respectively, is true.
• false, otherwise.

We denote the value of $\sigma$ in $(S,\leq^S,\geq^S)$ as $\sigma^S$.

So here is the dualization process:

Fact: If $(S,\leq^S,\geq^S)$ is a poset, then $(S,\geq^S,\leq^S)$ is a poset as well.

The second poset in the fact above is called the dual of $S$: $S^*=S$ as a set, but the interpretations of the "smaller than" and "greater than" orders are reversed: $\leq^{S^*}=\geq^S$ and $\geq^{S^*}=\leq^S$.

Here's a nice restatement for your theorem:

Theorem: Let $(S,\leq^S,\geq^S)$ be a poset, and consider it's dual $(S^*,\leq^{S^*},\geq^{S^*})$. Let $\sigma$ be a statement in the language of posets (possibly involving elements of $S$).

Then $\sigma^S=(\sigma^*)^{S^*}$, that is, the value of $\sigma$ in $S$ is the same as the value of $\sigma^*$ in $S^*$.

What about your example? $\sigma=0\leq 1$; The dual is $\sigma^*=0\geq 1$.

When we interpret $\sigma$ in $S=\left\{0\leq^S 1\right\}$, we get a true statement: $0\leq^S 1$, which is true by definition.

When we interpret $\sigma^*$ in $S^*=\left\{1\leq^{S^*}0\right\}$, we also get a true statement: $$(\sigma^*)^{S^*}\iff(0\geq 1)^{S^*}\iff 0\geq^{S^*}1\iff 0\leq^S1$$ where the last equivalence follows from the definition of the dual. The last statement is true, so the first one is as well. No problem here as well.