MHB Do we have to show that the relation is irreflexive?

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The discussion centers around the properties of relations, specifically focusing on whether a strict order \( S \) must be shown to be irreflexive. It is established that \( S \) is defined as \( R - I_A \), indicating that it is derived from an order \( R \) by removing reflexive pairs. The participants explore the implications of this definition, particularly questioning the necessity of proving \( S \) is irreflexive alongside its transitivity. Clarifications are sought regarding the relationships between elements in \( S \) and the implications for antisymmetry. The conversation emphasizes the importance of clearly articulating logical reasoning in mathematical proofs.
evinda
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Hello! (Smile)

  • Let $R$ be an order of the set $A$. Then $R$ induces a strict order $S$ at the set $A$.
    $$$$
  • Let $S$ be a strict order of the set $A$. Then $S$ induces an order $R$ at the set $A$.

The first sentence is proven like that in my notes:

We define $S:=R-I_A$ and we can see that $S$ is a strict order at $A$.

$$x \in A\\y \in A$$

$$xSy \leftrightarrow \langle x,y \rangle \in S=R-I_A\\ \langle x,y \rangle \in R \wedge \langle x,y \rangle \notin I_A$$

We want to show that $\langle y,x \rangle \notin S$.
If not then $\langle y,x \rangle \in S \wedge \langle y,x \rangle \in I_A \leftrightarrow \langle x,x \rangle \notin I_A$.In that way we have shown that $S$ isn't weak antisymmetric.
But don't we also have to show that $S$ is irreflexive?
We know that $S$ is transitive as a subset of a transitive relation, right? (Thinking)
 
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evinda said:
We define $S:=R-I_A$ and we can see want to show that $S$ is a strict order at $A$.

evinda said:
$$x \in A\\y \in A$$
For all $x,y$? For some $x,y$? Which ones? Please use words.

evinda said:
$$xSy \leftrightarrow \langle x,y \rangle \in S=R-I_A\\ \langle x,y \rangle \in R \wedge \langle x,y \rangle \notin I_A$$
OK, I am being picky, but what is the relationship of the second line to the first?

evinda said:
We want to show that $\langle y,x \rangle \notin S$.
Why do we want to show that?

evinda said:
If not then $\langle y,x \rangle \in S \wedge \langle y,x \rangle \in I_A \leftrightarrow \langle x,x \rangle \notin I_A$.
There are too many ways to parse this.

(1) $\langle y,x \rangle \in S$, and also $\langle y,x \rangle \in I_A$ is equivalent to $\langle x,x \rangle \notin I_A$ (but both sides of the equivalence may be false).

(2) $\langle y,x \rangle \in S$ and $\langle y,x \rangle \in I_A$ are together equivalent to $\langle x,x \rangle \notin I_A$ (but both sides may be false).

(3) We have $\langle y,x \rangle \in S$ and $\langle y,x \rangle \in I_A$, and since these statements together are equivalent to $\langle x,x \rangle \notin I_A$, we have the latter.

There are possibly other ways. Please stop abbreviating the flow of reasoning with arrows. This is not helping. Please write words.

evinda said:
In that way we have shown that $S$ isn't weak antisymmetric.
I am not familiar with this concept.

evinda said:
But don't we also have to show that $S$ is irreflexive?
Yes.

evinda said:
We know that $S$ is transitive as a subset of a transitive relation, right?
Not every subset of a transitive relation is transitive.
 

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