An isomorphism from a poset $(S_1,R_1)$ to a poset $(S_2,R_2)$ is a bijection $f: S_1 \rightarrow S_2$ such that, for all $x,y \in S_1$
$(x,y) \in R_1 \leftrightarrow (f(x), f(y)) \in R_2$
When such an isomorphism exists, we say that $(S_1,R_1)$ is isomorphic to $(S_2,R_2)$.
Questions:
- Show that every poset is isomorphic to itself.
Attempt: Every poset is isomorphic to itself...it's like the reflexive definition for equivalence relations which is $(\forall x \in S)[(x,x) \in R]$
So maybe I could apply the definition?
$(\forall x \in S)[(x,x) \in (S_1,R_1)]$
$(\forall x \in S)[(x,x) \in (S_2,R_2)]$
2.Prove that if $f$ is an isomorphism then so is $ f^{-1}$
Attempt: Definition 6.3.6 states that we let $R$ be any relation on a set $S$. The inverse relation of R, denoted $R^{-1}$ is defined by the condition $(x,y) \in R \leftrightarrow (y,x) R^{-1}$
Using the definition on $(x,y) \in R_1 \leftrightarrow (f(x),f(y)) \in R_2$, we have
$(x,y) \in R_1 \leftrightarrow (y,x) \in R_1^{-1}$
$(f(x),f(y)) \in R_2 \leftrightarrow f(y),f(x) \in R_2^{-1}$
$(y,x) \in R_1^{-1} \leftrightarrow f(y),f(x) \in R_2^{-1}$
It appears that they're kind of symmetrical in a way.
Prove that the composition of two isomorphisms is an isomorphism.
Attempt: Definition 6.3.9 states that we let $R_1$ and $R_2$ be relations on a set S. The composition of $R_2$ with $R_1$ is the relation
$R_2 \circ R_1 =[(x,y) \in S \times S:( \exists v \in S)[(x,y) \in R_1 \land (v,y) \in R_2]$
Maybe we should have a composition like $(S_1,R_1) \circ (S_2, R_2)$. Then ,
$(S_2, R_2) \circ (S_1,R_1) =[(x,y) \in S \times S:( \exists v \in S)[(x,y) \in (S_1,R_1) \land (v,y) \in (S_2, R_2)]$
I'm kind of confused and lost...maybe the transitive property would work better?
Definition: R is transitive if $(\forall x, y, z \in S)[((x,y) \in R \land (y,z) \in R) \rightarrow (x,z) \in R]$
so maybe let $x = (S_1,R_1), y = (S_2, R_2)$ and $z = (S_3,R_3)$. Then
$(\forall (S_1,R_1), y, z \in S)[(((S_1,R_1),(S_2, R_2)) \in R \land ((S_2, R_2),(S_3,R_3)) \in R) \rightarrow ((S_1,R_1),(S_3,R_3)) \in R]$...
If the original poset definition was being used here I could understand it a little bit better, but this is a different form. Any hints?