To be continued....
#Aryan help - 2
steep steeple
tiny oceanBOT
steep steeple
@smoky slate I get that what I'm asking here is when we take the equivalence class as [y] = { x | (x,y) € R } how are we saying that in the set [y] everything will be related to each other or such a relation must have existed in the original equivalence relation
This
smoky slate
That is by transitivity
Let a and b be in [y]
Then we know (y,a) and (y,b) are in the relation
By symmetry and transitivity, (a,b) is in the relation
steep steeple
Yes because symmetry
smoky slate
So you see everything in [y] is related to each other
steep steeple
So since we know that it's an equivalence relation we know that for every (x,y) that formed that equivalence class so in the relation there must exist (x,x) and (y,y) and (y,x) among others due to the reflexive symmetric and transitive properties of the equivalence relation
Sorry for bad english
In that way we can say that everything in the equivalence class must be related to each other
pulsar jewel
The way you worded this seems overly complicated.
Try and rerun herzogs proof a second ago.
If I pick a and b from [y] how can I show a relates to b?
steep steeple
Ok so if I pick a and b from [y] then we can say that (a,y) and (b,y) is a relation that exists right. Since it's an equivalence relation therefore (y,b) also must exist using transitive property on (a,y) and (y,b) we can say that (a,b) also must exist @smoky slate
smoky slate
Yes
steep steeple
Lessgoo
smoky slate
So you proved that any two elements must be related if they are in the same equivalence class
Type .solved if your question is solved
steep steeple
How do u make the graph of the equivalence relation? I thought only poset had hasse diagram
Is it just a directed graph showing relations
smoky slate
Yes
steep steeple
Ok thanks 🙏🏼😋👍🏼💯
smoky slate
All edges will be bidirectional, so sometimes people just put undirected edges since it is understood
steep steeple
Bidirectional edges self loops
Showing reflexive and symmetry
And fully connected graph showing transitivity
smoky slate
Fully connected? What does that mean
steep steeple
Every node connected to every other
Uh like a complete graph my bad
Not connected
smoky slate
Do you agree that {(a,a),(b,b)} is an equivalence relation on {a,b}
steep steeple
Yes it's sufficient
smoky slate
Will its graph be fully connected?
steep steeple
Ooooooo
smoky slate
It will just be two separate nodes with self loops
smoky slate
But all the vertices in an equivalence class will be fully connected with each other
That is true
steep steeple
But they weren't in ur example
steep steeple
Like a partition was fully connected
smoky slate
Yes
steep steeple
But not every partition was interconnected and they wouldn't be by definition because they are infact partition
smoky slate
Yes