Showing a relation is a partial order on a set

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SUMMARY

The discussion focuses on proving that the relation defined by X ≤ Y if X ⊆ Y is a partial order on a universal set U. A relation qualifies as a partial order if it is reflexive, antisymmetric, and transitive. The user initially struggled with the proof of reflexivity but later clarified their understanding with the help of their professor. The second part of the problem raises concerns about defining this relation on the set of all sets, which leads to paradoxes such as Russell's Paradox.

PREREQUISITES
  • Understanding of set theory and universal sets
  • Knowledge of partial order definitions: reflexivity, antisymmetry, transitivity
  • Familiarity with logical implications in mathematical proofs
  • Awareness of Russell's Paradox and its implications in set theory
NEXT STEPS
  • Study the properties of partial orders in set theory
  • Research Russell's Paradox and its impact on set definitions
  • Learn about other types of relations, such as total orders and equivalence relations
  • Explore formal proof techniques in mathematics, particularly in set theory
USEFUL FOR

Students of mathematics, particularly those studying set theory, logic, and proof techniques. This discussion is also beneficial for educators seeking to clarify concepts related to partial orders and their applications.

ironspud
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Homework Statement



Okay, so here's the problem:

(a) Let U be a universal set and suppose that X,Y\in U. Define a relation,\leq, on U by X\leq Y iff X\subseteq Y. Show that this relation is a partial order on U.

(b) What problem occurs if we try to define this as a relation on the set of all sets?


Homework Equations



A relation R is a partial ordering if R is a reflexive, antisymmetric, and transitive relation.

A relation R on a set A is reflexive if, for all x\in A, x R x.

A relation R on a set A is antisymmetric if, for all x,y\in A, x R y\wedge y R x\Rightarrow x=y.

A relation R on a set A is transitive if, for all x,y,z\in A, x R y\wedge y R z\Rightarrow x R z.


The Attempt at a Solution



I'm really lost here. On part (a), I thought I was doing fine at first, but the more I think about it, the more I feel I'm way off base. Here's what I mean:

Proof that R is reflexive:
Let a\in X.
Since a\in X, then a\in X.
Thus, X\subseteq X.
Therefore, (a,a)\in R.

I think this would make sense if I was trying to prove the relation was a subset of X\times X (right?), but I'm trying to show the relation on U with X,Y\in U. With that, I think, being the case, I really have no idea how to proceed.

Anyway, clearly I'm over my head here. If anyone could help me out, I'd really appreciate it.

Thanks!
 
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Nevermind. I just spoke with my professor. Seems I was just over thinking everything.

Still, if anyone would like to provide some insight to part (b), I'd appreciate it.
 

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