Showing a relation is a partial order on a set

ironspud
Messages
10
Reaction score
0

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!
 
Physics news on Phys.org
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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top