greatest least bound and least upper bound proof

tae1466 · #1 $Old$ November 4th, 2009, 02:21 PM

I've been struggling with this one.

Let A be a partially ordered set. Suppose X⊆Y⊆A.

1. Assuming that all the least upper bounds and greatest lower bounds exist, prove that glb(Y) ≤ glb(X) ≤ lub(X) ≤ lub (Y)

2. Find two subsets X and Y of Real Numbers for which X is a proper subset of Y and yet glb(Y) = glb(X) and lub(X) = lub(y)

Drexel28 · #2 $Old$ November 4th, 2009, 06:44 PM

Quote:

Originally Posted by tae1466 $View Post$

I've been struggling with this one.

Let A be a partially ordered set. Suppose X⊆Y⊆A.

1. Assuming that all the least upper bounds and greatest lower bounds exist, prove that glb(Y) ≤ glb(X) ≤ lub(X) ≤ lub (Y)

2. Find two subsets X and Y of Real Numbers for which X is a proper subset of Y and yet glb(Y) = glb(X) and lub(X) = lub(y)

1. I'll do part of this. See if you can do the rest.

Problem: Let $X\subseteq Y$ . Prove that $\sup\left(X\right)\le\sup\left(Y\right)$

Proof: We know that since $X\subseteq Y$ that $x\in X\implies x\in Y$ . But $\forall x\in Y\quad x\le\sup\left(Y\right)$ , so $\forall x\in X\quad x\le \sup\left(Y\right)$ . Therefore $\sup\left(Y\right)$ is an upper bound for $X$ . And by definition $\sup\left(X\right)$ is at least as small as any upper bound of $X$ . The conclusion follows.

2. What about $X=(0,1)$ and $Y=[0,1]$ ?