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Thread: greatest least bound and least upper bound proof

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    greatest least bound and least upper bound proof

    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)
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    MHF Contributor Drexel28's Avatar
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    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 $\displaystyle X\subseteq Y$. Prove that $\displaystyle \sup\left(X\right)\le\sup\left(Y\right)$

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

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