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This is one of the question I'm working on:

Suppose $A$ is a set, $F \subseteq \mathbb{P(A)}$, and $F \neq \emptyset$. Then prove that the greatest lower bound of $F$ (in the subset partial order) is $\cap F$.

Now this is my attempt at this problem:

We know that $\cap F$ is a lower bound of F since $\forall X \in F (\cap F \subseteq$ X). Now we need to prove that this is the greatest lower bound of $F$.

Now I'm stuck here. How to show that it is the greatest lower bound ?

Sibi
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2 Answers2

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HINT: Show it directly from the definition of greatest lower bound. Suppose that $L$ is a lower bound for $F$. Then $L\subseteq X$ for each $X\in F$. What can you say about the relationship between $L$ and $\bigcap F$?

Brian M. Scott
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  • Thanks, I think I have to show that $L \subseteq \cap F$. But I cannot understand why it has to be true. Let's take an example of $A = {1,2}, F = { {2 }, {1,2}}$. Now $L \subseteq \cap F$ is not true because ${1,2}$ is not a subset of ${2}$. – Sibi Apr 14 '15 at 21:19
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    @Sibi: But ${1,2}$ is not a lower bound for $F$ at all, so it’s irrelevant. The only lower bounds for that $F$ are $\varnothing$ and ${2}$. – Brian M. Scott Apr 14 '15 at 21:22
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∩F is the smallest element in the subset partial order ( as you pointed out ∀X∈F(∩F⊆ X)). Therefore it is the greatest lower bound.

alexgiorev
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  • How is it smallest ? That's the thing I'm not able to understand. – Sibi Apr 14 '15 at 21:21
  • Well, by definition: Suppose R is a partial order on a set A, B ⊆ A, and b ∈ B. Then b is called an R-smallest element of B (or just a smallest element if R is clear from the context) if ∀x ∈ B(bRx). Let S = {(X,Y)∈P(A)×P(A)| X⊆Y}. Then by the definition above, Y∈F is the smallest-S element of F if ∀X∈F(YSX), or in other words∀X∈F(Y⊆X) . But you have already pointed out that ∀X∈F(∩F⊆ X). – alexgiorev Apr 14 '15 at 21:30