Finding Lower Bounds for {2, 4} using Partial Order

[BubblesAreUs]

Since I'm not sure if posting assignment questions is allowed, I'm just going to ask specific questions just to be safe.

1. Homework Statement

Find all the lower bounds of given pair. Say (a, b) in T.

Homework Equations

Proof for greatest lower bound:

∀g,a,b ∈ T ⇔ ( g ≺ a) ^ (g ≺ b) ^ ( ∀l ∈ T [ (l ≺ a ) ^ ( l ≺ b)] ⇒ (l ≺ g)

by "≺", I meant Partial Order, ≤

The Attempt at a Solution

Since g are ordered before a and b. Can we assume for pair {2, 4}, g will be equivalent to

{empty set}, {1,0}, {1,1}, {1,2}, {1,3}, {1,4}, {2, 1}, {2,2}, {2,3}, {2,4},

I am not sure if I'm on the right track, but that's what it seems to be according to the proof.

 

[fourier jr]

In part 1 you write (a, b) which could be an ordered pair or an open interval in ℝ but then you use set notation in part 3. Those could have greatest lower bounds if there's a partial ordering defined on them but I think I need more info. Part 2 makes sense though, a lower bound different from g is less than g but what are they & how are they being compared?

edit: for example with {2, 4} ⊆ (ℕ, ≤) (meaning the set is ℕ & its partial order is ordinary < or =) the greatest lower bound is 2 & the other lower bound is 1. If {2, 4} ⊆ (ℤ, ≤) then the glb is still 2 but the other lower bounds are all the integers less than 2 .

 

[BubblesAreUs]

So if the proof for lower bound was applied to the pair {2,4} in T.

∀g,2,4 ∈ T ⇔ ( g ≺ 2) ^ (g ≺ 4) ^ ( ∀l ∈ T [ (l ≺ 2 ) ^ ( l ≺ 4)] ⇒ (l ≺ g)

In other words, g comes before 2 and 4 while l comes before g. I'm trying to compute the lower bound of the given pair. I also forgot to mention that T = ( X₈, ≺ ) where ≺ is defined on X₈ = { 1, 2, 3, 4,5, 6, 7, 8} by the rule x ≺ y ⇔ 5x ≤ 3y.

Therefore I was able to produce the pairs:

{empty set}, {0,0}, {0,1}, {0,2}, {0,3,}, {0,4}, {1,2}, {1,3}, {1,4}, {2,4},

Each of pairs produced go before their respective elements.

Is this correct?

 

[fourier jr]

Now that I understand how you got those pairs, once you know that $2 \prec 4$ because 5×2 ≤ 3×4 you don't need to compare 4 with any lesser elements because it obviously isn't a lower bound for {2, 4}. The way I understood the problem was that you want $x \in X_{8}$ such that $x \prec 2$, in other words x such that 5×x ≤ 3×2 = 6 which I think you got but you need to give the elements (not subsets) of $X_{8}$ that are lower bounds of {2, 4}, which are 0, 1 & 2. The glb is 2 because it's the lower bound which is greater than the other ones.