# Given is a binary relation. Determine the Hasse diagram for it and two upper bounds

Given is the set $A = \left\{1,2,3,4,5,6,7\right\}$ with relation $R = \left\{(1,2),(3,2),(2,4),(5,4),(4,6),(4,7)\right\}$

Determine the Hasse diagram for the reflexive-transitive closure of $R$ and determine two upper bounds of $\left\{2,5\right\}$ in this reflexive-transitive close of $R$.

So I've read about Hasse diagrams and the good thing is that you can ignore edges that are transitive and reflexive. So basically, if my understanding is correct, it's sufficient if I just draw the Hasse diagram for the given relation $R$:

 6     7
\   /
\ /
4
/|
/ |
5  2
/ \
/   \
1     3


Is the Hasse diagram correct like that?

But I'm not sure how I solve the problem with those upper bounds now. It's mentioned to find two lower bounds of $\left\{2,5\right\}$. But now I don't know where to look at in the Hasse diagram? I understand it like that that $\left\{2,5\right\}$ means that we need to look at the "nodes" $2$ and $5$ in the Hasse diagram and check their children. $5$ has no child and thus no lower bound, so we look at $2$. It has as children nodes $1$ and $3$ thus the two lower bounds are $1$ and $3$?

I hope you can tell me because if they ask such a question in an exam I feel pretty helpless : /

You are confused about upper and lower bounds, I think. A lower bound of a set $S$ is just an element $b$ such that $b\le s$ for every $s\in S.$ Similarly, an upper bound of a set $S$ is just an element $b$ such that $b\ge s$ for every $s\in S.$ Here I'm writing $a\le b$ instead of $aRb$ because it's more intuitive.
So an upper bound of $\{2,5\}$ is an element $\ge 2$ and $\ge 5.$ In the diagram, there are $3$ to choose from: $4,6,7.$ As an aside, in the case, $4$ is the least upper bound. Not only is it a lower bound, but it's smaller than all the other lower bounds.
• Thank you, I have edited the Hasse diagram and I believe it's correct now because it seems to "create" every element which belongs to reflexive-transitive closure of $R$ but I still don't understand the second part of the question with the lower bounds, do you? I understand it like that that $\left\{2,5\right\}$ means that we need to look at the "nodes" $2$ and $5$ in the Hasse diagram and check their children. $5$ has no child and thus no lower bound, so we look at $2$. It has as children nodes $1$ and $3$ thus the two lower bounds are $1$ and $3$? – roblind Mar 12 '18 at 21:13
• Thank you for the edit and help! :) Just one last question to your last paragraph the first sentence. The element really needs to be $\geq 2$ AND $\geq 5$? – roblind Mar 12 '18 at 21:57
• Yes, that's right. That's typical usage in math, not just in Hasse diagrams. If there are 1000 elements in a set, an upper bound for the set is $\ge$ every one of them. – saulspatz Mar 12 '18 at 22:07