Hasse Diagram - lower upper bound

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• Jun 15th 2008, 10:45 AM
robocop_911
Hasse Diagram - lower upper bound
Please have a look at the Hasse diagram -

How come the upper bounds of {a,b,c} are e,f,j and h and lower bound is a
Why doesn't it include d and g also in its upper bound?

• Jun 15th 2008, 11:03 AM
Plato
It does not include d because the pair (c,d) is not in the diagram.
There is no g in the diagram.
• Jun 15th 2008, 11:09 AM
robocop_911
Quote:

Originally Posted by Plato
It does not include d because the pair (c,d) is not in the diagram.
There is no g in the diagram.

I've added "g" in the diagram Now could you please explain me why upper bound of {a,b,c} are e,f,j,h and Not d and g
• Jun 15th 2008, 11:19 AM
Plato
As I said before, the pairs (c,d) & (c,g) are not in the graph.
The elements g & d are not comparable to c in the poset.
There is neither an “upward path” from c to d nor one from c to g.
• Jun 15th 2008, 11:30 AM
robocop_911
Quote:

Originally Posted by Plato
As I said before, the pairs (c,d) & (c,g) are not in the graph.
The elements g & d are not comparable to c in the poset.
There is neither an “upward path” from c to d nor one from c to g.

Also for "Lower bound" How come the lower bounds of {j,h} are a,b,c,d,e and f and not "g".?
• Jun 15th 2008, 11:33 AM
Plato
Are elements g & j are comparable?
• Jun 15th 2008, 11:42 AM
robocop_911
Quote:

Originally Posted by Plato
Are elements g & j are comparable?

Hi Plato!

Had it been only, "h" then the lower bound would have been a,b,c,d,e,f,g right?

Also, How come lower bounds of {a,c,d,f} are f,h and j? and it lower bound is a only?

• Jun 15th 2008, 11:46 AM
Plato
Quote:

Originally Posted by robocop_911
Had it been only, "h" then the lower bound would have been a,b,c,d,e,f,g right?

Yes that is correct.
You know, you really should learn the definitions.
• Jun 15th 2008, 11:54 AM
robocop_911
Quote:

Originally Posted by Plato
Yes that is correct.
You know, you really should learn the definitions.

Definition doesn't answer the following question...
How come lower bounds of {a,c,d,f} are f,h and j? and it lower bound is "a" only?

There is no upward connection between "c" and "d", also how come "f" is coming again in the lower bounds? How can that be?
• Jun 15th 2008, 11:59 AM
Plato
Quote:

Originally Posted by robocop_911
Definition doesn't answer the following question...
How come lower bounds of {a,c,d,f} are f,h and j? and it lower bound is "a" only? There is no upward connection between "c" and "d", also how come "f" is coming again in the lower bounds? How can that be?

Well you have just made the point for me.
Yes they do. Just apply them carefully.
• Jun 15th 2008, 12:04 PM
robocop_911
Quote:

Originally Posted by Plato
Well you have just made the point for me.
Yes they do. Just apply them carefully.

Hello Plato,
It seems as if you are having good knowledge about Hasse diagram, do you mind if I ask you questions one by one, so that it can help me clarify my doubts?
• Jun 15th 2008, 12:48 PM
robocop_911
Okay, here are my questions...

1) {b,d,g} upper bounds - g and h
my question - why not only "h" why include g in it?

2) {b,d,g} least upper bound - g
my question - How come it is "g" why not "h" since h>g

3) {b,d,g} lower bounds - a and b
my question - why not only "a" why include b in it?

4) {b,d,g} greatest lower bound - b
my question - why not "a" since a<b

5) {a,c,d,f} upper bounds - f,h and j
my question - why include "f" in upper bound? Also, there is no "upward" connection between a and d.
• Jun 15th 2008, 01:12 PM
Plato
1) {b,d,g} upper bounds - g and h
my question - why not only "h" why include g in it?
Because both succeed every term in {b,d,g}. Don’t forget that partial orders are reflexive, therefore (g,g).

2) {b,d,g} least upper bound - g
my question - How come it is "g" why not "h" since h>g
Because there is only one LUB for any given set and g is the first or least.

3) {b,d,g} lower bounds - a and b
my question - why not only "a" why include b in it?
See 1). It is the same reason: b proceeds each element in the set.

4) {b,d,g} greatest lower bound – b my question - why not "a" since a<b
See 2). It is the same reason: b is the greatest or last upper bound.

5) {a,c,d,f} upper bounds - f,h and j
my question - why include "f" in upper bound? Also, there is no "upward" connection between a and d.
"a-b-d is an upward" connection between a and d.
There is an "upward" connection between any term in {a,c,d,f} to f [recall (f,f)]..

• Jun 15th 2008, 01:59 PM
robocop_911
1) {b,d,g} upper bounds - g and h
my question - why not only "h" why include g in it?
Because both succeed every term in {b,d,g}. Don’t forget that partial orders are reflexive, therefore (g,g).
Then how come upper bounds of {a,b,c} is e,f,j and h and not something like c,e,f,j,h? (c,c) (b,b) that's also reflexive!

2) {b,d,g} least upper bound - g
my question - How come it is "g" why not "h" since h>g
Because there is only one LUB for any given set and g is the first or least.

So least upper bound should be from the set given only right? In this case it is {b,d,g} so here it is "b". Is it also the greatest lower bound?
• Jun 15th 2008, 02:15 PM
Plato
I wish you would learn the definitions and how to use them.
In any case, good luck and good night.
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