MHB Graded poset definition trouble

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A graded poset is defined as a poset with a function that establishes a hierarchy based on covering relations. The discussion highlights a specific example, P={a,b,c,d}, where the first definition classifies it as a graded poset, while the alternative characterization suggests it is not due to lacking a least and greatest element. The confusion arises from the distinction between graded and bounded posets, as the latter requires extremal elements. Ultimately, the example is confirmed to be a graded poset but not a bounded one. Understanding these definitions is crucial for clarity in poset classification.
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Graded poset on wiki: Graded poset - Wikipedia, the free encyclopedia

Wikipedia defines a 'graded Poset' as a poset $P$ such that there exists a function $\rho:P\to \mathbb N$ such that $x< y\Rightarrow \rho(x)< \rho(y)$ and $\rho(b)=\rho(a)+1$ whenever $b$ covers $a$.

Then if you go to the 'Alternative Characterizations' on the page whose link I gave above you would see that the first line reads:
A bounded poset admits a grading if and only if all maximal chains in $P$ have the same length.
Here's the problem. Consider $P=\{a,b,c,d\}$ with $a<b,b<d,a<c,a<d$. All other pairs are incomparable. Then according to the first definition $P$ is a graded poset while the second definition says otherwise.

Maybe I am committing a very silly mistake but just can't find it.

Please help.
 
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Re: graded poset definition trouble

caffeinemachine said:
Consider $P=\{a,b,c,d\}$ with $a<b,b<d,a<c,a<d$. All other pairs are incomparable. Then according to the first definition $P$ is a graded poset while the second definition says otherwise.
This is indeed a graded poset, but it is not a bounded poset. The latter has to have a least and a greatest elements.
 
Re: graded poset definition trouble

Evgeny.Makarov said:
This is indeed a graded poset, but it is not a bounded poset. The latter has to have a least and a greatest elements.

Thanks! Guess it will take some time for the definitions to sink in.
 
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