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I am new in the field and I am a bit confused about the definition of the RDFS vocabulary.

Specifically, the vocabulary defines that rdfs:XMLLiteral, which is a class, is the subclass of rdfs:Literal (rdfs:XMLLiteral and rdfs:Literal are connected using rdfs:subClassOf).

This is straightforward and is easily understood. However, the vocabulary also said the class rdfs:XMLLiteral is an instance of rdfs:DataType (linked by rdf:type). So, in this case, why the rdf:type is used instead of rdfs:subClassOf, given both the rdfs:XMLLiteral and the rdfs:DataType are classes.

My personal thought is that, because subclass relationship between Class A and B (suppose A ∈ B) implies that every individual that belongs to A also belongs to B.

So suppose we have an "x" which belongs to the rdfs:XMLLiteral class, if there was a subclass relationship between rdfs:XMLLiteral and rdfs:DataType, then "x" is also a rdfs:DataType, which is not the fact (because "x" is just an individual literal). Furthermore, because the rdfs:DataType and rdfs:Class are connected by rdfs:subClassOf according to the vocabulary, then "x" is also a class, if the subclass relationship between rdfs:XMLLiteral and rdfs:DataType existed. Therefore, such a subclass relationship should not exist.

I do not know if my thought here is right, and I hope someone can give some suggestions to help me understand the subclass relationship and instance relationship in the RDFS vocabulary.

unor
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Chengke Wu
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1 Answers1

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If you understand well the set-theoretic relations ∈ and ⊆, then you have a good basis for understanding rdf:type and rdfs:subClassOf. What you describe about rdf:XMLLiteral is pretty accurate. However:

suppose A ∈ B

This means that A belongs to B, not that every individual that belongs to A also belong to B.

A rdf:type B could be understood (roughly) as "A ∈ B", while A rdfs:subClassOf B could be understood (roughly) as "A ⊆ B".

A set can be an element of another set, as well as included in yet another one, or the same. E.g., {a} ⊆ {a,{a}} and {a} ∈ {a,{a}}.

However, the analogy with set theory is only roughly correct because RDF allows classes to be instances of themselves, whereas set theory forbids sets to be elements of themselves (assuming the axiom of regularity). The reason why RDF still "works" is due to the way its semantics is defined, where rdf:type is not really interpreted as the set membership relation, and rdfs:subClassOf is not really interpreted as set inclusion.

Antoine Zimmermann
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