julypraise
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Let \mathcal{S} = \{S_{i}:i \in \mathbb{R} \} where S_{i} is a set. Then \mathcal{S} is a set? Or, can this notation make sense in some way?
julypraise said:Let \mathcal{S} = \{S_{i}:i \in \mathbb{R} \} where S_{i} is a set. Then \mathcal{S} is a set? Or, can this notation make sense in some way?
Because the S_i are sets, this is valid set builder notation defining a class \mathcal{S}. And by the axiom of replacement and the fact \mathbb{R} is a set, the class \mathcal{S} is indeed a set.julypraise said:Let \mathcal{S} = \{S_{i}:i \in \mathbb{R} \} where S_{i} is a set. Then \mathcal{S} is a set? Or, can this notation make sense in some way?
Not anything. There isn't, for example, a set of all sets that don't contain themselves!Whovian said:A set can be of anything. Even {Lincoln, Charizard, {Fish Fingers, Custard}} is a set. Your set is, therefore, valid. Note that its cardinality is one of the alephs, I don't know which one.
Hurkyl said:Not anything. There isn't, for example, a set of all sets that don't contain themselves!
julypraise said:Let \mathcal{S} = \{S_{i}:i \in \mathbb{R} \} where S_{i} is a set. Then \mathcal{S} is a set? Or, can this notation make sense in some way?
DonAntonio said:I can't see why you think S couldn't be a set, as long as each S_i is...What did you have in mind?
DonAntonio
julypraise said:You know, the concept of indexing in my mind (in my intuition) is kind of a countable process. But then now the index set is a continuum. So I thought it might not be possible; I mean this kind of indexing might not be possible by ZFC.