# Math Help - Indexed Union of Countably Infinite Sets

1. Originally Posted by MoeBlee
What do you mean by that? An isomorphism involves two structures, not just two sets. Do you mean isomorphic with regard to the the structure <set membership>? Then, no, not all such structures where the sets are countable are isomorphic.

What we do have is that any two denumerable (i.e., countably infinite) sets are equinumerous.
In Model Theory, an isomorphism is a bijection which preserves all constants, functions and relations in the language. When talking about sets without any further structure, an isomorphism is simply a bijection since there are no other symbols in the language.

2. Originally Posted by Deveno
a set-isomorphism is just a fancy word for bijection
I've not seen that terminology. I'll take your word for it, though I find it unnecessary to conflate bijection with isomorphism.

Originally Posted by Deveno
membership induces additional structure on sets (making a poset out of P(X)). "isomorphism" is essentially a categorical notion, not limited just to "sets with structure".
Of course we can always ask

whether <X membership_on_X> and <Y membership_on_Y> are isomorphic.

And the partial ordering of PX by the subset relation is a separate matter. Yes, of course, if X and Y are equinumerous then

<X subset_relation_on_PX> and <Y subset_relation_on_PY> are isomorphic.

But you've not shown how that serves toward proving that a countable union of countable sets is countable.

Originally Posted by Deveno
explicitly, if you have a bijection between set A and set B, then saying whether of not A and B are the "same set" depends on your notion of "same". the typical categorical notion is that two sets of the same cardinality are indistiguishable.
Not in ordinary set theory. X and Y are the same if and only if X and Y have the same members. X and Y may have the same cardinality, but that does not entail that X and Y are the same. Moreover that X and Y have the same cardinality does

NOT entail that <X membership_on_X> and <Y membership_on_Y> are isomorphic.

Originally Posted by Deveno
are you aware of an example of two countably infinte sets with no lattice isomorphism of power sets between them?
I didn't say anything about that in my previous post. But in this post I have said that yes, of course, if X and Y are equinumerous then

<X subset_relation_on_PX> and <Y subset_relation_on_PY> are isomorphic.

How that serves toward proving that a countable union of countable sets is countable is not stated by you.

3. Originally Posted by DrSteve
When talking about sets without any further structure, an isomorphism is simply a bijection since there are no other symbols in the language.
Isomorphism of models (structures for a language) is one kind of isomorphism. But in set theory there is the more general notion of a isomorphism comparing (1) a set X and relations and/or operations on X with (2) a set Y and relations and/or operations on Y. If you wish to say that where there are no specified operations or relations along with said set, then X and Y are isomorphic if and only if X and Y are equinumerous, then you're free to use the terminology that way, though personally I don't find that it adds clarity to the matter.

4. Originally Posted by MoeBlee
Isomorphism of models (structures for a language) is one kind of isomorphism. But in set theory there is the more general notion of a isomorphism comparing (1) a set X and relations and/or operations on X with (2) a set Y and relations and/or operations on Y. If you wish to say that where there are no specified operations or relations along with said set, then X and Y are isomorphic if and only if X and Y are equinumerous, then you're free to use the terminology that way, though personally I don't find that it adds clarity to the matter.
I agree - certainly using the word "bijection" is much less likely to cause confusion, especially since the language of set theory does have a relation symbol.

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