I have been given that the answer is 2^(aleph null) but i am struggling to see where it comes from. Any help would be appreciated.What is the cardinality of the set of order-isomorphisms of the rational numbers to itself?
2^(aleph null) = aleph one
I am not sure what an order-isomorphism is, maybe a mapping that preserves order? If so, then 2^(aleph null) sounds about right.What is the cardinality of the set of order-isomorphisms of the rational numbers to itself?
I don't even know what subject you're talking about!!
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