FOM: Re: CH and 2nd-order validity

[Harvey Friedman]

Reply to Black 10L37PM 10/15/00:

>>>For example:
>>>    Is CH independent of true arithmetic?
>>
>>The answer is no
>
>Don't you mean yes????

Of course. This proves that you read my posting.

>Also (though it doesn't matter) for second-order ZFC shouldn't one really
>just say second-order ZF (because C is - presumably - a sematic
>consequence)?

Yes, but I can imagine that someone working with second-order logic does
not accept the axiom of choice. In fact, there are interesting to do in
comparing the theory of second-order logic with and without the axiom of
choice - either in the theory or in the metatheory. I am now stopping
myself from listing such questions...