- #1
xwolfhunter
- 47
- 0
So I'm reading Naive Set Theory by Paul Halmos. He asks:
Isn't this mind-numbingly faulty reasoning? The answer is "no ##x##es are specified by this sentence."" You don't even have to try and quibble with the details of the sentence, the quibbling of which is faulty the way I see it anyway. Neither the ##X##es nor the ##x##es specified in the sentence exist.
Besides the issues inherent with trying to perform logical operations on literally nothing, his logic seems to be vacuous anyway - sure, if an ##x## is not a member of all ##X##es in a set then there must be some ##X## which does not contain that ##x##, but why does he act as though this sentence is logically useful in this context? While it's true that all the ##x##es or none of the ##x##es in the framework of the sentence meet the conditions, either way no ##x##es are specified, because there are no ##x##es in the framework. The operations are on unknown ##x##es - his result is legitimate for an unspecified set rather than ##\phi##. But since we know the set is ##\phi##, we know that there are no ##x##es in existence in this reference frame, so the argument he used is rendered moot. His result is entirely of unspecified ##x##es of unspecified ##X##es of an unspecified set. He's talking about a hypothetical scenario, but the one we have is defined, and then he's applying the results of the hypothetical scenario to the actual scenario.
My question is what the heck is the point of this? He implies that you need to take caution to avoid scenarios like this - my question is why? Why would you avoid scenarios like this? It's cut and dry, there are no ##x##es specified by the sentence. There's no mathematical uncertainty here that I see.[tex] \{ x \in X~ \mathrm{for ~ every} ~X~ \mathrm{in} ~ \phi \} = \phi[/tex]
His response is that no ##x## fails to meet the requirements, thus, all ##x##es do. He reasons that if it is not true for a given ##x## that ##x \in X~ \mathrm{for ~ every} ~X~ \mathrm{in} ~ \phi##, then there must exist an ##X## in ##\phi## for which ##x \notin X## is true. He then says that since there are no ##X##es in ##\phi##, this result is absurd. So no ##x## fails to satisfy the stated condition, thus every ##x## does.Which ##x##'s are specified by the sentence[tex]x \in X~ \mathrm{for ~ every} ~X~ \mathrm{in} ~ \phi ~\mathrm{?}[/tex]
Isn't this mind-numbingly faulty reasoning? The answer is "no ##x##es are specified by this sentence."" You don't even have to try and quibble with the details of the sentence, the quibbling of which is faulty the way I see it anyway. Neither the ##X##es nor the ##x##es specified in the sentence exist.
Besides the issues inherent with trying to perform logical operations on literally nothing, his logic seems to be vacuous anyway - sure, if an ##x## is not a member of all ##X##es in a set then there must be some ##X## which does not contain that ##x##, but why does he act as though this sentence is logically useful in this context? While it's true that all the ##x##es or none of the ##x##es in the framework of the sentence meet the conditions, either way no ##x##es are specified, because there are no ##x##es in the framework. The operations are on unknown ##x##es - his result is legitimate for an unspecified set rather than ##\phi##. But since we know the set is ##\phi##, we know that there are no ##x##es in existence in this reference frame, so the argument he used is rendered moot. His result is entirely of unspecified ##x##es of unspecified ##X##es of an unspecified set. He's talking about a hypothetical scenario, but the one we have is defined, and then he's applying the results of the hypothetical scenario to the actual scenario.
My question is what the heck is the point of this? He implies that you need to take caution to avoid scenarios like this - my question is why? Why would you avoid scenarios like this? It's cut and dry, there are no ##x##es specified by the sentence. There's no mathematical uncertainty here that I see.[tex] \{ x \in X~ \mathrm{for ~ every} ~X~ \mathrm{in} ~ \phi \} = \phi[/tex]