Some sets, such as the set of all teacups, are not members of themselves. Other sets, such as the set of all non-teacups, are members of themselves. Call the set of all sets that are not members of themselves 'R.' If R is a member of itself, then by definition it must not be a member of itself. Similarly, if R is not a member of itself, then by definition it must be a member of itself.It's not like I was ever attracted to logic. When it was studied on my undergraduate degree, I ignored the material (just as I ignored all ten novels on the English Novel course!). So I don't even know how I even heard of it. Maybe it was on Brian Magee's The Great Philosophers TV series in 1987 (featuring a young, striking Martha Nussbaum discussing Aristotle).
Wherever, the reason why the detail remains has almost certainly less to do with the substance of the paradox than with how I misunderstood it. For I've long asked, under the breath of reading: in what way is narrative implicated in the content it utilises in order to exist?
For example, how might a story of a journey repeat the experience of a journey rather than merely appropriate it? And, does the tale of addiction correlate to the comfort of storytelling? If so, what is a writer's and reader's cold turkey? The same question goes for accounts of descents into madness or obsessive love. Every narrative is implicated; even the narrative of narratives.
The issue might expand into and explain my a priori dissatisfaction with genre fiction. Why are science-fiction novels written in the most old-fashioned manner? Why do crime thrillers appear to be accomplices to the crimes they describe? Why does a Romance novel limit love to the movement of its own narrative rather than embody it? Is fiction really the sublimated denial of the abyss of experience?
Can Russell's Paradox help us understand this? Stanford obliges again:
The significance of Russell's paradox can be seen once it is realized that, using classical logic, all sentences follow from a contradiction. For example, assuming both P and ~P, any arbitrary proposition, Q, can be proved as follows: from P we obtain P or Q by the rule of Addition; then from P or Q and ~P we obtain Q by the rule of Disjunctive Syllogism.Well, that's that cleared up.