📂Set Theory

Russell's Paradox

Paradox ¹

If the set of all sets $\mathscr{U}$ exists, then some set $R$ both belongs and does not belong to $\mathscr{U}$.

Explanation

In the 6th century BC, the Cretan philosopher Epimenides said:

“All Cretans are liars!”

If Epimenides’s claim is true, then since Epimenides is also a Cretan, this claim is false. However, if this claim is false, then Epimenides is a liar, so it does not violate the claim and thus becomes true. In logic, the word ‘all’ is this dangerous.

From 1874 to 1884, Cantor’s works were the prototype of what would later come to be called set theory. At the time, the backlash from academia was so severe that it drove him to the point of mental illness, but by the time Bertrand Russell announced this paradox in 1902, Cantor’s set theory had already become central as a foundation across all of mathematics. Cantor’s life was unhappy, but it took only 20 to 30 years for his theory to engulf the entire academic world. The great mathematician Hilbert said, ‘No one shall expel us from the paradise that Cantor has created.’ It had become difficult even to imagine rigorous mathematics without the concept of a set.

Russell’s paradox, announced in such a situation, literally shook the very foundations of the mathematics of the time. What if the contradiction Russell found also existed in the very sets one was using? If the conditions under which such a paradox arises cannot be found, then all research would inevitably carry perpetual anxiety. The rigor that mathematics loves and takes pride in would break down from the very first line of a proof, and no matter what result was produced, there would be a danger that Russell’s paradox lay hidden within it.

Let us examine how this $R:= \left\{ S \in \mathscr{U} : S \notin S \right\}$ that Russell found causes problems:

• If $R \in R$, then since it is not the case that $R \notin R$, $S = R \in \mathscr{U}$ fails to satisfy the condition $(S \notin S)$ and thus cannot be included in $R$. That is, $R \notin R$.
• If $R \notin R$, then $R$ satisfies the condition $(S \notin S)$ and is thus included in $R$. Therefore $R \in R$.
• However, by the law of the excluded middle², it cannot be that $R \notin R$ and $R \in R$ at the same time.

Through the short argument above, we can see that the very premise that something like ’the set of all sets’ exists is wrong. To guarantee this, one can say that the axiom schema of specification or the axiom of regularity is needed, and scholars come to move away from the existing naive set theory and to seek a more rigorous axiomatic system.