One of the most well-known examples is Frege's axiom of general comprehension, which asserts that for any definite property $P$, one may form the set $$\{x\mid\ P(x)\ \},$$ consisting of all objects $x$ with property $P$. This natural-seeming principle is one of the main axioms of what is now known as naive set theory, and formed a central axiom in Frege's Begriffsschrift and later the Grundgezetze (which I see now that you said you weren't interested in, oh well), intended as a formal logical foundation of arithmetic and all mathematics. But the axiom was famously refuted by Betrand Russell with the Russell paradox, showing that there can be no set $R=\{x\mid\ x\notin x\ \}$, consisting of the sets $x$ that are not members of themselves, since then $R\in R\iff R\notin R$, a contradiction.
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One of the most well-known examples is Frege's axiom of general comprehension, which asserts that for any definite property $P$, one may form the set $$\{x\mid\ P(x)\ \},$$ consisting of all objects $x$ with property $P$. This axiom formed an important part natural-seeming principle is one of the basis main axioms of what is now known as naive set theory, and formed a central axiom in Frege's Begriffsschrift, intended as a formal logical foundation of arithmetic and all mathematics. But it the axiom was famously refuted by Betrand Russell with the Russell paradox, which shows showing that there can be no set $R=\{x\mid\ x\notin x\ \}$, consisting of the sets $x$ that are not members of themselves, since then $R\in R\iff R\notin R$, a contradiction. |
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One of the most well-known examples is Frege's axiom of general comprehension, which asserts that for any definite property $P$, one may form the set $$\{x\mid\ P(x)\ \},$$ consisting of all objects $x$ with property $P$. This axiom formed an important part of the basis of Frege's Begriffsschrift. But it was famously refuted by the Russell paradox, which shows that there can be no set $R=\{x\mid\ x\notin x\ \}$, consisting of the sets $x$ that are not members of themselves, since then $R\in R\iff R\notin R$, a contradiction. |
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