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 Mar 14 awarded Popular Question Jul 2 awarded Curious Apr 30 revised Distinct polynomials with exactly one shared root Clarify meaning of question. Apr 30 comment Distinct polynomials with exactly one shared root I guess that's right. I didn't state the question exactly right though. I'll post an edit. I meant to be looking at polynomials with rational coefficients. I'm not sure how much harder that problem will be. Apr 30 asked Distinct polynomials with exactly one shared root Mar 24 awarded Custodian Mar 24 reviewed Approve Is there an isomorphism between $\{\frac{m}{2^n}:m,n \in \mathbb{N} \text{ and } m < n\}$ and $\mathbb{Q}$ Mar 24 asked Is there an isomorphism between $\{\frac{m}{2^n}:m,n \in \mathbb{N} \text{ and } m < n\}$ and $\mathbb{Q}$ Mar 3 comment Uncountable Cardinals without AC Actually, I think I got it. Every set of ordinals has an upper bound because we can just take the union of the set. Mar 3 comment Uncountable Cardinals without AC That clears up the questions I had. Another question... what justifies the use of the supremum in the proof. $ON$ is well ordered, so any set of ordinals with an upper bound should have a least upper bound. What guarantees that the set of order types of well orderings of subsets of $\omega$ has an upper bound at all? Mar 3 accepted Uncountable Cardinals without AC Mar 3 revised Uncountable Cardinals without AC added 2 characters in body Feb 27 asked Uncountable Cardinals without AC Feb 22 comment Proving Separation from Replacement In Kenneth Kunen's The Foundations of Mathematics, the Replacement Scheme is listed as $\forall x \in A \exists ! y \varphi(x,y) \rightarrow \exists B \forall x \in A \exists y \in B \varphi(x,y)$ (with $A$ universally quantified implicitly). I'm guessing that this is probably equivalent to the version used in Gaisi Takeuti & Wilson Zaring, Introduction to Axiomatic set theory. But it makes the desired proof slightly more difficult. There is more constraint on $\varphi$, so that in addition to being functional, its domain must be the entire set $A$. Feb 18 asked Proving Separation from Replacement Feb 18 comment How do the separation axioms follow from the replacement axioms? If no element of $A$ satisfies $\phi$ then $X = \emptyset$. If we have the ability to use comprehension, then the existence of $\emptyset$ is straightforward. But without comprehension, I'm having a hard time seeing how we can show the existence of $\emptyset$. Feb 18 comment How do the separation axioms follow from the replacement axioms? The usual way I've seen the existence of $\{x\}$ established is through the use of pairing and separation. Since we're trying to show that these two axioms follow from replacement, they can't be used in that process. So how do we know that $\{x\}$ is a set? Feb 18 comment The comprehension axioms follows from the replacement schema. The version of the replacement axiom I've been using is this: $\forall x \in A \exists ! y \varphi(x,y) \rightarrow \exists B \forall x \in A \exists y \in B \varphi (x,y)$. With that I'm having difficulty seeing how the $\psi$ that you're using satisfies the axiom. Say for instance $\varphi(u)$ is false for some members of $A$. These would be counterexamples to $\forall x \in A \exists ! \varphi (x,y)$. Dec 11 asked Cardinality of the set of all well formed formula in propositional logic? Sep 28 accepted What are some reasonable things to prove about the Collatz Conjecture?