# Tagged Questions

For questions on axioms, mathematical statements that are accepted as being true, usually without controversy.

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### Proper definition of addition and multiplication

I recently started to study maths at university and in the analysis course we started, as usual, by looking at the axioms of $\mathbb R$ as a field. I think, I've understood the underlying intuition ...
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### ZF: Difference between POW and SEP?

In ZF, the Power Set Axiom (POW) says that given any set $A$, there exists a set $\mathcal{P}(A)$ such that $$\forall a(a\in\mathcal{P}(A)\leftrightarrow a\subseteq A)\tag{1}$$ Questions: ...
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### Do Kolmogorov's axioms really need only disjointness rather than pairwise disjointness?

According to 1 2, the third Kolmogorov axiom is for disjoint sets $(A_n)_{n \in \mathbb{N}}$ $P(\cup_n A_n) = \sum_n P(A_n)$ Is that really disjoint rather than pairwise disjoint? If we ...
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### Axiom of determinancy in action

I was wondering if anybody could give examples of actual works which utilized the axiom of determinancy. My issue is that I have heard of it, and read the Wikipedia, but have had trouble actually ...
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### Axiomatic definition of sin and cos?

I look for possiblity to define sin/cos through algebraic relations without involving power series, integrals, differential equation and geometric intuition. Is it possible to define sin and cos ...
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### Why do we have Axiom of Pairing but we don't have its generalisation, i.e a collection exists instead of pairing axiom?

For unions we have the generalised axiom, not just union for pairs. But for pairing we don't have generalisation, that a collection exists for any number of sets. If we have some axiom that says ...
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### is axiom of powers required?

we can form any subset of a set using axiom of specification. then we can form collection of subsets using axiom of pairing repeatedly. we have for every condition S , (A={x belongs to X & S(x)} =...
Question Why are the axioms for vector space independent? More precisely $1x=x$ seems redundant... (I take the axioms from: Wikipedia) Explanation One has for zero vector: \lambda0+\lambda0=\...
Say we are in a first order theory, and one of our inference rules is the associative rule saying that we can infer $(A \vee B) \vee C$ from $A \vee (B \vee C)$. Using the other logical inference ...