# Tag Info

2

The proof is simple enough that you don't need a book. You just prove it by induction. The base case (one set) is trivial, and the inductive case is not much harder, using the fact that $$\prod_{i = 1}^{n+1} X_i \cong \left (\prod_{i=1}^n X_i\right ) \times X_{n+1}.$$

1

Even without measure theory: If $A$ has occurred, the also, either $B$ has occurred or or $B$ has not occurred. The probablility that $A$ has occurred and $B$ has not occurred is by definition $P(A\cap B^c)$ and similarly the probablility that $A$ has occurred and $B$ has is $P(A\cap B)$. So $$P(A) = P(A\cap B)+P(A\cap B^c) \\ P(A\cap B^c) = P(A) - ... 1 The reason for the identity P(A^c∩B)=(P(B)−P(A∩B)) is that the two sets A^c∩B and A∩B are disjoint and make up B, that is (A^c∩B) \cup (A∩B) = B. The probability of the left hand side is thus equal to the probability on the right hand side, and since the union on the left hand side is disjoint, it is equal to the sum of the probabilities of ... 1 You could think of it like$$P(B) \cup P(B^{c})=1$$so$$P(A)=P(A)\cap(P(B) \cup P(B^{c}))$$and therefore$$P(A \cap B^c) = (P(A) − P(A ∩ B)) I don't know if that is formally written correctly, but maybe conveys the right logic.

1

Are you familiar with measure theory? If so, $P$ is a measure, and $P(A) = P(A \cap B) + P(A \cap B^{c})$ (since measures are countably additive as long as the sets are pairwise disjoint), which implies $P(A) - P(A \cap B) = P(A \cap B^{c})$. You can apply a similar argument for $P(A^{c} \cap B)$.

1

The reals can either be defined axiomatically or constructed from (usually) a model of the rationals, in various ways. Axiomatically, the reals are an ordered field, which means that translation invariance property you refer to is an axiom. So, if your approach to the reals is that they are simply a model (one of many, but all isomorphic) of the axioms of ...

1

I think the answer should be no. I'm not a researcher in this field (or even in math), so I can't say if my opinion is in any way aligned to some "mainstream math" community, but the more general Wikipedia article on non-well-founded set theory (warning: to which I have contributed myself some edits in the past) lists some alternatives to Aczel's axiom and ...

1

Well, there is more than one axiomatic set theory. The domain of discourse of an axiomatic set theory is a collection of objects that may or may be not a set of the axiomatic set theory in question; In the case of ZFC, the collection of all its sets isn't one of the sets of ZFC, but it could be a set according to some other theory. To clarify; Let's prefix ...

5

The question is what do you consider a "logical argument". All the arguments that you brought up are philosophical and based on the idea that mathematics should somehow describe our physical reality, and must therefore be compatible with our physical intuition. Personally I find this argument to be lacking, since our physical intuition changes all the time ...

11

It is even better not to think of 'the domain' as a single object at all (whether 'set', 'class' or 'collection'). Instead of talking of a domain (singular) of individuals, just talk directly of the relevant individuals (plural). As logicians very often do when they forget about their official story! Thus even the great Alonzo Church talks of the range of ...

4

One seeming-paradox is that any consistent first-order theory has a denumerable model. Therefore, the "sets" in at least some models of Zermelo–Fraenkel set theory can indeed be placed into a set. It's just that the set containing the "sets" in the model cannot be described in the formalization of the theory. For details, see the link. In other words, we ...

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