# Tag Info

1

I believe $6$ is indeed indispensible, and may be equivalent to $1$, here's my reasoning: what $6$ actually says is we have an action of $(F,+)$ upon $(V,+)$, that is, the map $v \mapsto a\cdot v$ (let's call this map $\phi_a$) induces a group homomorphism (the operation being $+$): $F \to V$ via $a \mapsto \phi_a(v)$ for any fixed $v \in V$. Indeed, we ...

2

You won't be able to entirely remove axiom 3, since otherwise $V=\emptyset$ would (vacuously) satisfy the other axioms. However, you can remove axiom 4 if you replace axiom 3 with this slightly stronger version (which I will call axiom 3*): (Axiom 3*) There exists an element $\mathbb{0'} \in V$ such that for all $x \in V$, $0_K \cdot x = \mathbb{0'}$. ...

14

I think they are redundant after all! Here's a proof that axiom 1 is redundant. Let $a,b\in V$, and consider $(1+1)\cdot (a+b)$. By axiom 7 and 8, this is equal to $(a+b)+(a+b)$; on the other hand by axiom 6 it is $(1+1)\cdot a + (1+1)\cdot b$, or $(a+a)+(b+b)$ by axiom 7 and 8. We can then use axioms 2, 3, 4 to show that \begin{align*} a^{-1} + (a+b) + ...

3

No. Just because something isn't mentioned in the axiom does not mean that it is necessarily not true. When formalizing logic to include equality, something which is the modern standard in mathematics, one of the axioms included is the following schema: If $x=y$, then for every formula, ...

0

A group action is indeed nothing else than a homomorphism into the group of bijection on the set $X$. However, the first axiom does not garantee that the image of g is actually a bijection. You could fix an element $x_0\in X$ and define $g.x:=x_0$ for every $g\in G$. This satisfies the first axiom, but does not give a bijection of $X$ (except for the trivial ...

1

Consider the trivial group $E = \{ e \}$. Forgetting about the "identity" axiom, we can let $E$ act on $\mathbb{R}$ by $e \cdot x = 0$ for all $x \in \mathbb{R}$. This satisfies the "compatibility" axiom, as clearly $ee \cdot x = 0$ always. But it is not an actual group action, and it is not a homomorphism to the group of bijections of $\mathbb{R}$.

1

The "homomorphism from a group $G$ into $\operatorname{Sym}(X)$" interpretation is equivalent to these two axioms; you need both axioms to imply the homomorphism interpretation and vice versa. The Wikipedia page that you linked states this pretty clearly: From these two axioms, it follows that for every $g$ in $G$, the function which maps $x$ in $X$ to ...

3

The continuum hypothesis says $2^{\aleph_0} = \aleph_1$, right? The cardinal of $[0,1]$ is $2^{\aleph_0}$. By definition, $\aleph_1$ is the cardinal of the least uncountable ordinal. That is, a well-ordered set that is uncountable, and each initial segement is countable. The last step: if two sets have the same cardinal, then there is a one-to-one ...

1

Let me give a different reference from the one Noah gave (although the mathematics is more or less the same). Jech, "The Axiom of Choice", Chapter 7. The last part of the chapter deals with all sort of various implications and separations of choice from finite sequences, including a nontrivial condition which guarantees some sort of control on when one type ...

3

Not quite a complete answer to your question, but too long for a comment: http://projecteuclid.org/download/pdf_1/euclid.pjm/1102983625 shows that e.g. the axiom of choice for all familes of sets with exactly two elements, does not prove the axiom of choice for all families of sets with exactly 3 elements. (see page 235). This strongly suggests that the ...

2

In most ways to formalize logic with equality, "$t=t$" is an axiom for every expression (or "term", in the jargon of logic) $t$. In other words, the validity of the claim $t=t$ is part of the conventions about what the symbol $=$ means. It's not really about any deep or philosophical truth -- simply an agreement among mathematicians that if you want to ...

1

$1/(-a)+1/(a)=0$ is equivalent to what you want to prove, so your argument is not an improvement. Here is an argument: $$1/(-a)=1/((-1)\cdot(a))=1/(-1) \cdot 1/a = (-1) (1/a) = -(1/a)$$ For that to work, you need to prove that $-a = (-1)(a)$ $1/(ab)=(1/a) \cdot (1/b)$ $1/(-1)=-1$ all of which are easy because they follow from uniqueness: $x=-a$ iff ...

2

A formula of the form $\phi \land \lnot\phi$ has no models (other than in the model theory of some rather obscure substructural logics). This is unconnected with Russell's paradox and not at all deep. There is no such thing as an "empty model" for a non-empty language, e.g., a language including he membership symbol $\not\in$. [I should point out that this ...

11

Working in classical logic, the rules of logic dictate that $x\in x\land x\notin x$ is a false statement. Therefore the only structure satisfying $\forall x(x\in x\land x\notin x)$ is the empty structure. However due to superficial reasons, we choose not to accept the empty structure as a valid interpretation of a first-order language, so it is not a model ...

4

In this case, the fact that the emptyset is not "open" is the only barrier to your definition yielding a topology. Closure under arbitrary nonempty unions is obvious; to see that the intersection of two "opens" is "open," let $S_X=\sum_{n\not\in X}{1\over n}$; then note that $$S_{X\cap Y}\le S_X+S_Y<\infty$$ if $S_X, S_Y<\infty$. In such situations as ...

16

In axiomatic set theory, we usually take the position that everything is a set, and in particular everything we need to care about as elements of a set are themselves sets. The expectation is then that things like "$1$" are to be interpreted as abbreviations for particular sets that represent the numbers. A common choice of representative for the natural ...

1

For any questions about coin tossing that involve an unbounded number of tosses the proper sample space $\Omega$ consists of all infinite binary sequences ${\bf b}=(b_1,b_2,\ldots)$. This space obtains a bona fide probability measure $\mu$ through a measure-theoretic construction starting from the requirement that the events $$A_n:=\{{\bf b}\in\Omega\>| ... 1 A collection of events satisfying axioms 1-3 is called a \sigma-algebra. You, no doubt, encountered this in the context of probability space$$(\Omega,\mathcal{A},P)$$where the second member of the tiple \mathcal{A}, the collection of events on which the probability measure P is defined, must be a \sigma-algebra. Both \{\emptyset, \Omega\} and ... 0 The axioms are right, and one cannot say that A = \{1\} is an event. It may be that your \sigma-algebra of events is simply \{\emptyset, \Omega\}. So to answer the exercise, one should say: "Based on the axioms one cannot say whether A is an event". 1 Certainly, the Continuum Hypothesis (simple or generalised) can be formulated in the language of pure second-order logic. This is done explicitly in Stewart Shapiro's wonderful book on second-order logic, Foundations without Foundationalism: see pp. 105-106. The full story is a bit too long to give here. But as a taster, you start off by essentially using ... 2 3 comes from the first two.$$(a/b)/(c/d)\\=d/(c/(a/b))\\=d/(b/(a/c))\\=(a/c)/(b/d)\\=(a/(b/d))/c\\=(d/(b/a))/c\\=(d/c)/(b/a)

1

I'm not sure it is shorter. To say a non-empty set you need $\exists x (x \in S)$ which adds to your statement. To prove equivalence, you define a function $f(x)=\emptyset, f(\{x\})=\{\emptyset \}\dots$ and use replacement on your $S$ to show the existence of the canonical $\omega$.

0

In the absence of the Axiom of Foundation they are not equivalent. Without Foundation it is consistent that x={x} for some x. The Axiom of Foundation is that any non-empty x has a member y which is disjoint from x.

0

The same idea as other answers, but a different explanation (which makes more sense to me, at least): there is no such thing as true, period. A statement can only be true or false in some model, which consists of a certain set of "original" statements whose truth is assumed, and everything provable from those statements. The "original" statements are the ...

29

You're treating the word "axiom" as you were probably taught in high school. That an axiom is something which is "simply true as an assumption". Modern mathematics has changed that definition to "an assumption made in a certain context". Not every axiom is called an axiom, some axioms are proved as theorems, and sometimes lemmas are used for axioms. And not ...

6

The existence of a model of a statement does not mean that statement is "true" (whatever that means; see below). For example, the Poincare disk is a model of Euclid's first four postulates plus the negation of the parallel postulate; this does not mean that the parallel postulate is "false." What having a model of a set of statements does mean, is: that set ...

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