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5

This is not that incredible if you understand what's going on. Given a second-countable space $X$, fix a countable basis $\{U_n\mid n\in\Bbb N\}$. Now you can observe that every open set is uniquely determined by the basis elements it contains, namely the function $U\mapsto\{n\in\Bbb N\mid U_n\subseteq U\}$ is an injective function. And since the real ...


5

$\bf 1.$ The Axiom of Choice Given a set $S$, to say that $S$ is not empty is to say that $\exists x(x\in S)$ (in English: there exists some $x$ such that $x$ is an element of $S$). First-order logic has an inference rule which allows us to move from $\exists x(x\in S)$ to some [new] constant symbol $s$ such that $s\in S$. This process, called existential ...


4

The proof is not trivial to come by. Originally given by Hodges, the proof shows a variant of Zorn's lemma can be proved from the assertion that every commutative ring with a unit has a maximal ideal. Wilfrid Hodges, Krull implies Zorn, J. London Math. Soc. (2) 19 (1979), no. 2, 285--287. Some decades later, Banaschewski gave a somewhat different ...


4

That is correct, if the real numbers are a countable union of countable sets, then measure theory as we know it goes out the window. The same can be said if the real numbers are a countable union of countable union of countable sets, and so on. Essentially for the same reason. To your question, the Lebesgue measure is the completion of the Borel measure. So ...


3

Nope. It does not imply that. This is a result due to Derrick and Drake. The idea, in its essence, is to add $\aleph_n$ Cohen reals, for each $n$, and take only the sets which are definable by a bounded part of the forcing.


3

The discrete topology on a countable set is second countable; it has a countable base consisting of the singleton sets. Thus, the assertion that all second countable spaces are Lindelöf implies that every cover of a countable set $N$ by subsets of $N$ has a countable subcover. I will show that the latter (ostensibly weaker) assertion implies ...


3

Since I am quite unfamiliar with schemes, I'll follow a hunch.If $i$ and $j$ are such that $p\in U_i\cap V_j$ then there is some affine open set $W\subseteq U_i\cap V_j$ such that $p\in W$. Or at least, for every $p$ there exists at least one pair of $i$ and $j$ such that the above is true. In its current form, the proof suggests choosing $i(p)$ and ...


3

Not every set of ideals has a maximal elements. It is true that with the axiom of choice, the set of all ideals has a maximal element, but it is not true that every set of ideals has a maximal element. For example, look at $\Bbb Z[\{x_n\mid n\in\Bbb N\}]$ and consider the set of ideals which are generated by $\{x_i\mid i<n\}$ for each $n\in\Bbb N$. This ...


2

Take the subset $E$ in $A\times \mathcal{P}(A)$ defined by $\{(a,B) : a\in B\}$. Then the second projection is a suejective map onto $\mathcal{P}^*(A)=\mathcal{P}(A) -\emptyset$. A right inverse, composed with the first projection, yields a selecting function as in AC.


2

Well. As with the failure of the axiom of choice in full, I should point out that the failure of any such arbitrary choice principle is not limited to "interesting sets". The axiom of choice could be in full force when considering all the sets relevant for most mathematicians. Then, it might fail badly somewhere up in the clouds of the set theoretic ...


2

Yes. Assuming the axiom of choice the answer is positive. You can find the proof in W.R. Scott's paper: Scott, W. R. "Groups and cardinal numbers." Amer. J. Math. 74, (1952). 187-197. The axiom of choice is used there for all manner of cardinal arithmetics. Without the axiom of choice it is no longer necessary that the proof can go through. Because ...


2

Since you seem to already have considerable information about ordinal numbers, you might use the following proof of Hartogs's theorem, which seems slightly more efficient than Asaf's because you don't need to well-order equivalence classes. Given $X$, there is a set $W$ of all well-orderings of subsets of $X$. Each such well-ordering is isomorphic to a ...


2

I don't know anything about sheaves, but I can give a "no" answer to what Hanno mentions in his comment as a special case. Namely, it is consistent with $\mathsf{ZF}$ that there are families of abelian groups $(A_i : i \in \mathbb{N})$ and $(B_i : i \in \mathbb{N})$ and a family of maps $(\varphi_i : i \in \mathbb{N})$ such that each $\varphi_i$ is a ...


2

Here's the thing about $\Bbb R$. It's linearly ordered. And this means that finite subsets are well-ordered canonically by that linear ordering. In particular, this means that the countable union of finite sets of real numbers is always countable. So the usual proof works, with the minor modification that now we need to appeal to the properties of $\Bbb R$ ...


1

Jyotirmoy writes that: [To prove the Banach-Tarski theorem] requires the Axiom of Choice. This statement needs some qualification. A bit of background: the standard collection of set-theoretic axioms is called ZFC, where "C" stands for the axiom of choice. Deleting the axiom of choice yields a smaller collection of axioms called ZF. Here's the key ...


1

This statement allready follows from the addition theorem, which states that $A \times \{0,1\}$ is equipotent with $A$ for every infinite $A$ (as on $A \times \{0,1\}$, $(a,x) \mapsto (a, 1+x)$ is fixed-point-free). The addition theorem is known to be strictly weaker than the axiom of choice (a result by Sageev).


1

If you look at the fourth section of Hodges paper, Wilfrid Hodges, Six impossible rings, J. Algebra 31 (1974), 218--244. He proves (Section 3, Th. 1) that the implication that you are asking for is not provable without the axiom of choice. Namely, the following chain of implications is irreversible without using Zorn's lemma: $$\text{Every non-empty ...


1

Yes, it's fine. Note that the use of the axiom of choice is only in proving that a choice function $g$ exists, and we fix this choice function before starting the recursion. Afterwards it's just recursion. Also note that you don't choose from $f(n)$, but rather from $\{y\mid f(n)<y\}$, which is nonempty since $f(n)$ is not maximal.


1

Yes, this is fine. In fact you don't even need the full strength of the axiom of choice: the axiom of dependent choice is all that you're really using.


1

This seems to be a typo (or a small mistake, if you prefer), and the proof goes along just fine if you also require that $\{\Lambda\}$ is open in each $X_a$. The space is still compact. You can find the same proof, minus the typo, on the Wikipedia page of the Tychonoff's theorem.


1

Assuming $\sf ZF$ we can prove Hartogs theorem. Namely, for every set $X$ there exists a set $A$ and a well-ordering of $A$ such that there is no injection from $A$ into $X$. The proof does not use the axiom of choice. It goes by considering all the well-orders whose domain is a subset of $X$, then considering the equivalence classes under the order ...


1

Revised, since there was an oversight in my original argument. I’ve now looked at Tarski’s paper; the very old-fashioned notation is a worse problem than the French. I’ve updated it and filled in some details. The proof in question is on pages $94$ and $95$. In essence he starts by assuming that there is an injection $\varphi:\wp(X)\to\wp(X)$ such that ...


1

This can be obtained with the following theorem of Kuratowski: $\omega\leq^*\mathfrak a\iff\omega\leq2^\mathfrak a$ Namely, there is a surjection from $A$ onto $\omega$ if and only if there is an injection from $\omega$ into $\mathcal P(A)$, which in turn is to say that there is an injection from $\mathcal P(A)$ into itself which is not a surjection. ...


1

Something is not clear to me about your question. Global choice implies the axiom of choice for sets. Of course it does. Every set is a subclass of $V$, and if $V$ can be well-ordered, then every set can be well-ordered. Requiring that there is an injection from the class of ordinals into $V$ is provable in $\sf ZF$. This function is the identity function. ...


1

If you look at the fourth section of Hodges paper, Wilfrid Hodges, Six impossible rings, J. Algebra 31 (1974), 218--244 he proves (Section 3, Th. 3) that the implication that you are asking for is not provable without the axiom of choice. Note that the prove is in $\sf ZF+DC$ in which ACC is equivalent to the statement "Every ideal is finitely ...



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