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As others have pointed out, your statement of Zorn's lemma is ambiguous, and the one reasonable interpretation in English is not what is wanted (and in fact makes the statement false). This might be the source of your confusion. However, to address your questions at face value: A maximal element of $P$ is an element such that no other is greater than it. ...

6

The statement depends on the axiom of choice. It is consistent, modulo large cardinals, that all (well-ordered) infinite cardinals have cofinality $\omega$. This was first proved by Gitik, around 1980, from a proper class of strongly compact cardinals. The result is significant, and rather involved. It is actually an interesting construction. In a forcing ...

5

No. Your proposed order on $S$ is not a well-order. For example consider the set defined by these functions, $$f_n(x)=\begin{cases} 0 & x\leq n\\ x-n & x>n\end{cases}$$ If considering this as $A_0$, then in your proof you can note that $A_n=\{f_k\mid k\geq n\}$ and $B=\varnothing$. Your argument, if so, did not give a well-ordering of $\Bbb R$ at ...

4

There are two basic ideas: Given $X$, if it is finite then pick any bijection with $\Bbb Z/(n)$, and you have a finite group; otherwise consider $\Bbb Z[X]$, the ring of polynomials whose free variables are elements of the set $X$. We can prove, using the axiom of choice, that $\Bbb Z[X]$ has the same cardinality as $X$. Therefore there exists a bijection ...

4

Let me add on Andres' wonderful answer, something which I think is missing. Despite the necessity of large cardinals, it was in fact one of the first uses of the technique of forcing (and symmetric extensions) to show that in fact $\omega_1$ and $\Bbb R$ can be countable unions of countable sets. This is a classical result due to Feferman and Levy. In her ...

4

First of all, a word of terminology. Cardinals may or may not apply to the general term meaning "sets which represent an equivalence class of sets in the relation 'there exists a bijection between two sets'", and not just a particular class of ordinals. In the broader sense, it is easy to prove there are uncountable cardinals because the power set of the ...

2

On (1): see Asaf or Ryan's answer. Since on Kunen's definition $R$ only well-orders $X$ if $X$ is the field of $R$ (at least when $X$ has at least two elements), it's clearly a typo. On (2): When $X\subseteq \omega$, the type of $(X; R)$ can be bigger than $\omega$. For instance, take $R$ to be the natural ordering on $\omega\setminus\{0\}$ with $0$ added ...

2

For (1) After edit: After actually reading the problem, I think Asaf is correct and that $\alpha \preceq X$ should be $\alpha \preceq A$ (that was how I originally read it). So if this is the case, there is a problem, and so your problem is fixed by taking $X = 2 = \{0,1 \}$ with the usual ordering. For (2), you are getting confused on ordinals and ...

2

The general statement "If $S$ is a spanning set, and $L$ is linearly independent then $|L|\leq|S|$" is unfamiliar to me, and I'm not sure it was investigated enough to merit an answer either. It does imply that every two bases have the same cardinality, so it does require some weak form of the axiom of choice. My guess would be that it requires a bit more ...

2

I wouldn't think so. A complete graph should always have a spanning tree. Pick your favorite vertex $u$ from the vertex set $V$, infinite or not, and construct a giant star tree with $u$ as the center. That is, construct the tree $T = (V, F)$ where the edge set is $F = \{uv : v \in V \setminus \{u\} \}$. This has to be a spanning tree (and as pointed out ...

2

Yes, very much. Because when switching to $|A_i|$ you need to effectively choose canonical representatives from each equivalence class, and bijections from $A_i$ to that set. Neither of these processes is well-defined without the axiom of choice. We can have the following situation: There exists a set $S$ which can be partitioned into countably many ...

2

The reason this is not "obvious" and instead it is a rather deep mathematical truth is that partial orders, even ones where every chain has an upper bound, can get pretty hairy and intangible, mathematically speaking. When we think, and try to visualize structure in our heads, and a partial order is structure, we are limited by our imagination, which is ...

1

Some things to point out. It suffices to well-order the interval $[0,1)$, since it has the same cardinality as $\Bbb R$. $\omega_1$ is the least uncountable ordinal, but it is consistent that $|\Bbb R|=\omega_2$, or other values. As Mauro notes, this covers only finite decimal expressions. Which in fact is a countable set. Your plan for the infinite ...

1

In your particular case, you can get reasonably explicit by transferring the group structure over from a bijection between the irrationals and the reals. It's not hard to write down an explicit enumeration $q_n$ of $\mathbb Q$, and then we can map $\mathbb{R}\setminus \mathbb{Q}$ to $\mathbb{R}$ by $\sqrt{2}/2^{2k+1}\mapsto q_k$ and $\sqrt{2}/2^{2k}\mapsto ... 1 The reason you can't just prove$\sf DC$is that you are making infinitely many choices at once. There is, generally, no well-defined mechanism for choosing the specific$x_{n+1}$. In some cases you can in fact prove that such sequence exist, but those are the exception, not the general rule. As Andres notes in the comments,$\sf DC\$ is vastly weaker than ...

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