Let $(x,\le)$ be well-ordered set and let $f: \ x \rightarrow x$ be monotonically increasing function. Prove that $\forall a \in x$ $$a \le f(a)$$ Find an example of set $x$ linearly ordered such that the statement doesn’t hold. My try: Assume $X=\{a \in x \ : \ a \ge f(a) \}$ is non-empty set […]

I could use a hand understanding a proof from Jech’s Set Theory. Firstly, note that Jech defines that an increasing function $f : P \rightarrow Q$ is a function that preserves strict inequalities (where $P$ and $Q$ are partially ordered sets). We then have Lemma 2.4. If $(W,<)$ is a well-ordered set and $f : […]

This question already has an answer here: List of explicit enumerations of rational numbers [closed] 1 answer

So, to explain the title, I’m referring to the necessity of the axiom of choice in the existence of a well ordering on reals, or any uncountable set. Now, while tweaking some sets, I came across this : We start with the natural numbers, $N$. We take the power set of the naturals, $P(N)$. Then […]

I would like to define a well-order on $\mathbb R$. My first thought was, of course, to use $\leq$. Unfortunately, the result isn’t well-founded, since $(-\infty,0)$ is an example of a subset that doesn’t have a minimal element. My next thought was to use that $P(\mathbb N)$ is in bijection with $\mathbb R$ and then […]

Is there an explicit well-ordering of $\mathbb{N}^{\mathbb{N}}:=\{g:\mathbb{N}\rightarrow \mathbb{N}\}$? I’ve been thinking about that for awhile but nothing is coming to my mind. My best idea is this: Denote by $<$ the usual “less than” relation on $\mathbb{N}$. Since $\mathbb{N}^{\mathbb{N}}$ is the set of infinite sequences ${\{x_{n}\}}_{n\in \mathbb{N}}$ with $x_{n}\in \mathbb{N}$, we can define ${\{x_{n}\}}_{n\in \mathbb{N}}\leq […]

Without appealing to the axiom of choice, it can be shown that (Proposition:) if $X$ is well-orderable, then $2^X$ is totally-orderable. Question: can we show the stronger result that if $X$ is well-orderable, then so too is $2^X$? Proof of Proposition. Pick any well-ordering of $X$. Then the lexicographic order totally orders $2^X$. More explicitly: […]

So, from what I understand, the axiom of choice is equivalent to the claim that every set can be well ordered. A set is well ordered by a relation, $R$ , if every subset has a least element. My question is: as anyone constructed a well ordering on the reals? First, I was going to […]

Let’s say I take a set $S$, where $S$ can be well ordered. From what I understand, one can use that well ordering to totally order $\mathscr{P}(S)$. How does a body actually use the well ordering of $S$ to construct a total ordering of $\mathscr{P}(S)$? Perhaps construct is too strong a word. Is there a […]

How is well-ordering in real line possible? I know that the axiom of choice provides possible well-ordering, but intuitively, this does not seem to make sense. How can you compare 1.111111…. and 1.111111… when we can never know the full digits of each number?

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