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0 votes
0 answers
31 views

Set representation of (real) numbers [duplicate]

Using the Von Neumann representation we can represent the non-negative whole numbers using the empty set, e.g. $1$ as {$\emptyset$}. How do we represent with this notation numbers like $\sqrt2, -1, \...
user8363722's user avatar
3 votes
1 answer
116 views

In ZF, is it possible that there is no cardinal such that Reals injects into?

Working in ZF, is it possible that there is no cardinal number such that $\mathbb{R}$ can inject into? For if there exists a cardinal number $\kappa$ such that $\mathbb{R}$ injects into $\kappa$, then ...
mathlearner98's user avatar
0 votes
0 answers
105 views

Can every perfect set that is not a closed interval, or $(-\infty,a]$, or $[b,\infty)$ be written as a union of these types of intervals?

I have been reading the book "Introduction to Set Theory" by Jech and Hrbacek and have come to the following exercise in the chapter on sets of real numbers : Every perfect set is either an ...
scipio's user avatar
  • 595
3 votes
1 answer
216 views

Can you prove that $\Bbb R$ is uncountable using the Lebesgue measure?

I have been studying measure theory from the ground up, and am quite excited by the seeming power it holds. I thought of this last evening, and I wish to ask if the following proof of uncountability ...
FShrike's user avatar
  • 42.5k
4 votes
3 answers
446 views

Possible interpretation of real numbers as functions? [duplicate]

At the end of the day, a real number can be viewed simply as a function over the integers —> the naturals which tells you the digit as that ten’s place (assuming base ten)? You could augment this ...
Alex's user avatar
  • 161
1 vote
1 answer
143 views

Defining real numbers to exclude incomputable numbers

The real numbers are normally constructed via Dedekind cuts or similar approaches, which result in incomputable numbers: numbers that no finite algorithm can produce to arbitrary precision. This is ...
SRobertJames's user avatar
  • 4,450
1 vote
2 answers
165 views

Continuum family of continuum subsets of $\mathbb R$ which are not pairwise order isomorphic

I need to construct a continuum family of pairwise order inequivalent subsets of $\mathbb R$, such that cardinality of intersection of every constructed subset and every nontrivial interval is also ...
Gleb Chili's user avatar
  • 1,133
0 votes
1 answer
63 views

Exist strictly increasing $f \colon \omega _1 \mapsto \mathbb{R}$ [duplicate]

Does there exist a strictly increasing injective function $f \colon \omega _1 \mapsto \mathbb{R}$, where $\omega _1$ denotes the first uncountable ordinal?
Richard22's user avatar
3 votes
0 answers
74 views

How high in the constructible hierarchy do you need to go to see Dedekind-incompleteness?

This is a follow-up to my questions here and here. Let $X= (A,+,*,<)$ be an ordered field. Let us define a constructible hierarchy relative to $X$ as follows. Let $D_0(X)=A\cup A^2 \cup \{+,*,&...
Keshav Srinivasan's user avatar
5 votes
0 answers
138 views

Where is the first gap in the constructible hierarchy relative to a real closed field?

This is a follow-up to my question here. Let $X= (A,+,*)$ be a real closed field. Let us define a constructible hierarchy relative to $X$ as follows. Let $D_0(X)=A\cup A^2 \cup \{+,*\}$. For any ...
Keshav Srinivasan's user avatar
2 votes
1 answer
105 views

Where is the copy of $\mathbb{N}$ in the constructible hierarchy relative to a real closed field?

Let $X$ be a real closed field. Let us define a constructible hierarchy relative to $X$ is defined as follows. (This is slightly nonstandard terminology.). Let $L_0(X)=X$. For any ordinal $\beta$, ...
Keshav Srinivasan's user avatar
-2 votes
2 answers
34 views

Countability of the set [closed]

Let $f$ be differentiable function from $\mathbb{R}$ to $\mathbb{R}$. Consider the set $$A_y=\{x \in \mathbb{R} : f(x)=y \}$$ I want to know whether $A_y$ is countable for each $y\in \mathbb{R}$. I ...
Madhan Kumar's user avatar
3 votes
2 answers
599 views

Why does axiom of choice not imply the set of real numbers is countable?

The axiom of choice implies all sets can be well ordered. If that is true, you can well order the set of real numbers and the set of the integers. Now, why can one not just pair the set of real ...
Bquantump's user avatar
0 votes
1 answer
605 views

How is the Continuum Hypothesis equivalent to the existence of a well-ordering on $\Bbb R$ whose bounded initial segments are countable?

There exists an well-ordering $(<)$ on $\Bbb R$ such that the set $\{x \in \Bbb R\mid x < y \}$ is countable for every $y \in \Bbb R.$ How to prove that the above statement is equivalent to ...
math maniac.'s user avatar
  • 2,013
3 votes
1 answer
188 views

A closed countable set with Cantor-Bendixson rank of $\omega +1$

I'm looking to find a closed countable set that has a Cantor-Bendixson Rank of $\omega +1$. I know that $\{0\}\cup\{\frac{1}{x+1}|x\in\omega\}$ has a Cantor-Bendixson Rank of $2$ because we take out ...
Ari Fiorino's user avatar

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