All Questions
Tagged with real-numbers set-theory
79
questions
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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, \...
3
votes
1
answer
116
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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 ...
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105
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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 ...
3
votes
1
answer
216
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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 ...
4
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3
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446
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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 ...
1
vote
1
answer
143
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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 ...
1
vote
2
answers
165
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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 ...
0
votes
1
answer
63
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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?
3
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74
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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 \{+,*,&...
5
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138
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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 ...
2
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1
answer
105
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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 $\beta$, ...
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2
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34
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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 ...
3
votes
2
answers
599
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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 ...
0
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1
answer
605
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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 ...
3
votes
1
answer
188
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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
...