All Questions
8
questions
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Prove: every finite cover $\mathcal U$ of $M\subseteq\mathbb R$ by open intervals contains two sets of disjoint intervals whose union covers $M$
It's not hard to show that if three open intervals in $\mathbb R$ have a non-empty
intersection, then one of the intervals is contained in the union of the other two. The simplest way to show this is ...
- 3,800
0
votes
0
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92
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Is axiom of choice required to throw away repeated intervals in a constructive argument?
I am looking at this answer to this question:
Let $U \subseteq \mathbb{R}$ be open and let $x \in U$. Either $x$ is rational or irrational. If $x$ is rational, define
\begin{align}I_x = \bigcup\...
- 743
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 ...
- 33
3
votes
1
answer
448
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Partition of positive reals with each part closed under addition without choice
It is an easy exercise using transfinite recursion to prove the following (in ZFC):
There exists sets $S,T$ that partition $\mathbb{R}_{>0}$ such that each of $S$ and $T$ is closed under addition.
...
- 59.1k
4
votes
0
answers
125
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Existence of two subsets of $\mathbb R$ with a certain property, undecidable in $ZF$?
Inspired by Question # 2420627 "Prove that there is a bijection $f:\mathbb R\times \mathbb R\to \mathbb R$ in the form of $f(x,y)=a(x)+b(y)$" and the answer and comments by Thomas Andrews.
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- 58.4k
14
votes
2
answers
554
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Can $\mathbb{R}^{+}$ be divided into two disjoint sets so that each set is closed under both addition and multiplication?
Can $\mathbb{R}^{+}$ be divided into two disjoint nonempty sets so that each set is closed under both addition and multiplication?
I know if we only require both sets to be closed under addition then ...
- 169
0
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1
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195
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Axiom of choice and an example of a Well-ordered $\Bbb R$
From the axiom of choice we get that every set can be ordered in a way that will make it a well ordered set, including $\Bbb R$. However, since the ordinal of such a well-ordered set of $\Bbb R$ will ...
- 392
9
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4
answers
2k
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Using Zorn's lemma show that $\mathbb R^+$ is the disjoint union of two sets closed under addition.
Let $\Bbb R^+$ be the set of positive real numbers. Use Zorn's Lemma to show that $\Bbb R^+$ is the union of two disjoint, non-empty subsets, each closed under addition.
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