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Results tagged with set-theory
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user 7206
forcing, large cardinals, descriptive set theory, infinite combinatorics, cardinal characteristics, forcing axioms, ultrapowers, measures, reflection, pcf theory, models of set theory, axioms of set theory, independence, axiom of choice, continuum hypothesis, determinacy, Borel equivalence relations, Boolean-valued models, embeddings, orders, relations, transfinite recursion, set theory as a foundation of mathematics, the philosophy of set theory.
53
votes
1
answer
4k
views
When does $A^A=2^A$ without the axiom of choice?
Assuming the axiom of choice the following argument is simple, for infinite $A$ it holds: $$2\lt A\leq2^A\implies 2^A\leq A^A\leq 2^{A\times A}=2^A.$$
However without the axiom of choice this doesn't …
- 39.8k
53
votes
1
answer
6k
views
Does $2^X=2^Y\Rightarrow |X|=|Y|$ imply the axiom of choice?
The Generalized Continuum Hypothesis can be stated as $2^{\aleph_\alpha}=\aleph_{\alpha+1}$. We know that GCH implies AC (Jech, The Axiom of Choice, Theorem 9.1 p.133).
In fact, a relatively weak form …
- 39.8k
50
votes
0
answers
2k
views
How many algebraic closures can a field have?
Assuming the axiom of choice given a field $F$, there is an algebraic extension $\overline F$ of $F$ which is algebraically closed. Moreover, if $K$ is a different algebraic extension of $F$ which is …
- 39.8k
30
votes
Accepted
How to construct a basis for the dual space of an infinite dimensional vector space?
It is consistent with the axioms of $\sf ZF$ that this is impossible. Specifically, if you consider $\Bbb R[x]$, then its dual space is just $\Bbb{R^N}$. And it is consistent with $\sf ZF$ that $\Bbb{ …
- 39.8k
27
votes
1
answer
3k
views
If $V$ is a vector space with a basis. $W\subseteq V$ has to have a basis too?
Suppose $V$ is a vector space, we say that $\mathcal B$ is a basis for $V$ if:
Every $v\in V$ can be written as a linear combination of elements of $\mathcal B$;
If $\sum\alpha_i b_i = 0$, where $\a …
- 39.8k
27
votes
1
answer
2k
views
How hard is it to destroy a diamond? (with a real)
If we start with $V\models\lozenge$, it is not hard to force the failure of diamond. You can blow up the continuum, or destroy all the Suslin trees. You can blow up the continuum of $\aleph_1$, and th …
- 39.8k
26
votes
Accepted
How much of GCH do we need to guarantee well-ordering of continuum?
Yes, you are right. This is a theorem of Specker. If there are no intermediate cardinals between $A,\mathcal P(A)$ and $\mathcal{P(P(}A))$, then $A$ can be well-ordered.
You can find nice details in:
…
- 39.8k
26
votes
2
answers
3k
views
Sizes of bases of vector spaces without the axiom of choice
Assuming the axiom of choice does not hold we have that there is a vector space without a basis. The situation can be, in some sense, worse. It is consistent that there are vector spaces that have two …
- 39.8k
24
votes
2
answers
2k
views
Short proof of $\frak p=t$
It is known for a while now that $\frak p=t$, as a result of Malliaris-Shelah. The original paper draws from model theoretic methods.
I've heard rumors that there was a proof which was purely set the …
- 39.8k
23
votes
Does ZF+AD settle the original Suslin hypothesis?
No. Because of silly reasons.
Recall that the powers of $\sf AD$ are quite limited to the world below $\Theta$. In particular, the proof that $\sf AD$ does not imply countable choice goes through addi …
- 39.8k
23
votes
1
answer
4k
views
A recommended roadmap into inner models
A friend of mine and myself (both grad students with a relatively decent set theoretic background) want to venture into the universe of inner models. [pun intended :-)]
I would very much like to get …
- 39.8k
23
votes
When was the continuum hypothesis born?
Finally, a good use for the newly purchased copy of "Zermelo's Axiom of Choice".
Moore writes that Cantor formulated the following problem in 1878:
Every infinite subset of $\Bbb R$ is either den …
- 39.8k
22
votes
1
answer
1k
views
When will the real numbers be Borel?
In set theory Borel sets are important, but we don't actually care about the sets. We can about the Borel codes. Namely, the algorithm to generate a given Borel set starting with the basic open sets ( …
- 39.8k
22
votes
Hahn's Embedding Theorem and the oldest open question in set theory
I don't know the answer to (1), and would be glad to give it some thought later this week. Regardless to (1) the answer to (2) is semi-negative.
There are two conjectures which seem to be slightly ol …
- 39.8k
22
votes
3
answers
3k
views
Half Cantor-Bernstein without choice
I had a discussion with one of my teachers the other day, which boiled to the following question:
Assume ZF. Let $A,B$ be sets such that there exist $f\colon A\to B$ which is injective and $g\colon A …
- 39.8k