Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with lo.logic
Search options questions only
not deleted
user 7206
first-order and higher-order logic, model theory, set theory, proof theory, computability theory, formal languages, definability, interplay of syntax and semantics, constructive logic, intuitionism, philosophical logic, modal logic, completeness, Gödel incompleteness, decidability, undecidability, theories of truth, truth revision, consistency.
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
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
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
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
22
votes
2
answers
2k
views
If ZFC has a transitive model, does it have one of arbitrary size?
It is known that the consistency strength of $\sf ZFC+\rm Con(\sf ZFC)$ is greater than that of $\sf ZFC$ itself, but still weaker than asserting that $\sf ZFC$ has a transitive model. Let us denote t …
- 39.8k
22
votes
1
answer
2k
views
The Continuum Hypothesis and Countable Unions
I recently edited an answer of mine on math.SE which discussed the implication of the two assertions:
$AH(0)$ which is $2^{\aleph_0}=\aleph_1$, and
$CH$ which says that if $A\subseteq 2^{\omega}$ and …
- 39.8k
22
votes
4
answers
3k
views
What sort of large cardinal can $\aleph_1$ be without the axiom of choice?
Assuming the axiom of choice it is very easy to see that $\aleph_1$ is a regular Joe of a successor cardinal. It is not very large in any way except the fact that it is the first uncountable cardinal. …
- 39.8k
20
votes
2
answers
2k
views
What is $\omega_1^{CK}(\mathsf{Ord})$?
We know that if $\alpha<\omega_1^{CK}$ then there is some recursive $R$ such that $(\omega,R)$ has order type $\alpha$.
Let's consider now the ordinals, $\mathsf{Ord}$ with their natural order. This …
- 39.8k
20
votes
2
answers
3k
views
What is a Choice Principle, really?
This question is quite soft, and I apologize in advance if it borderline off-topic.
When working in theories between ZF and ZFC the term "choice principle" is heard quite often. For example:
$\quad$ …
- 39.8k
19
votes
0
answers
897
views
What examples of existence forcing proofs are there?
Forcing proofs tend to be fairly constructive, in the sense that if I claim that there is a forcing that does something, I usually prove this by constructing that forcing.
There are only a handful of …
- 39.8k
19
votes
1
answer
812
views
If all reals are generic, is the set of reals generic?
Let $W\subseteq V$ be two models of $\sf ZFC$ with the same ordinals. Is the following situation consistent:
For every $x\in\Bbb R^V$ there is some $P_x\in W$ such that for some $G\subseteq P_x$ whi …
- 39.8k
19
votes
6
answers
3k
views
Sierpinski's construction of a non-measurable set
In the early 20th century there was a lot of fuss over the axiom of choice implying that there are Lebesgue non-measurable sets of reals. In his book about The Axiom of Choice, Gregory Moore points to …
- 39.8k
18
votes
1
answer
4k
views
Countable unions and the axiom of countable choice
Let us denote by ACC the axiom of countable choice, namely the assertion that the product of countably many non-empty sets is non-empty, and denote by UCC the assertion that a countable union of count …
- 39.8k