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 descriptive-set-theory
Search options answers only
not deleted
user 454
Descriptive Set Theory is the study of definable subsets of Polish spaces, where definable is taken to mean from the Borel or projective hierarchies. Other topics include infinite games and determinacy, definable equivalence relations and Borel reductions between them, Polish groups, and effective descriptive set theory.
5
votes
Accepted
$\sigma$-algebra generated by analytic sets
"Inverse image of an analytic set is analytic" would imply ${\cal B}^1/ {\cal B}^1$-measurable, so let's try that. [Check my argument.]
Let $f : \mathbb R \to \mathbb R$ be continuous. Let $B \subse …
- 41.1k
2
votes
Accepted
Borel $\sigma$-algebras on paths of bounded variation
I think there is a problem with a mere semi-norm. The constant functions have variation distance $0$ from each other. Any variation-open set contains either all the constants, or none of them. There …
- 41.1k
7
votes
Accepted
A Baire subset of reals that is not Suslin measurable
While Gabe's answer is deleted ("One shouldn't try to work in ZF at 5am"), let me work in ZFC.
(a) The usual middle-thirds Cantor set $C$ is nowhere dense in $\mathbb R$. It has cardinal $\mathfrak c …
- 41.1k
2
votes
Accepted
Can there be an upper bound on the Borel rank of the preimages of Borel sets under a surject...
How about this: Construct a Cantor set $E \subseteq X$ on which $f$ is bijective. Then $f$ is a homeomorpism of $E$ onto $f(E)$, and $f(E)$ has Borel subsets of arbitrarily high rank.
- 41.1k
1
vote
How to give a Borel set whose projection is not Borel?
I like the presentation of descriptive set theory found in Chapter 8 of
Cohn, Donald L., Measure theory., Boston, MA: Birkhäuser. ix, 373 p. (1993). ZBL0860.28001.
- 41.1k
29
votes
A rare property of Hausdorff spaces
How about this example: $[0,1]^A$ with the product topology, where $A$ is uncountable. Then every nonempty $G_\delta$ set is uncountable. Since your sets $f^{-1}(x)$ are $G_\delta$ sets, this has yo …
- 41.1k
2
votes
Measure on hyperspace of compact subsets
In some cases a "natural" measure may be a Hausdorff measure (if it exists) that is positive and finite. So we want not just a Polish topology, but a specific metric. Given a metric for $X$ itself, …
- 41.1k
0
votes
Does every separated measurable space embed into a power of $\{0,1\}$?
How do you like this construction:
Let $X$ be a set and $\Sigma$ a sigma-algebra on $X$ that separates points of $X$. For each $E \in \Sigma$, define $\phi_E \colon X \to \{0,1\}$ by
$\phi_E(x) = 1$ …
- 41.1k
2
votes
Continuity on a measure one set versus measure one set of points of continuity
second answer:
Suppose we add some good conditions for $X$ and $Y$. Complete or Polish or something. Then it would suffice to prove this:
LEMMA. Let $X, Y$ satisfy (conditions to be determined). …
- 41.1k
7
votes
Continuity on a measure one set versus measure one set of points of continuity
first answer.
As stated ($X, Y$ merely metric spaces), NO.
(Remark: we may as well take $D=Y$ and $f$ the identity on $D$: if we can do that case, then we can apply it to get the general case.)
Let …
- 41.1k
2
votes
When can the one-one continuous image of a perfect set fail to be perfect?
Your argument is fine. Perhaps someone didn't say "Polish" or something.
- 41.1k