bio | website | math.nju.edu.cn/~yuliang |
---|---|---|
location | China | |
age | ||
visits | member for | 3 years |
seen | 2 hours ago | |
stats | profile views | 730 |
A mathematical logician.
Apr 13 |
awarded | Yearling |
Apr 4 |
comment |
Martin-Löf randomness relative to a $\Delta^0_2$-representation of a real
The witness must not be random, were it exists. For example, if $x=0^{''}$ is random and $r$ is 3-random. Then $y=x$ must not below $r'$. Actually, I believe it is not true. |
Apr 3 |
comment |
Martin-Löf randomness relative to a $\Delta^0_2$-representation of a real
Is it true that if $x\equiv_T0″$ is random and $z$ is low for $x$, then $z\oplus 0′\not\geq_T0‴$? |
Apr 2 |
comment |
Interaction between Turing and many-one reducibility
The answer is no. Let $Y$ be the halting problem and $X$ be a 1-generic set below $Y$. |
Apr 2 |
comment |
Absolutely algorithmically random infinite sequence
Recursion theory people call what you called as bi-immuness. It contains all the weakly-random and weakly-generic reals. |
Feb 16 |
answered | Are there two computable binary trees such that each has a branch not computing any branch through the other? |
Oct 22 |
awarded | Good Answer |
Oct 22 |
awarded | Mortarboard |
Oct 22 |
awarded | Enlightened |
Oct 22 |
awarded | Nice Answer |
Oct 22 |
revised |
Can one cover the plane with less than continuum of lines?
added 1 characters in body |
Oct 22 |
answered | Can one cover the plane with less than continuum of lines? |
Oct 22 |
comment |
Can one cover the plane with less than continuum of lines?
@ToddTrimble, I think you are right. |
Oct 22 |
comment |
Can one cover the plane with less than continuum of lines?
For your question 1, the answer is no. If $\{X_{\alpha}\}_{\alpha\in A}$ cover a circle, then $|A|=2^{\aleph_0}$ (just because each line can cover at most two points on the circle). |
Oct 8 |
awarded | Caucus |
Aug 27 |
comment |
Analytic uniformization
You are right. I corrected the typo. The existence of $A$ follows from a well known fact that there is a $\Sigma^1_1$ set which does not contain a hyperarithmetic real. |
Aug 27 |
revised |
Analytic uniformization
added 1 characters in body |
Aug 18 |
accepted | Concerning Silver's result |
Aug 18 |
revised |
Concerning Silver's result
added 5 characters in body |
Aug 18 |
comment |
Concerning Silver's result
Sorry. I made a mistake. I am not sure your question. What I know is that every $0^{\sharp}$-admissible ordinal is an $L$-cardinal |