1,037 reputation
812
bio website math.nju.edu.cn/~yuliang
location China
age
visits member for 3 years
seen 2 hours ago
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