Mathematical Logic 2020

Lecture: HGX308, M 9:55-11:35
Section: HGX307, R 18:30-20:10

Slides 01 (handout) Slides 02 (handout) Slides 03 (handout) Slides 04 (handout) Slides 05 (handout) Slides 06 (handout) Slides 07 (handout) Slides 08 (handout) Slides 09 (handout) Slides 10 (handout) Slides 11 (handout) Slides 12 (handout) Slides 13 (handout) Slides 14 (handout) Slides 15 (handout) Slides 16 (handout) Slides 17 (handout)



第13页第3-4行:“并且读写头停在……读写头停在0上)”改为“并且读写头停在\(y\)­ 串的 最后一个 1 (如果有的话,不然 \(y = 0\) ,停在隔开原 \(x + 1\)­ 串与 \(y + 1\)­ 串的 0 ) 的右侧。”

How to define membership relation from subset relation and power set operation

In my talk at 2018 Chinese Mathematical Logic Conference, I asked if \((V,\subset,P)\) is epsilon-complete, namely if the membership relation can be recovered in the reduct. Professor Joseph S. Miller approached to me during the dinner and pointed out that it is epsilon-complete. Let me explain how.


Let \((V,\in)\) be a structure of set theory, \((V,\subset,P)\) is the structure of the inclusion relation and the power set operation, which are defined in \((V,\in)\) as usual. Then \(\in\) is definable in \((V,\subset,P)\).


Fix a set \(x\). Define \(y\) to be the \(\subset\)-least such that

\[\forall z \big((z\subset x\wedge z\neq x)\rightarrow P(z)\subset y\big).\]

Actually, \(y=P(x)-\{x\}\), so \(\{x\}= P(x) – y\). Since set difference can be defined from subset relation and \((V,\subset,\{x\})\) can define \(\in\), we are done.


Here is another argument figured out by Jialiang He and me after we heard Professor Miller’s Claim.

Since \(\in\) can be defined in \((V,\subset,\bigcup)\) (see the slides). Fix a set \(A\), it suffices to show that we can define \(\bigcup A\) from \(\subset\) and \(P\).

Let \(B\) be the \(\subset\)-least set such that there is \(c\), \(B=P(c)\) and \(A\subset B\). Note that
\bigcap\big\{P(d)\bigm|A\subset P(d)\big\}= P\big(\bigcap\big\{d\bigm|A\subset P(d)\big\}\big).
Therefore, \(B\) is well-defined. Next, we show that
\bigcap\big\{d\bigm|A\subset P(d)\big\}=\bigcup A.
Clearly, \(A\subset P(\bigcup A)\). This proves the direction from left to right. For the other direction, if \(x\) is in an element of \(A\), then it is in an element of \(P(d)\) given \(A\subset P(d)\), i.e. it is an element of such \(d\).

Therefore \(\bigcup A\) is the unique set whose power set is \(B\).