自然语言

我说的这句话是假话

集合论

Russell's paradox

$A=\{A|A\notin A\}$, then does $A\in A$?
if $A\in A$: $A\notin A$
if $A\notin A$: $A\in A$
Thus, such set doesn't exist.

停机问题

The halting problem is undecidable because assuming a solution leads to a logical contradiction, as shown by Alan Turing in 1936. Here’s a brief proof sketch:

1. Assume there exists a program （or Turing machine） H（P, I） that can decide whether any program P halts when given input I. It outputs "halts" or "does not halt".

2. Using H, construct a new program D that takes a program P as input and does the following:

   • Call H（P, P） to determine whether P halts when given itself as input.
   • If H says "halts", then D goes into an infinite loop （does not halt）.
   • If H says "does not halt", then D immediately halts.

3. Now ask: what happens when D is run with itself as input, i.e., D（D）?

   • If D（D） halts, then H（D, D） would say "halts", causing D to loop forever — contradiction.
   • If D（D） does not halt, then H（D, D） would say "does not halt", causing D to halt — contradiction.

Therefore, the program H cannot exist. No universal halting decider is possible.

A Graph to help you understand

flowchart TD
    A[Assume H（P, I） exists<br>decides if P halts on I] --> B[Construct D（P）:<br>1. Run H（P, P）<br>2. If H says 'halts' → loop forever<br>3. If H says 'does not halt' → halt]
    B --> C[Run D（D）]
    C --> D{H（D, D） result?}
    D -- "halts" --> E[D loops forever<br>Contradiction]
    D -- "does not halt" --> F[D halts<br>Contradiction]
    E --> G[Therefore H cannot exist]
    F --> G

Curry’s Paradox (self-referential implication)

Consider the sentence: “If this sentence is true, then Santa Claus exists.”
Let $S$ be “$S\to P$” where $P$ is any proposition.

• Assume $S$ is true. Then $S\to P$ gives $P$ (by modus ponens).
• Thus $S\to P$ holds (we assumed $S$ and derived $P$).
• But $S$ is $S\to P$, so $S$ is true.
• Therefore $P$ is true — any statement can be proved.

Cantor’s Diagonal Argument ( Valid in $\text{ZFC}$ )

Proves that the set of all infinite binary sequences is uncountable.

• List all sequences $s_1,s_2,\dots$.
• Construct a new sequence where the $n$-th bit is the opposite of the $n$-th bit of $s_n$.
• This new sequence differs from every listed sequence, contradicting the assumption that the list was complete.

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