I am having trouble understanding dedekind cuts. I know what they are and could find it if it was in a real line. but i am having difficulty understanding these 2 problems

Let $(A_1, A_2)$ be a Dedekind cut of $Q$. Prove that if $x < a$ and $a \in A_1$ then $x \in A_1$.

Prove that if $x > a$ and $a \in A_2$ then $x \in A_2$


Let $(A_1, A_2)$ and $(B_1, B_2)$ be Dedekind cuts of $Q$. Prove that either $A_1 \subseteq B_1$ or $A_1 \supseteq B_1$ .


1 Answer 1


Assume $x\notin A_1$. Then $x\in A_2$ because $A_1\cup A_2=\mathbb Q$. Then $a<x$ because $a\in A_1$ and $x\in A_2$. This contradicts $x<a$, hence x\in A_1$.

The second question works the same way.

Assume neither $A_1\subseteq B_1$ nor $A_1\supseteq B_1$. Then there exists $x\in A_1\setminus B_1$ and $a\in B_1\setminus A_1$. If $x<a$ then $a\in B_1$ implies $x\in B_1$ by the previous result - contradiction. If $x>a$, then $x\in A_1$ implies $a\in A_1$ by the previous result - contradiction. Therefor $x=a$, even more contradiction.

  • $\begingroup$ amazing. I did part a by my self and i got it by contradiction it was actually just question 2 that bothered me. Thanks a lot $\endgroup$
    – MathGeek
    Feb 26, 2013 at 0:42
  • $\begingroup$ (+1) can you recommend any book in which the basic algebraic operations has been defined and explained using dedkind cuts ? $\endgroup$ Mar 10, 2019 at 4:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .