Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is part of a bigger problem I'm working on.
To construct a a decreasing sequence of sets, $A_{n}\supseteq A_{n+1}$, I did the following: Let $B=\cup_{n=1}^{\infty} B_n$ and set $A_n= B\setminus (B_1\cup\ldots\cup B_n).$ Is this right? Do the $B_j's$ have to be disjoint?

How do I construct the increasing counterpart?


share|cite|improve this question
If you have a decreasing sequence, taking complement usually works for getting an increasing sequence. – user21436 Jan 14 '12 at 19:26

To have that $A_n\subseteq A_{n-1}$ you don't have to have the $B_n$'s disjoint.

We have that that $x\in A_n$ if and only if $x\in B_k$ for some $k>n$. This means that if $x\in A_{n+1}$ then it is in $B_k$ for some $k>n+1>n$ therefore $x\in B_k$. If you want an increasing sequence $C_n\subseteq C_{n+1}$ we can simply take $C_n = B_1\cup\ldots\cup B_n$.

If the sets $B_n$ are not disjoint, or strictly increasing in $\subseteq$ then it is fairly possible to have $C_k=C_{k+1}$ for some $k$ (and similarly with the $A_k$'s).

share|cite|improve this answer
Thanks. how would I construct the increasing counterpart? – jojo Jan 14 '12 at 19:43
@jojo: Did you post this after the edit? – Asaf Karagila Jan 14 '12 at 19:43
before the edit. – jojo Jan 14 '12 at 19:52
@jojo: Did the edit answer your question? – Asaf Karagila Jan 14 '12 at 20:12
Yes! Thank you again. – jojo Jan 14 '12 at 20:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.