Mathematics Asked on December 26, 2020

To prove:$$text{If } A_1subseteq A_2 subseteq … subseteq A_ntext{ ,then } bigcup_{i=1}^n A_i = A_n$$ using the axioms of ZFC Set Theory.

Honestly, this statement is **very obvious**, but I do not want to take that for granted. How can I prove this from the axioms, especially the axiom of unions?

I’m fairly new to ZFC Set Theory, so I don’t know where to start. I do know that the Axiom of Unions (and Specification) gives rise to the definition of $bigcup_{i=1}^n A_i$ for sets $A_1,…,A_n$ though. The uniqueness follows from the Axiom of Extensionality.

*My thoughts:*

We already know that the union exists, and it is defined such that for every element $xin bigcup_{i=1}^n A_i$, there exists at least one set $A_j$ such that $xin A_j$. What’s next?

Thank you!

**Edit:**

Induction works if the chain is finite or countably infinite. What if the chain length is **uncountable**? Does this result still hold, and if yes, how do we prove it?

**Edit 2:**

So now I wish to prove (or disprove) the following using the axioms of ZFC Set Theory:

Consider an infinite set $A$, and an index set $I$ such that $$A_1subseteq A_2subseteq … subseteq A_i subseteq … subseteq A$$

where $iin I$. The index set $I$ is uncountable. We have $$bigcup_{iin I}A_i = A$$

This is regarding your second edit. First of all denoting $A_1 subseteq A_2 .... subseteq A$ is not correct when the index set $I$ is uncountable and moreover you don't know the elements of $I$.

Take $A_i =(0,i)$ where $iin (0,1)$ and $A = [0,1]$. These are nested in the sense that $i leq j$ implies that $A_i subseteq A_j$ and observe that for $i$ in index set $A_i subset A$. Clearly union of all $A_i$ is the interval $(0,1)$ which is not $A$.

Correct answer by Infinity_hunter on December 26, 2020

To be completely formal, you'd could first prove by induction that $A_isubseteq A_n$ for every $ileq n$. Then prove that $bigcup_{i=1}^n A_isubseteq A_n$ and that $A_nsubseteqbigcup_{i=1}^n A_i$. This last one should be trivial by definition of the union.

As for the first one, if $xin bigcup_{i=1}^nA_i$, then $xin A_j$ for some $j$, and then we know (from the proof by induction) that $xin A_n$.

Answered by Luiz Cordeiro on December 26, 2020

Get help from others!

Recent Questions

- How can I transform graph image into a tikzpicture LaTeX code?
- How Do I Get The Ifruit App Off Of Gta 5 / Grand Theft Auto 5
- Iv’e designed a space elevator using a series of lasers. do you know anybody i could submit the designs too that could manufacture the concept and put it to use
- Need help finding a book. Female OP protagonist, magic
- Why is the WWF pending games (“Your turn”) area replaced w/ a column of “Bonus & Reward”gift boxes?

Recent Answers

- Joshua Engel on Why fry rice before boiling?
- Peter Machado on Why fry rice before boiling?
- Jon Church on Why fry rice before boiling?
- haakon.io on Why fry rice before boiling?
- Lex on Does Google Analytics track 404 page responses as valid page views?

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP