0
$\begingroup$

Question:

Given an infinite increasing sequence of events $E_1 \subset E_2 \subset ... \subset E_n ....$ show that $$\bigcup_{k=1}^n E_k = E_n$$ to verify “=”, you need to use the argument “A = B if A ⊂ B and B ⊂ A"

My Attempt:

Obviously $E_n \subset \bigcup_{k=1}^n E_k$ as when expanded we can see that $E_n$ is a part of $\bigcup_{k=1}^n E_k$

However I am stuck on showing that $ \bigcup_{k=1}^n E_k \subset E_n $.

My current proof involves the fact that since each event ($E_k$) includes its previous events ($E_1$ to $E_{n-1}$) the union of a event and its previous events is just the event($E_k \cup E_{k-1} = E_k$ ). However, this is effectively the statement I am trying to prove leaving me stuck.

Any help would be appreciated, thank you.

$\endgroup$
2
  • 1
    $\begingroup$ If $x\in\bigcup_{k=1}^n E_k$ then for some $1\leqslant j\leqslant n$, $x\in E_j\subset E_n$. $\endgroup$
    – Math1000
    Sep 30, 2023 at 21:15
  • $\begingroup$ Better phrased as a statement about sets. Using "Events" implies there is something else from probability theory at play at first look. $\endgroup$
    – AlgTop1854
    Sep 30, 2023 at 22:26

1 Answer 1

1
$\begingroup$

If $x$ is in your union, by definition $x$ belongs to some $E_k$ with $k\le n$. But any such $E_k$ is a subset of $E_n$. That is all there is to it.

$\endgroup$
1
  • $\begingroup$ Thank you that's a very elegant way of explaining it. $\endgroup$
    – AkAn
    Sep 30, 2023 at 22:54

You must log in to answer this question.

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