Let $\{A_i\}_{i=0}^{i=n}$ a series of events such that $\forall i$ $P(A_i)=1$. Show that $\bigcap\limits_{0 \leq i \leq n}A_i=1$.

My attempt:

Let $0\leq k\ne j\leq n$, so $P(A_j)=1, P(A_k)=1$.

From exclusion inclusion principle:

$P(A_j \cup A_k)=P(A_j)+P(A_k)-P(A_j\cap A_k)$

$1 = 1 + 1 - P(A_j \cap A_k) \Rightarrow P(A_j\cap A_k)=1$

Could anyone confirm/be ashamed of my "work"?

Thanks in advance!

  • $\begingroup$ Please also edit the title to reflect that you want to show $P\left(\bigcap_i A_i\right) = 1$. $\endgroup$ – Viktor Glombik Feb 13 at 8:34

It ,is not enough to prove that intersection of two of the sets has probability $1$. You are asked to prove that $P(\cap_i A_i)=1$. For this note that $P(A_i^{c})=0$ for all $i$. This implies that $P(\cup_i A_i^{c}) \leq \sum_i P(A_i^{c})=0$. Hence $P(\cup_i A_i^{c})^{c})=1$. But $(\cup_i A_i^{c})^{c}=\cap_iA_i$ and this completes the proof.

  • $\begingroup$ Thanks, but why isn't that enough? if I take 2 arbitrary events and show that they intersect to 1, I can do that with any 2 events, no? $\endgroup$ – superuser123 Feb 13 at 7:28
  • $\begingroup$ Do you mean that I can't assume the same thing for more than 2 events? $\endgroup$ – superuser123 Feb 13 at 7:30
  • $\begingroup$ Yes, infinite intersections are , in general, more complicated than finite intersections and you are asked to prove the result for an infinite intersection. @superuser123 $\endgroup$ – Kabo Murphy Feb 13 at 7:33
  • $\begingroup$ Got it, thanks! $\endgroup$ – superuser123 Feb 13 at 7:33
  • $\begingroup$ @KaviRamaMurthy Isn't $\bigcap_{0\le i\le n}$ a finite intersection? $\endgroup$ – Viktor Glombik Feb 13 at 8:36

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