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

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!

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

## 1 Answer

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.

• 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? – superuser123 Feb 13 at 7:28
• Do you mean that I can't assume the same thing for more than 2 events? – superuser123 Feb 13 at 7:30
• Yes, infinite intersections are , in general, more complicated than finite intersections and you are asked to prove the result for an infinite intersection. @superuser123 – Kabo Murphy Feb 13 at 7:33
• Got it, thanks! – superuser123 Feb 13 at 7:33
• @KaviRamaMurthy Isn't $\bigcap_{0\le i\le n}$ a finite intersection? – Viktor Glombik Feb 13 at 8:36