Infinite intersections and infinite unions

Azzajazz · March 9th, 2016, 01:15 AM

I've started my study in a Bachelor of Science recently, and this question popped up on the second assignment for the maths course. I already know the answer, but I thought I'd post it for people to try.

Suppose we are given a set $A_n$ for every $n\in\mathbb{N}$.
Fill in the blank spaces with two words that make a true statement.

$\displaystyle x\in\bigcap_{k = 1}^\infty\bigcup_{n = k}^\infty A_n$ if and only if $x\in A_n$ for _______ _______ $n\in\mathbb{N}$.

The question didn't require a formal proof, but it did require some level of justification.

mathman · March 9th, 2016, 05:41 PM

You need "an infinite number of". I don't know how to reduce that to two words.

v8archie · March 9th, 2016, 06:45 PM

"infinitely many"?

Azzajazz · March 9th, 2016, 07:37 PM

Can you justify it?

mathman · March 10th, 2016, 07:01 PM

Let $S_k=\cup_{n=k}^{\infty}{A_n}.$
a) If $x\in{A_n}$ infinitely often, then $x\in{S_k}$ for all k, and therefore $x\in {\cap_{k=1}^{\infty}{S_k}}$ .
b) If $x\in{A_n}$ for only a finite number of n's, there will be a largest n, say m so that $x\notin{S_k}$ for k > m, and therefore $x\notin{\cap_{k=1}^{\infty}{S_k}}$ .

Azzajazz · March 10th, 2016, 09:01 PM

Yep. Spot on.

March 9th, 2016, 01:15 AM	#1
Azzajazz Math Team Joined: Nov 2014 From: Australia Posts: 672 Thanks: 239	Infinite intersections and infinite unions I've started my study in a Bachelor of Science recently, and this question popped up on the second assignment for the maths course. I already know the answer, but I thought I'd post it for people to try. Suppose we are given a set $A_n$ for every $n\in\mathbb{N}$. Fill in the blank spaces with two words that make a true statement. $\displaystyle x\in\bigcap_{k = 1}^\infty\bigcup_{n = k}^\infty A_n$ if and only if $x\in A_n$ for _______ _______ $n\in\mathbb{N}$. The question didn't require a formal proof, but it did require some level of justification. Thanks from greg1313 and Joppy
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March 9th, 2016, 05:41 PM	#2
mathman Global Moderator Joined: May 2007 Posts: 6,216 Thanks: 493	You need "an infinite number of". I don't know how to reduce that to two words. Thanks from greg1313
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March 9th, 2016, 06:45 PM	#3
v8archie Math Team Joined: Dec 2013 From: Colombia Posts: 6,696 Thanks: 2176 Math Focus: Mainly analysis and algebra	"infinitely many"?
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March 9th, 2016, 07:37 PM	#4
Azzajazz Math Team Joined: Nov 2014 From: Australia Posts: 672 Thanks: 239	Can you justify it?
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March 10th, 2016, 07:01 PM	#5
mathman Global Moderator Joined: May 2007 Posts: 6,216 Thanks: 493	Let $S_k=\cup_{n=k}^{\infty}{A_n}.$ a) If $x\in{A_n}$ infinitely often, then $x\in{S_k}$ for all k, and therefore $x\in {\cap_{k=1}^{\infty}{S_k}}$ . b) If $x\in{A_n}$ for only a finite number of n's, there will be a largest n, say m so that $x\notin{S_k}$ for k > m, and therefore $x\notin{\cap_{k=1}^{\infty}{S_k}}$ .
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March 10th, 2016, 09:01 PM	#6
Azzajazz Math Team Joined: Nov 2014 From: Australia Posts: 672 Thanks: 239	Yep. Spot on.
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