Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm not really sure how to type this in notation...but I hope this makes sense. I help in proving that:

$$F \cap \bigcup\limits_{i=1}^\infty E_i = \bigcup\limits_{i=1}^\infty (F\cap E_i)$$


$$F \cup \bigcap\limits_{i=1}^\infty E_i = \bigcap\limits_{i=1}^\infty(F\cup E_i)$$

I really don't know where to begin with this one.

share|cite|improve this question

Probably the easiest way to prove these statements is by what I call element-chasing: show that every element of the left-hand side is also an element of the right-hand side and vice versa. Let’s look at the first one. Suppose that $x \in F \cap \bigcup\limits_{i=1}^\infty E_i$. The definition of intersection tells you that $x \in F$ and $x \in \bigcup\limits_{i=1}^\infty E_i$. Since $x \in \bigcup\limits_{i=1}^\infty E_i$, the definition of union tells you that $x$ belongs to at least one of the sets $E_1,E_2,E_3,\dots$, so there is some positive integer $k$ such that $x \in E_k$. (There might be more than one such integer, but all we need is one.) Now we know that $x \in F$ and $x \in E_k$, so $x \in F \cap E_k$. But $x \in F \cap E_k$ is one of the sets that goes into the union $\bigcup\limits_{i=1}^\infty (F\cap E_i)$ on the right-hand side, so automatically $F \cap E_k \subseteq \bigcup\limits_{i=1}^\infty (F\cap E_i)$, and therefore $x \in \bigcup\limits_{i=1}^\infty (F\cap E_i)$. We’ve now shown that every member of $F \cap \bigcup\limits_{i=1}^\infty E_i$ is also a member of $\bigcup\limits_{i=1}^\infty (F\cap E_i)$, or in other words that $$F \cap \bigcup\limits_{i=1}^\infty E_i \subseteq \bigcup\limits_{i=1}^\infty (F\cap E_i).$$ To finish the argument, you have to show that $$\bigcup\limits_{i=1}^\infty (F\cap E_i)\subseteq F \cap \bigcup\limits_{i=1}^\infty E_i.$$ You do this the same way, by starting with an arbitrary $x \in \bigcup\limits_{i=1}^\infty (F\cap E_i)$ and showing that it must belong to $F \cap \bigcup\limits_{i=1}^\infty E_i$ as well.

You can handle the second statement the same way: first show that every member of $F \cup \bigcap\limits_{i=1}^\infty E_i$ is a member of $\bigcap\limits_{i=1}^\infty(F\cup E_i)$, and then show the opposite inclusion. I’ll get you started. Suppose that $x \in F \cup \bigcap\limits_{i=1}^\infty E_i$; then $x \in F$ or $x \in \bigcap\limits_{i=1}^\infty E_i$ (or both), and you’ll have to distinguish two cases.

(1) If $x \in F$, then $x \in F \cup E_i$ for every positive integer $i$, so $x \in \bigcap\limits_{i=1}^\infty(F\cup E_i)$ by the definition of intersection.

(2) If $x \in \bigcap\limits_{i=1}^\infty E_i$, then ... what? I’ll leave this case to you, as well as the opposite inclusion.

share|cite|improve this answer
Element chasing. Nice. I will use that in the future. – Asaf Karagila Sep 21 '11 at 5:40
@Asaf Karagila: Element-chasing is actually unfolding of definitions from set theory: definitions of set equality, $\subseteq, \bigcap, \cap, \bigcup, \cup$, \. And this method gives formal proofs as opposed to Venn diagrams. – beroal Sep 26 '11 at 14:04
@beroal: I am not 100% clear on what you were trying to say in that comment of yours. – Asaf Karagila Sep 26 '11 at 14:07
@Asaf: I think that beroal didn’t realize that you were commented on the term, not on the method. – Brian M. Scott Sep 26 '11 at 17:49
@Brian: Oh, that sounds like a reasonable thought. Albeit somewhat amusing (reminiscing how once I explained to a barfly about quantum theory later to find out he has a Ph.D. in physics). – Asaf Karagila Sep 26 '11 at 17:54

For the first one:

We prove that the left hand side is contained in the right hand side, and conversely.

Suppose that $x\in F\cap\bigcup\limits_{i=1}^{\infty}$. That means that $x\in F$, and $x\in \bigcup\limits_{i=1}^{\infty}E_i$; therefore, $x\in F$ and there exists $n$ such that $x\in E_n$. Therefore, there exists $n$ such that $x\in F$ and $x\in E_n$, so $x\in F\cap E_n$. Thus, there exists at least one $n$ such that $x\in F\cap E_n$, so $x\in \bigcup\limits_{i=1}^{\infty}(F\cap E_i)$, as desired.

For the converse inclusion, suppose that $y\in \bigcup\limits_{i=1}^{\infty}(F\cap E_i)$. That means that there exists an $n$ such that $y\in F\cap E_n$. Therefore, $y\in F$, and there is an $n$ such that $y\in E_n$; the latter condition implies that $y\in\bigcup\limits_{i=1}^{\infty}E_i$. Since we also have $y\in F$, then $y$ lies in the intersection $F\cap\bigcup\limits_{i=1}^{\infty}E_i$.

Thus, we have shown that $$F\cap\bigcup_{i=1}^{\infty}E_i \subseteq \bigcup_{i=1}^{\infty}(F\cap E_i)\text{ and } \bigcup_{i=1}^{\infty}(F\cap E_i) \subseteq F\cap\bigcup_{i=1}^{\infty}E_i.$$ Hence, the two sets are equal.

A similar argument holds for the second equality, remembering that lying in the intersection means lying in every one of the sets.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.