I was doing a problem involving limits of sets, and I wanted to figure out what

$$\bigcup_{k=N}^{\infty} E_k - \bigcap_{k=N}^{\infty} E_k$$

is. I got that it is empty, which is obviously false (for instance, $E_k = \{k \}$ is a counterexample). The following is my "proof." Could you please take a look and point out the mistake? Thanks.

We begin with the original:

$$\bigcup_{k=N}^{\infty} E_k - \bigcap_{k=N}^{\infty} E_k.$$

Definition of set difference

$$\left(\bigcup_{k=N}^{\infty} E_k \right) \bigcap \left( \bigcap_{k=N}^{\infty} E_k\right)^c$$

DeMorgan's Law

$$\left(\bigcup_{k=N}^{\infty} E_k \right) \bigcap \left(\bigcup_{k=N}^{\infty} E_k^c \right) $$

Distribution Law

$$\bigcup_{k=N}^{\infty} \left[ E_k \bigcap \left(\bigcup_{k=N}^{\infty} E_k^c \right) \right] $$

Distribution Law again

$$\bigcup_{k=N}^{\infty}\left[ \bigcup_{k=N}^{\infty} E_k \cap E_k^c\right]$$

$$=\bigcup_{k=N}^{\infty}\left[ \bigcup_{k=N}^{\infty} \emptyset\right] $$

$$= \emptyset$$


1 Answer 1


You used the same variable $k$ for both the union and the intersection that makes a big difference. Use two different variables and you will see where the mistake is.

  • $\begingroup$ Thanks, I will give it a try now. $\endgroup$
    – Ovi
    Aug 8, 2019 at 0:16
  • $\begingroup$ Ahh I see haha thanks. $\endgroup$
    – Ovi
    Aug 8, 2019 at 0:17

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