# checking a proof regarding finiteness of sets

If $X$ set and $U_i$, $i \in I$ a set family.

I know that $X \setminus \bigcup_{i\in I}$ is finite.

I want to prove that $X \setminus U_1 \cap U_2$ is also finite:

Here is my attempt:

$$X \setminus U_1 \cap U_2 = \{X \setminus U_1 \cup U_2\} \setminus\{\{U_1 \cup U_2\}\setminus\{U_1 \cap U_2\}\}$$

We notice that $\{X \setminus U_1 \cup U_2\}$ is finite and obviously you reduce the size of the set by "taking off" $\{\{U_1 \cup U_2\}\setminus\{U_1 \cap U_2\}\}$

Therefore $X \setminus U_1 \cap U_2$ is finite

• By $X \setminus U_1 \cap U_2$ do you mean $X\setminus (U_1\cap U_2)$ or do you mean $(X\setminus U_1)\cap U_2\text{ ?} \qquad$ – Michael Hardy Sep 24 '16 at 18:00
• BTW, my edit to the question should make some things clear about proper MathJax usage. $\qquad$ – Michael Hardy Sep 24 '16 at 18:03

Suppose $X=\{1,2,3,\ldots\}$ and $U_i= \{i\}$ for $i=1,2,3,\ldots$ Then $X \setminus \bigcup_{i=1}^\infty U_i=\varnothing,$ so that is a finite set, but $X\setminus(U_1 \cap U_2) = X\setminus\varnothing = X,$ and that is very far from finite.
Consider $X=[-1,1]$ and $U_i=(-1+1/i,1-1/i), i=2,3,\ldots.$ The set $X\setminus \bigcup_{i\in I}U_i=\{-1,1\}$ is finite. But $X \setminus U_2 \cap U_3$ is not.