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 started to get excited about mathematical analysis. So I bought a mathematical analysis book and started to study. But because of the reason that, book do not have solutions I do not have an idea where to start and how to prove the following:

Let $A_t$, $t \in T$, be a family of sets, and let $X$ be a set. Prove the identities:

$$X \setminus \bigcup A_t = \bigcap (X\setminus A_t)$$

$$X \setminus \bigcap A_t = \bigcup(X\setminus A_t) $$

Could you please help me? Also I need a book recommendation about mathematical analysis which goes like a theorem and its proof, a theorem and its proof.. Do you have any suggestions?


share|cite|improve this question
Usually to prove $C=D$ for sets $C$ and $D$, you show that $C\subseteq D$ and $D\subseteq C$. Try doing that with $C=X\setminus\bigcup A_t$ and $D=\bigcap(X\setminus A_t)$ by unwrapping the definitions of $\bigcup,\bigcap,\setminus$. – yunone Sep 28 '12 at 22:59
@yunone thank you so much! I really did not have an idea how to solve this. – Amadeus Bachmann Sep 28 '12 at 23:27
De Morgan's Laws at ProofWiki – Martin Sleziak Oct 3 '12 at 13:12
up vote 0 down vote accepted

Let's look at the case there $|T| = 1$, i.e. that the family of sets has only one element.

The $X\setminus A$ is trivially $X \setminus A$.

What about when $|T| = 2$?

Then $X \setminus (A_1 \cup A_2)$ is the set of $x \in X$ such that $x \not \in A_1, A_2$.

$X \setminus A_1$ is the set of $x \in X$ such that $x\not \in A_1$, and $X \setminus A_2$ is the set of $x \in X$ such that $x \not \in A_2$. This means that $\bigcap (X \setminus A_t)$ is the set of $x \in X$ such that $x \not \in A_1$ and such that $x \not \in A_2$. Equivalently, this is the set of $x \in X$ such that $x \not \in A_1, A_2$. My gosh, this is the same as the set above!

Now you might look back at this and say, where did we use our ability to enumerate the sets?

Do you see how to finish from here?

share|cite|improve this answer
so for the first one, it this true: from left to right: if m∈ X∖⋃At => m ∈ X and m∉⋃At. so m∉At for all t∈At. if m∈X and m∉At for all t∈At, then X\At={m} for al t∈T then, m∈⋂(X∖At) from right to left: if m∈⋂(X∖At) => m ∈ X\At for all t∈T. then, m∈X, m∉At for all t∈At. if m∉At for all t∈At => m∉⋃At. if m∈X and m∉⋃At => m∈X∖⋃At therefore X∖⋃At=⋂(X∖At) – Amadeus Bachmann Sep 28 '12 at 23:25

For example:

$$x\in X\setminus \bigcup A_t\Longleftrightarrow \,\,\,\text{for no}\,\,t\,\,,\,\,x\in A_t\Longleftrightarrow \,\,\forall\,t\,\,,\,x\in X\setminus A_t\Longleftrightarrow x\in\bigcap(X\setminus A_t)$$

The other equality is very similar.

share|cite|improve this answer
thank you so much for your help. Now I know what to do. – Amadeus Bachmann Sep 28 '12 at 23:29

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.