Can't understand parts of a question

Question

I am working on a math exercice : Let $E$ be a non-empty set and $f : E \to \mathcal{P}(E)$, a function.
a) Let $A = \{x \in E \mid x \notin f(x)\}$. Let $x \in E$. Show that $x \in f(x) \cup A$ and that $x \notin f(x) \cap A$. Deduce from that that $f(x) \ne A$.

I don't really understand what $f(x)$ is here, is it just a single number or a set ? Is this correct to make a union or intersection between a set and a number as in $f(x) \cup A$ and $f(x) \cap A$ ?

Thank you

EDIT:

$f$ is a function that maps elements of E to the power set of E... A has to be the empty set... if $x$ is an element of the union of $f(x)$ and $A$ but not of their intersection... it should be clear from the definition of a power set — jrs, May 30 '17 at 11:32
I think you meant to write that $;A:=\left{,x\in E;/;x\notin \color{red}{f(x)},\right};$ , didn't you? — DonAntonio, May 30 '17 at 11:32
@Omnomnomnom Of course. Was editing when your comment arrived. Thanx. — DonAntonio, May 30 '17 at 11:34
@DonAntonio, No, It rally is $A := {x \in E | x \notin f(x)}$ — Pop Flamingo, May 30 '17 at 11:35
@DonAntonio it's a typo or an empty set... but it would be a weird way to define an empty set... — jrs, May 30 '17 at 11:35
@TrevörAnneDenise Exactyl...yet you wrote something different... — DonAntonio, May 30 '17 at 11:35
@DonAntonio Oh, sorry, even in my comment I forgot the {} around, I posted the original exercice in edit. — Pop Flamingo, May 30 '17 at 11:38

score 2 · Answer 1 · answered May 30 '17 at 11:39

2

Regarding your question. As you can see $f(x)$ maps into a powerset. The elements of a powerset are sets. So here the unions and intersections are operations between sets.

answered May 30 '17 at 11:39

jrs

1,518

Oh ! thank you I understand !! :) – Pop Flamingo May 30 '17 at 11:40

score 0 · Answer 2 · answered May 30 '17 at 11:37

$f$ is a function taking an element $x\in E$ to a set which is a subset of $E$ that is for every $x$, $f(x)\subseteq E$.

Now we want to show that $x\in f(x)\cup A$. This is true because if $x\not\in f(x)$ then $x\in A$ by definition. Also if $x\in f(x)$ then $x$ is not in $A$ and therefore the intersection is empty. Now if by contradiction $f(x)=A$ then $f(x)=f(x)\cap A$ but then $f(x)=A=\phi$ it follows that $f(x)\cup A=\phi$ which is a contradiction to that $x\in f(x)\cup A$.

Can't understand parts of a question

2 Answers2