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

Let $f$ be a function from $X$ to $Y$, and let $A$, $B$ be subsets (non-proper) of $X$. For each of the following statements, either prove the statement or else give a counter example:

a.) $f(X\setminus A)=Y\setminus f(A)$

b.) $f(X\setminus A) \subseteq Y\setminus f(A)$

c.) $Y\setminus f(A) \subseteq f(X\setminus A)$

d.) $f(A\cup B) = f(A)\cup f(B)$

e.) $f(A) \cap f(B) = f(A \cap B)$

I have an exam tomorrow and have been lagging on the set theory.

Much appreciated.

share|cite|improve this question

closed as off-topic by Lord_Farin, Bruno Joyal, user7530, azimut, Dennis Gulko Nov 16 '13 at 18:50

  • This question does not appear to be about math within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

@Mathemagician1234 You can help through MSE, can't you? – Pedro Tamaroff Jun 15 '12 at 4:22
@Peter Ok,fine.Gee,a guy's gotta eat in these tough times-can't fault me for trying............LOL – Mathemagician1234 Jun 15 '12 at 4:24
@Mathemagician1234 No comments. – Pedro Tamaroff Jun 15 '12 at 4:26
@Mathemagician1234 : Advertising for your professional services is usually called "spam". – Adam Smith Jun 15 '12 at 20:35
This question appears to be off-topic because it consists of multiple, largely unrelated questions at once. – Lord_Farin Nov 16 '13 at 17:59
up vote 3 down vote accepted

a) False If $A \subsetneq X$, consider the constant function.

b) False Use $(a)$

c) False. Suppose $f$ is not surjective. Suppose $Y$ contains more than one element and $f$ is a content function.

d) True. Suppose $x \in f(A \cup B)$, then $x = f(y)$ for $y \in A$ or $y \in B$ so $x \in f(A)$ or $x \in f(B)$. Suppose $x \in f(A) \cup f(B)$. Then $x \in f(A)$ or $x \in f(B)$. So there exists a $y \in A$ or a $y \in B$ such that $f(y) = x$ so $x \in f(A \cup B)$.

e) True. You wrote $f(A) \cap f(B) = f(A) \cap f(B)$. Do you mean $f(A) \cap f(B) = f(A \cap B)$??? If you meant the latter, this is false. Suppose $f$ is a constant function and $A$ and $B$ are disjoint.

share|cite|improve this answer
Damn,beat me to it,had a feeling someone would. Nice job,BTW. – Mathemagician1234 Jun 15 '12 at 4:30
For (c), what happens if instead of being content, $f$ is unhappy with its lot in life? (-; – Arturo Magidin Jun 15 '12 at 4:31
e.) Yes! That is what I meant. – user5262 Jun 15 '12 at 4:35
@ArturoMagidin Sometimes the autocorrect gets it wrong. I am going to leave it like that so your comment makes sense. – William Jun 15 '12 at 4:36
@Mathemagician1234 You should just post answer anyway. A quality answer is sometimes more desirable than the first answer. – William Jun 15 '12 at 4:40

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