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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.

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

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What have you tried? Which ones are you stuck on? – Alex Becker Jun 15 '12 at 4:16
@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
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

1 Answer 1

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.

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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

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