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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.

  • 5
    This question appears to be off-topic because it consists of multiple, largely unrelated questions at once.2013-11-16

1 Answers 1

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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.

  • 0
    @Joshua: Look at the answers to [this question](http://meta.math.stacke$x$change.com/questions/1773/do-we-have-an-equation-editing-howto).2012-06-15