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Consider the following statement: $$\forall \epsilon > 0,\space\exists\delta>0:(|x-a|\lt\delta\implies|f(x)-L|\lt\epsilon).$$ (a) Write the converse of the statement.
(b) Write the contrapositive of the statement.

I am stuck on how to complete these problems because I do not understand the notation

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Have you read you course notes ? –  Siméon Dec 6 '12 at 18:08
    
Yes, I still just do not understand where to begin. –  CHZ Dec 6 '12 at 18:09
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Do you understand the notation $$\forall x,\exists y:(P\implies Q)\;?$$ That’s all the notation that you need to understand in order to do the problem. –  Brian M. Scott Dec 6 '12 at 18:25
    
So for the converse do I want to say: ∀p,∃q: (X>Y)? And I know that the contrapositive is just the negation of both sides. So, For all x there does not exist a Y where P yields Q? or is it: For all x there does not exist a Y where P does not yield Q? –  CHZ Dec 6 '12 at 18:29
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2 Answers 2

The usual definitions of "converse" and "contrapositive" used in logic only apply to implications, which are statements of the form $A \Rightarrow B$. The converse of $A \Rightarrow B$ is $B \Rightarrow A$, and the contrapositive is $(\lnot B) \Rightarrow (\lnot A)$, where $\lnot A$ is the negation of $A$.

Because the statement you wrote has two quantifiers at the front, it is not an implication, and the usual definitions do not apply to it. Therefore, you should ask your instructor, or consult your notes, to learn what the instructor wants you to do. It will be difficult to find much help in the usual textbooks or reference sources because these do not give any definition for the "converse" or "contrapositive" of statements that are not implications.

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$$\forall \epsilon > 0,\space\exists\delta>0:(|x-a|\lt\delta\implies|f(x)-L|\lt\epsilon).$$

Perhaps what you are asked to do is to replace the quantified (internal) implication with it's contrapositive:
$$\forall \epsilon > 0,\space\exists\delta>0:(|x-a|\lt\delta\rightarrow|f(x)-L|\lt\epsilon)$$ $$\iff \forall \epsilon > 0,\space\exists\delta>0:[\lnot(|f(x)-L|\lt\epsilon) \rightarrow \lnot(|x-a|\lt\delta)]$$ $$\iff \forall \epsilon >0, \; \exists \delta >0 :[\lnot\lnot(|f(x)-L|\lt\epsilon) \lor \lnot(|x-a|\lt\delta)]$$ $$\iff \forall \epsilon > 0, \; \exists \delta > 0:\lnot[\lnot (|f(x)-L|\lt\epsilon) \land (|x-a|\lt\delta)]$$ $$\iff \lnot \exists \epsilon > 0,\;\forall\delta>0: [(|f(x) - L| \geq \epsilon) \land (|x - a|\lt\delta)]$$

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