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Prove that a function $f:X \to Y $ is injective if and only if $\forall x_1, x_2 \in X$ where $f(x_1) = f(x_2)$ implies that $x_1 = x_2$

Taking the contrapositive we get (this is the step I'm a little hazy on)

For $x_1, x_2 \in X$ where $x_1 \neq x_2$ implies $\exists x_1, x_2 \in X$ such that $f(x_1) \neq f(x_2)$ (Do I put the there exists here because that is the negation of for all?)

Because this is the definition of injective this is true. QED

wdacda
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RhythmInk
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2 Answers2

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The contrapositive will be the following whenever $$x1,x_2 \in X,x_1 \neq x_2 \implies f(x_1) \neq f(x_2)$$

You don't need there exists symbol

Actually your original definition is not concise. An injective function is a function such that if $x_1 = x_2$ then $f(x_1) = f(x_2)$, you don't need $\forall$

Check this website here it has both the original and the contrapositive

defintion Link

alkabary
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Suppose not. Then it follows that $x_1\not=x_2$ and $f(x_1)=f(x_2)$. This contradicts with $f(x)$ is an injection.

Paul
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  • This was my first instinct but we aren't assuming the conclusion here? We can start by assuming it is injective and $x_1 \neq x_2$? – RhythmInk May 25 '15 at 02:49