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My book uses 2 equivalent definitions of injectivity, the first being $$x\neq y \Rightarrow g(x)\neq g(y)$$ and the second being $$g(x)=g(y) \Rightarrow x=y$$

Now as $f$ has $\emptyset$ as its domain I cannot make sense of either of these definitions as i cannot put in a variable to actually test either of these.

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  • $\begingroup$ I'd say $f$ is injective, vacuously. $\endgroup$ – user307169 Aug 5 '17 at 2:07
  • $\begingroup$ @tilper Thanks for your opinion, the definition also says it's injective btw $\endgroup$ – Jens Renders Aug 5 '17 at 2:08
  • $\begingroup$ @JensRenders I dont understand how the definition leads to the injectivity. If you could explain why in really simple terms that would be wonderful $\endgroup$ – B.Martin Aug 5 '17 at 2:11
  • $\begingroup$ Wouldn't the function have to be to only one element X or else the relation wouldn't be a function? $\endgroup$ – Michael McGovern Aug 5 '17 at 2:16
  • $\begingroup$ @JensRenders I don't understand your comment. The definition of what says $f$ is injective? $\endgroup$ – user307169 Aug 5 '17 at 2:19
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Note that in your definition, you implicitly quantify your variables $x$ and $y$ over your domain. That is, your definition is:

$$ \forall x,y \in \emptyset , x \neq y \implies g(x) \neq g(y). $$

Since there are no elements in the null set, this statement holds for every element in the null set (none), and so the function is indeed injective.

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  • $\begingroup$ Thanks that was the logical step i was missing $\endgroup$ – B.Martin Aug 5 '17 at 2:15
  • $\begingroup$ Glad to help! Note that you can mark answer which are sufficient answers to your question as accepted. $\endgroup$ – Sambo Aug 5 '17 at 2:17
  • $\begingroup$ Yeah the website wont let me accept an answer too quickly but i will as soon as i can $\endgroup$ – B.Martin Aug 5 '17 at 2:19
  • $\begingroup$ Oh, good point, I forgot. $\endgroup$ – Sambo Aug 5 '17 at 2:20
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    $\begingroup$ And $\forall x,y\in\emptyset(P)$ really means: $\forall x,y((x\in\emptyset\land y\in\emptyset)\implies P)$. Then it follows because $Q\implies P$ is true when $Q$ is false... $\endgroup$ – Thomas Andrews Aug 5 '17 at 2:27
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Yes, the function is injective. One way to see this is to make the definition more explicit:

A function $f:X\rightarrow Y$ is injective if

$$\forall x \in X, \forall y \in X, (x \neq y \Rightarrow f(x) \neq f(y))$$

or equivalently (the contrapositive) if

$$\forall x \in X, \forall y\in X, (f(x) = f(y) \Rightarrow x = y).$$

Whenever $S$ is an empty set, the statement $\forall x \in S, \ldots$ is always true— vacuously true. Such a statement says "Whenever you can find points in $S$ such that …", and because you can't find any points in an empty set $S$, the statement doesn't need to be checked for any points; it automatically holds.

The definition of injectivity is like this when the domain of $f$ is empty. It says "For any two points in the domain, …". The domain is empty so the statement doesn't need to be checked for any points; it automatically holds.

$f:\varnothing\rightarrow Y$ is injective.

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    $\begingroup$ Thanks for your answer, the lack of the quantifier for the elements was really stuffing me up and i wasn't getting that it was implicit $\endgroup$ – B.Martin Aug 5 '17 at 2:22
  • $\begingroup$ Excellent detail in the answer. You make it really clear why the quantifiers make the statement true. $\endgroup$ – Sambo Aug 5 '17 at 2:24

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