Is $f:\emptyset \to X$ injective

Question

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.

@tilper Thanks for your opinion, the definition also says it's injective btw — Jens Renders, Aug 05 '17 at 02:08
@JensRenders I dont understand how the definition leads to the injectivity. If you could explain why in really simple terms that would be wonderful — Ben Martin, Aug 05 '17 at 02:11
Wouldn't the function have to be to only one element X or else the relation wouldn't be a function? — Michael McGovern, Aug 05 '17 at 02:16
@JensRenders I don't understand your comment. The definition of what says $f$ is injective? — , Aug 05 '17 at 02:19

score 10 · Accepted Answer · answered Aug 05 '17 at 02:14

10

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.

answered Aug 05 '17 at 02:14

Sambo

6,243

Thanks that was the logical step i was missing – Ben Martin Aug 05 '17 at 02:15
Glad to help! Note that you can mark answer which are sufficient answers to your question as accepted. – Sambo Aug 05 '17 at 02:17
Yeah the website wont let me accept an answer too quickly but i will as soon as i can – Ben Martin Aug 05 '17 at 02:19
Oh, good point, I forgot. – Sambo Aug 05 '17 at 02:20
It seems to be an all too common defect of texts and in-class presentations to give insufficient attention to the quantifiers that govern a statement. Your answer here points out precisely what was missing in the original formulation. – Lubin Aug 05 '17 at 02:20
1

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... – Thomas Andrews Aug 05 '17 at 02:27

score 6 · Answer 2 · answered Aug 05 '17 at 02:20

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.

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 — Ben Martin, Aug 05 '17 at 02:22
Excellent detail in the answer. You make it really clear why the quantifiers make the statement true. — Sambo, Aug 05 '17 at 02:24

Is $f:\emptyset \to X$ injective

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