# Why the function $f: \emptyset \rightarrow \emptyset$ exist? $0^0 = 1$

If $$f: \emptyset \rightarrow \emptyset$$, that is:

1. $$\forall a \in \emptyset \Rightarrow \exists b\in \emptyset, (a,b)\in \emptyset \times \emptyset$$. But how we can say, that for non-existing element exist another non-existing element?
2. $$\forall (a_1,b_1),(a_2,b_2) \in {\emptyset}^2$$, either $$a_1 \neq a_2$$ or $$b_1 \neq b_2$$. But if element non-existing, we can say, that $$\neg(a_1 \neq a_2$$ or $$b_1 \neq b_2)$$ is true too, no? And this is saying us, that this is $$f$$ is a function and not a function at one moment, no?
3. The subset of $$\emptyset \subset \emptyset \times \emptyset$$ exist and the $$G_f = \emptyset$$

So if this function exist $$\Rightarrow$$ that this function is only one, becouse $$G_f = \emptyset$$ and $$\emptyset$$ is only one in the Set Theory, yes? And because in power set of $$\emptyset$$ exist only one element -- $$\{\emptyset\} = 2^0$$.

So, after this we can say, that $$0^0 = 1$$, yes? Like that for all numbers in $$\mathbb{R}$$ will be true, that $$r^0 = 1$$, because for all elements of the family $$\{X_{\alpha}\}$$ of the sets with $$r$$-power : $$\forall X\in \{X_{\alpha}\} \Rightarrow |X| = r$$ exist only one function $$f_r: \emptyset \rightarrow X$$. But why this function only one? How can exist the element in $$\emptyset \times X$$?

And why the function $$f_{\emptyset} : X\rightarrow \emptyset$$ non-existing? The subset of $$G_{f_r}\subset \emptyset\times X$$ exist but the subset of $$G_{f_{\emptyset}}\subset X\times \emptyset$$ non-exist? I know, that $$0^r$$ always $$= 0$$ but cannot understand it in this situation

• Regarding 1. : $\forall a\in S$... (etc)..... (where $S$ can be anything) does not imply that any $a$ belongs to $S.$ Interpret it as : "For any $a,$ IF $a\in S$ THEN... (etc)..." – DanielWainfleet Nov 5 '18 at 7:11

A function $$f:X\to Y$$ is a subset of $$X\times Y$$ fulfilling certain properties. For instance, for each $$x\in X$$, there must be an element in $$f$$ which has that $$x$$ as its first component.

In the case $$X\to\varnothing$$, with $$X$$ non-empty, you cannot achieve that. So there are no such functions. For $$\varnothing\to Y$$ for any $$Y$$, on the other hand, the requirement above is vacuously true.

And yes, $$0^0=1$$ makes perfect sense. Especially in a combinatorics setting like this.

• Ye, thx, I must remember about that function requires that the $dom(f)$ always $= X$. But what about 1 and 2 points in my question? And can I say, that elements in $\emptyset\times X$ -- $\{(\cdot ,x)\}$? – Just do it Nov 5 '18 at 7:00
• @Arsenii They are vacuously true. Specifically, by convention, for any property $\phi(x)$, we say that $\forall x\in\varnothing, \phi(x)$ is true. If it were false, there should be counterexamples, but there are none, so it is true. – Arthur Nov 5 '18 at 7:04

When you say $$\forall a \in \emptyset$$ this alone is a false statement I see, because an empty set doesn't have any elements; and if you start from a false statement whatever is implied from this statement will be true i.e. when the reason itself is false, then any result will be true

• "$\forall a\in\emptyset$" alone is not a statement. It is neither true nor false. There is no verb in that phrase. But even if you complete it into a statement, like "$\forall a\in\emptyset$, $a$ is a prime number", that is true. It's a so-called vacuously true statement. Think of it this way: if that statement were false, that means $\exists a\in\emptyset$ such that $a$ is not prime". But there exists nothing in $\emptyset$. So this can't be right, and the opposite thing is true. – alex.jordan Nov 5 '18 at 7:10
• @alex.jordan so the $\forall a\in \emptyset, a$ is a prime number -- true and that $\forall a\in \emptyset, a$ is a non-prime number -- true too? – Just do it Nov 5 '18 at 7:28
• @Arsenii Exactly. In addition, "$\forall a\in\emptyset, a$ is both prime and not a prime" is also true. – Arthur Nov 5 '18 at 7:36
• @Arsenii "$\lnot(\forall a\in \emptyset, a$ is a prime number$)$" is not true. It's very different from "$\forall a\in\emptyset,\lnot( a$ is a prime number$)$", which is true. – Arthur Nov 5 '18 at 7:51
• So we can say that the first and second part of the question is true in a similar way, because their nagation (there exist an element that belongs to $\emptyset$ such that ....) is false. Am I right? – Fareed Abi Farraj Nov 5 '18 at 8:58