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

  • $\begingroup$ 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)..." $\endgroup$ – 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.

  • $\begingroup$ 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 $\emptyset\times X$ -- $\{(\cdot ,x)\}$? $\endgroup$ – Just do it Nov 5 '18 at 7:00
  • $\begingroup$ @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. $\endgroup$ – 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

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    $\begingroup$ "$\forall a\in\emptyset$" 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. $\endgroup$ – alex.jordan Nov 5 '18 at 7:10
  • $\begingroup$ @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? $\endgroup$ – Just do it Nov 5 '18 at 7:28
  • $\begingroup$ @Arsenii Exactly. In addition, "$\forall a\in\emptyset, a$ is both prime and not a prime" is also true. $\endgroup$ – Arthur Nov 5 '18 at 7:36
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    $\begingroup$ @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. $\endgroup$ – Arthur Nov 5 '18 at 7:51
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    $\begingroup$ 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? $\endgroup$ – Fareed Abi Farraj Nov 5 '18 at 8:58

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