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I have a flaw in my reasoning, could you help me figure out what I am doing wrong?

To proof that $\emptyset$ is closed under addition we need to evaluate the following logical statement: $$ \forall x,y \in \emptyset: Q(x,y) $$ where $Q(x,y)$ is "$x+y \in \emptyset $"

We can transform this into a conditional statement like so:

$$ \forall x,y \in A:P(x,y)\implies Q(x,y) $$ where $A$ is a non-empty set with at least two elements and $P(x,y)$ is "$x,y \in \emptyset$". Here we can see that the first statement will always be false since $\emptyset$ has no elements and then by conditional logic this whole statement evaluates to true and so we conclude that $\emptyset$ is closed under addition. This is a vacuous truth.

But can we not use the same conditional logic to proof that $\emptyset$ is not closed under addition by using $R(x,y)$ to be "$x+y \notin \emptyset$"? $$ \forall x,y \in A:P(x,y)\implies R(x,y) $$

Since the antecedent $P(x,y)$ will always be false the whole statement will always evaluate to true and then we can conclude that $\emptyset$ is not closed under addition.

monkey_learning_maths
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2 Answers2

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For addition not to be closed on the empty set, you would need to show $\exists \,x,y \in \emptyset: x+y \notin \emptyset$.

This is not a consequence of the vacuous truth $\forall x,y \in \emptyset: x+y \notin \emptyset$, so there is no contradiction with the other vacuous truth $\forall x,y \in \emptyset: x+y \in \emptyset$. Assuming you have a suitable definition of addition, it remains the case that addition is closed on the empty set.

Henry
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  • @monkey_learning_maths For closure of a function on a set, you need to show all applications of the function to the elements work. Showing at least one works would be an example, but would not be sufficient (and in the case of the empty set neither necessary nor possible) – Henry Jan 06 '23 at 13:04
  • Now I understood everything, sorry for my confusion. I think it all stems from my wrong and quick definition of "not closed under addition". Thanks a lot! – monkey_learning_maths Jan 06 '23 at 13:10
  • @monkey_learning_maths To clarify, the proposition "not closed under addition" $\exists x, y \in \varnothing \colon x + y \notin \varnothing$ is the negation of the proposition "closed under addition" $\forall x, y \in \varnothing \colon x + y \in \varnothing$ by De Morgan's laws. – L. F. Jan 06 '23 at 13:20
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You have almost "reinvented" Proof By Contradiction !

We want to know whether $P$ is true or not.

You have :
$ P \rightarrow Q $
$ P \rightarrow R $

You know that :
$Q = \lnot R$ [ Equivalently $R = \lnot Q$ ]

Putting these together , we have a Contradiction.

Hence , we know that $P$ can not be true !

Example :

$M$ is the Empty Set of Martians.

$P$ : "Martians Exist" : $\exists m \in M$

$Q$ : "All Martians are born on Mars" : $\forall m \in M : BornOnMars(m)$
$R$ : "All Martians are not born on Mars" : $\forall m \in M : \lnot BornOnMars(m)$

$Q = \lnot R$ [ Equivalently $R = \lnot Q$ ]

$P \rightarrow Q$
$P \rightarrow R$

We can conclude ( Proof By Contradiction ) that "Martians Exist" is not true : $P$ is not true.

Prem
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  • Thanks for your comment. I understand your reasoning and I can see that P is not true. But what I really do not understand is how P -> Q is used to proof that the empty set is closed under addition, because since P is false (there are no elements in the empty set) we can not use P -> Q nor P -> R to say whether the empty set is closed or not under addition. I have updated my question and added a link to the post where my confusion originated. – monkey_learning_maths Jan 06 '23 at 11:56