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This question is rather short, but I couldn't find the specific answer I was looking for anywhere.

My understanding so far: (Please correct me if I am wrong on some of this)

  • I can assume any statement $P(x)$ for all $x \in \emptyset$ to be true because the empty set doesn't contain any elements in the first place.
  • I can also assume any statement $Q(x)$ to be false for all $x \in \emptyset$ because of the same reason.
  • An example would be $ \forall M \subseteq \emptyset $ where $M \neq \emptyset$, it holds that $M = \emptyset$ . This statement has to be true for all subsets $M$ because the premise (there are no subsets of $\emptyset$ which are not the empty set themselves) is false.

My question: What is this good for? Is this behavior just a trivial property of the empty set? The only usage that comes to my mind is if I want to prove something specific for the empty set itself (for example that $\emptyset$ is an ordinal).

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    $\begingroup$ What do you mean with a function to be "true" or "false" ? $\endgroup$
    – Peter
    May 2 at 14:47
  • $\begingroup$ @Peter I assumed the word "function" here should actually be (logical) "predicate". $\endgroup$
    – aschepler
    May 2 at 14:52
  • $\begingroup$ I edited the post and changed the word function to statement. I also tried to make clear what I wanted to say in the example. $\endgroup$ May 2 at 14:53
  • $\begingroup$ No you can not assume any statement to be false for all $Q(x)$. You MUST assume all statements are true for all $x \in \emptyset$. That does mean that both a statement and the negation of a statement are both true for all $x \in \emptyset$ but an negation of a statement being true over an empty does not imply a statement is false over an empty set. $\endgroup$
    – fleablood
    May 2 at 16:10
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    $\begingroup$ It happens that quantifying over the empty set can lead to some jokes, e.g. "during all my years as King of England, I was a perfect ruler". $\endgroup$
    – FiMePr
    May 2 at 16:38
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Be careful: $P(x)$ is not a statement, unless $x$ is a constant. If $x$ is a variable, then, $P(x)$ is a propositional function and its truth value is, in general, undetermined.

The statement: $\forall xP(x)$, is false, if there are values of $x$, for which, $P(x)$ is false. The statement: "$\forall x\in\varnothing P(x)$", is always true, since $\varnothing$ has no members. But, this is actually an abbreviation for: $$\forall x\left[x\in\varnothing\rightarrow P(x)\right]$$ This is true, because, if $A$ is always false, $A\rightarrow B$ is always true. So, when you say:

I can assume any statement $P(x)$ for all $x\in\varnothing$ to be true because the empty set doesn't contain any elements in the first place.

This is not right. You should not assume $P(x)$ to be true. You should only assume "For all $x\in\varnothing$, $P(x)$" to be true. Similarly, you should not assume $Q(x)$ to be false. It has no truth value. Furthermore, if you assume that "For all $x\in\varnothing$, $Q(x)$" is false, your wrong. This is always a true statement.

Your example, formally expressed, gives the following: $$\forall M\left[M\subseteq\varnothing\wedge M\neq\varnothing\rightarrow M=\varnothing\right]$$

Since "$M\subseteq\varnothing\wedge M\neq\varnothing$" is always false (for every value of $M$), then "$M\subseteq\varnothing\wedge M\neq\varnothing\rightarrow M=\varnothing$" is always true. Thus, the above is a true statement. But from this, you could never conclude: "$M\neq\varnothing$ and $M=\varnothing$".

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  • $\begingroup$ Thank you for your answer. I now have a better understanding of using the statements P(x), Q(x), and so on. I just included the example as an extreme case of using the empty set quantification. $\endgroup$ May 2 at 15:47

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