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This question already has an answer here:

Can any number be said to be larger than every member of the empty set? I have been asked to compare a positive integer to the empty set, but I'm not sure how to compare...

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marked as duplicate by MJD, Harish Chandra Rajpoot, user223391, user99914, Claude Leibovici Jan 14 '16 at 8:11

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    $\begingroup$ Yes, the statement "$3>x$ for every $x$ in the empty set" is true. $\endgroup$ – user147263 Jan 4 '16 at 19:25
  • $\begingroup$ What @Normal said is true. It makes perfect sense, and it's true, to say that a number is larger than every member of $\emptyset$; similarly, it makes sense & it's true to say that a number is less than every member of $\emptyset$, or that it's equal to every member of $\emptyset$. But I don't know what it means to "compare a positive integer to the empty set": what could that mean?? $\endgroup$ – BrianO Jan 4 '16 at 21:01
  • $\begingroup$ Yes. I also makes sense to say any number is smaller than every member of the empty set. That every number is greener, wetter or smellier than every member of the empty set. There are no members of the empty set so in any actual number will have more of any possible measurable quality then every one of the non-existent members of the empty set. $\endgroup$ – fleablood Jan 5 '16 at 3:57
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There are two ways to say this. One is to compare an ordinal to the ordinal represented by the empty set, which is zero. Any non-zero ordinal is greater than the empty set. In the Von Neumann definition of ordinals we define $a \lt b$ with $a, b$ ordinals as $a \subset b$ or equivalently $a \in b$ (by the construction of the ordinals). $\emptyset \subset a$ for all ordinals (and even all sets) $a$.

The second sense is to read your question literally. You ask if a number can be said to be greater than every member of the empty set. As there are no members of the empty set, the statement is true. If $b$ is the number we are asking about, you are asking about $\forall a (a\in \emptyset \implies b\gt a)$ As the antecedent is always false, the implication is always true.

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  • $\begingroup$ Actually $\emptyset\subset a$ is only true for $a\ne\emptyset$ (assuming you mean $\subsetneq$ by $\subset$). $\endgroup$ – Mario Carneiro Jan 4 '16 at 20:21
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Yes it does make sense, but it does not mean that there is an element in the empty set smaller than your number. Further your number is also less than any element in the empty set.

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Yes, it makes sense. It also makes sense to say that, for any proposition $P$ whatsoever, whether $P$ is true or false, we can infer: $$\forall x: [x\in \emptyset \implies P]$$ See Understanding Vacuously True (Truth Table)

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