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I remember my high school days where subsets were defined in the following manner:

Given two sets A and B, if every element of B is an element of A, then B is called a subset of A.

A common confusion arising from this definition is that of an empty set being a subset of every set. 'But an empty set doesn't contain any elements which are in any other set, so how can it be a subset of every set?'

I feel the following definition removes this confusion:

Given two sets A and B, if by removing the elements of B we can form A, then A is a subset of B.

Since for any general set X, we can remove its elements to form the empty set, we can say that the empty set is a subset of every set.

Are there any flaws in this way of thinking about subsets?


I'm not a teacher, but I thought the question better suits this community.

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    $\begingroup$ Regarding the empty set being the subset of every set: The way we can verify the definition of subset (Given two sets $A$ and $B$, if every element of $B$ is an element of $A$, then $B$ is called a subset of $A$) is by asking whether there is any element in $B$ that is not an element of $A$. So, given any set $A$, since there is no element in $\varnothing$ that is not an element of $A$, it follows that $\varnothing$ is a subset of $A$. $\endgroup$ Commented May 30, 2023 at 14:23
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    $\begingroup$ Instead of "removing the elements of $B$", say "removing elements of $B$", or, if you want to be very precise, "removing zero or more elements of $B$". $\endgroup$ Commented May 30, 2023 at 15:08
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    $\begingroup$ As Steven Gubkin emphasizes in his accepted answer, this is just one case among many of a vacuous truth, and it's probably better to address that directly. $\endgroup$ Commented May 30, 2023 at 17:27
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    $\begingroup$ I think it's helpful to teach people to rephrase "Every X is a Y" into "If something is an X, then it is (also) a Y." $\endgroup$ Commented May 31, 2023 at 4:33
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    $\begingroup$ @Barmar yes I have, it's the one linked in my previous comment. Please don't criticise if you haven't read it. $\endgroup$
    – David
    Commented Jun 1, 2023 at 6:39

6 Answers 6

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In both formal and informal treatments of set theory, we need to specify which operations and relations are allowable and build from there.

Usually we take sets and set membership as primitives. We also usually build on top of first order logic.

So it is very natural to define $B \subset A$ as "For all $b$ in $B$, we have $b \in A$". This uses first order logic (the universal quantifier) and set membership.

Your definition requires us to also have a notion of "removing a set of elements from a set". This operation would generally be defined in terms of more primitive notions as follows: Let $X$ and $S$ be sets. We define the set $X - S$ as the collection of all elements of $X$ which are not elements of $S$. This requires "set formation" axioms, but it is a legit definition.

Now your proposed definition of the subset relation is "$B \subset A$ if and only if there exists a set $S$ so that $A - S = B$".

This definition is valid, but is a lot more logically complex than the usual definition. Try proving a few basic theorems using the usual definition and this modified definition. For example, try proving that if $A \subset B$ and $B \subset C$ then $A \subset C$. I think you will find that the new definition is much clunkier.

I think it would be fine to mention the equivalence of these two definitions to a student who was struggling to understand that the empty set is a subset of any other set, but at some point they are going to have to deal with understanding vacuously true statements if they want to understand this equivalency.

Personally I think your time would be better spent really working on understanding vacuously true statements. It will generalize better and enable them to understand the standard definition. Definitely hard work though!

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    $\begingroup$ Thanks for addressing every issue. $\endgroup$ Commented May 30, 2023 at 14:15
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    $\begingroup$ (+1) especially for Personally I think your time would be better spent really working on understanding vacuously true statements. -- This was my feeling as I was reading through the question. Anyone who continues in mathematics will encounter many other situations in which assertions having vacuously true cases will arise. For example, see the MSE question/answers $ f $ is continuous at any isolated points of $ E $. "Vacuously true"? @Harshit Rajput $\endgroup$ Commented May 30, 2023 at 16:26
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    $\begingroup$ @DaveLRenfro 'Vacuously true statements' is the best thing I'm taking from this answer. In my limited experience with mathematics, I have often thought about these 'end cases' but never knew there was a term for them. Thanks for the link. $\endgroup$ Commented May 30, 2023 at 16:36
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    $\begingroup$ @Harshit Rajput: Since this is a teaching oriented Stack Exchange group, I'll mention that my personal feeling about vacuously true cases is that if you think your student (or your intended reader, if writing something) might misunderstand a vacuously true situation, then you should go the extra mile and make the vacuous case explicit. (This is similar to my reasons following "These two formulas can be replaced with the single formula" in this MSE answer.) $\endgroup$ Commented May 30, 2023 at 17:37
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    $\begingroup$ I would also say that understanding vacuous truth is one of the real, tangible, applicable goals of mathematics education. My view is that it's that kind of thinking which can be used outside of mathematics that justifies requiring all students to have competence in mathematics - not trivial tasks like balancing a checkbook. Of course I know that not all students will complete their required math education with understanding of some of these more abstract but more applicable concepts, but that doesn't mean we shouldn't strive to teach them to as many students as possible. $\endgroup$ Commented Jun 1, 2023 at 2:40
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In addition to the very good answers that you've already received, it's probably worthwhile to also mention the following point:

The alternative that you suggest might lead to a similar type of confusion as the more common definition does: every set $B$ is a subset of itself and to see this from your definition, one needs to understand that "removing some elements of $B$" includes the case of not removing any element at all.

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  • $\begingroup$ Love how in mathematics challenges can't be evaded if they exist. $\endgroup$ Commented May 31, 2023 at 13:37
  • $\begingroup$ The phrasing "removing the/some elements of B" does indicate that B is nonempty. I can't decide whether "removing elements of B" is significantly more accurate to our intent, so, echoing Will's comment above, probably best to explicitly say "removing zero or more elements of B". $\endgroup$
    – ryang
    Commented Jun 1, 2023 at 3:32
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This answer is going to have a lot of overlap with Steve Gubkin's, but maybe a bit of a different focus. I agree with him that

(1) The key issue here is understanding how vacuous cases work.

(2) Introducing a notion of "removing" doesn't help, because it introduces a new primitive notion, where set theory is deliberately built on just the primitive $\in$, and because "removing" introduces its own ambiguities.

What I want to say is that there are two goals you might have here:

(a) Teaching students what $\subseteq$ means and

(b) Teaching students how to parse mathematical language.

For goal (a), I think rephrasing the definition in as many ways as you can think of (within time limitations) is great. For some students, "removing" will resonate where the usual language doesn't, and vice versa. Pointing out the vacuous case of the empty set is great here; it is usually good teaching (and writing) to point confusing cases. Just be clear that all of the things you are saying are equivalent, and clarify why they are equivalent.

But I think you also need to think about goal (b). Students need to be able to read standard mathematical texts, and standard texts will assume that the student is comfortable with the operation of mathematical logic, more and more so as you get further into the curriculum. An introductory set theory course is often the place for students to get used to the language of logic. So I don't think you should dodge the issue of finding another phrasing, I think you should take the time to explain why $\forall_{x \in \emptyset} \ P(x)$ is always considered true.

I haven't tried this, but studies on the Wason selection task find that students understand the mathematical meaning of "if" best if it is formulated as a rule to be enforced (e.g. "If you are drinking alcohol, then you must be over 18".) I suspect the same would work with statements of the form $\forall_{x \in \emptyset} P(x)$. For example: "A website has a rule that it will only host videos if all of the humans depicted have given their consent to be filmed. Does a video which shows only cats obey this rule?"

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  • $\begingroup$ How about more broadly saying that some concepts are defined by a demonstrable lack of counter-examples. The only way that a statement like "every X is a Y" can be false is if there exists an X that is not a Y. If there don't exist any X's at all, then the statement "every X is a Y" can't possibly be false, and it would thus be vacuously true. $\endgroup$
    – supercat
    Commented Jun 1, 2023 at 15:51
  • $\begingroup$ @supercat In terms of wording, that seems strange to me. How can "any X is Y" be true if there are no X's? From a logical wording standpoint, I dont quite follow that. You cant find any X that is not a Y, but you cant find any X that is a Y either. Why would it be any more true than false? $\endgroup$
    – JMac
    Commented Jun 1, 2023 at 17:05
  • $\begingroup$ @JMac: The statement "Any X is Y" is the opposite of "No X is Y", which would by default be true unless there exists an X which is a Y". Since the opposite of "Any X is Y" is true by default, "Any X is Y" is false by default." Another way of looking at it would be to observe that for any finite set, the statement "Every X is Y" implies that "The number of X that are Y is equal to the number of X that exist", while "Any X is Y" implies that the number of X that are Y is non-zero. The number of X that must be Y to satisfy the first statement may be less than, equal to, or greater than... $\endgroup$
    – supercat
    Commented Jun 1, 2023 at 17:13
  • $\begingroup$ ...the number required to satisfy the second, depending upon whether the number of X that exist is zero, one, or more than one. $\endgroup$
    – supercat
    Commented Jun 1, 2023 at 17:13
  • $\begingroup$ supercat used the word "every", rather than "any". The word "any" is often confusing, see math.stackexchange.com/a/402073/448 . Putting in the word any, would you agree that "every female US president has reduced the deficit"? I would, and hopefully your students will too -- that's the way "every" works in math. $\endgroup$ Commented Jun 1, 2023 at 17:19
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A potential objection: what does it mean to be able to make a set by removing elements from another set? Some students will think of removing the elements one at a time. Then there'll be misconceptions such as that $\{1, 2, 3\}$ is not a subset of $\mathbb{N}$, because every time you remove an element from $\mathbb{N}$, it is still infinite.

There is a theory that students learn abstract concepts in mathematics by generalising from actions (things you do) to processes (things that can be done but don't have to be) and objects (things that exist and can be reasoned about). For example, a student can compute $x + 12$ when $x = 5$ by substituting and simplifying, or recognise $x + 12$ as a process which transforms $x$ without having to know what $x$ is, or conceptualise $x + 12$ as a function which has certain properties and can be composed with other functions.

The traditional definition of a subset works well in this theory because we can first talk about a procedure for verifying $A \subseteq B$ by checking all of the elements of $A$ to see if they are elements of $B$, but then the definition doesn't say that you have to do this or even that it has to be possible to follow that procedure ─ the traditional definition doesn't rely on the procedure or refer to it at all, it says what a subset is rather than how to verify that something is a subset. On the other hand, the proposed alternative definition talks about a thing you have to do, i.e. remove elements from $A$ until only the elements of $B$ remain. So it's harder to see how to generalise from that definition to a concept of "subset" which doesn't depend on a procedure.

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    $\begingroup$ This is a very important point. Many people including some careless mathematics professors have made this mistake, even in published undergraduate textbooks. For example, you cannot obtain a vector space basis by iteratively removing one dependent vector at a time... It's ridiculously frustrating to see people trying to rephrase mathematical concepts in ways that make it muddier or even bogus. $\endgroup$
    – user21820
    Commented Jun 1, 2023 at 13:31
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The only flaw is that it doesn't really fix the perceived "problem" of the original definition.

What if $A$ and $B$ are both empty? What elements of $B$ are you going to remove to get $A$?

I don't think it sidesteps anything. The same issue just pops up elsewhere.

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'But an empty set doesn't contain any elements which are in any other set, so how can it be a subset of every set?'

Does "If there is no element in A that isn't in B, then A is a subset of B" fit your intuition any better?

Given two sets A and B, if by removing the elements of B we can form A, then A is a subset of B.

"The elements of B" is ambiguous. It could be taken to mean all of the elements. What you really mean is "if there is a subset C of B such that removing C from B results in A, then A is subset of B".

I'll leave it as an exercise for the reader what the problem with that is.

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