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It is said that null set is a subset of every set.but the definition for subset says that "if every elements of A is also an elements of B then $A \subset C$ ".

Therefore is it mean every set contains null element as its element? How is it possible a set contain void set as it subset?

Sathasivam K
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    It's not true that every set contains $\varnothing$ as an element. However, $\varnothing$ is a subset of every $A$, because (vacuously) every element of $\varnothing$ is in $A$. – Wojowu Jul 31 '16 at 15:56
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    NO! Every ELEMENT of the subset is in the set. The subset itself is NOT an element of the subset. Every element of the void set IS in every set because the void set doesn't have any elements at all. So every element, all zero of them, is in every set. (They are also all sitting in my bathtub at this moment.) – fleablood Jul 31 '16 at 15:56
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    The void set is NOT an element of the void set. So the void set does not need to be an element of every set. But every single $x \in \emptyset$ is an element of every set. How is that possible? Because there are NO $x \in \emptyset$. This is a little bit like saying "Every single methane breathing lion on earth is in my living room right now drinking tea". How is that possible? Because there are no methane breathing lions. – fleablood Jul 31 '16 at 16:01
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    Why the rude and discouraging downvote? This is a very legitimate and for students very confusing problem. The fact that we all figured it out years ago and get tired of explaining it again and again is no reason to downvote a new student's sincere concern. – fleablood Jul 31 '16 at 16:05
  • Very thanks @ fleablood – Sathasivam K Jul 31 '16 at 16:50
  • @fleablood: If we all got tired of explaining this, this means that this should really be closed as a duplicate, rather than receive four answers. In the name of procrastination, I'll see if I can find a suitable candidate. – Asaf Karagila Jul 31 '16 at 17:00
  • I mean, off the bat, the first link in the "Related" is http://math.stackexchange.com/questions/719554/how-is-it-possible-for-a-singleton-to-exist-if-∅-is-a-subset-of-every-set which seems somewhat fitting, but let's not stop there. – Asaf Karagila Jul 31 '16 at 17:01
  • Also, just a another few clicks away, http://math.stackexchange.com/questions/1218176/does-every-subset-start-with-an-empty-set?noredirect=1 and http://math.stackexchange.com/questions/631042/direct-proof-of-empty-set-being-subset-of-every-set – Asaf Karagila Jul 31 '16 at 17:03
  • And of course, http://math.stackexchange.com/questions/1103664/is-empty-set-element-of-every-set-if-it-is-subset-of-every-set and http://math.stackexchange.com/questions/656331/why-is-the-empty-set-a-subset-of-every-set and after finding five different threads that could, potentially, serve as a duplicate target, I'll stop. Please let me know what do you think. Or maybe I should post another answer here instead? – Asaf Karagila Jul 31 '16 at 17:05
  • (Oh, yeah, I should also add that all these links came from "Related" or "Related" or "Related". They were in a two-clicks distance for anyone who checked this question; they are not difficult to find either. I expect people to post questions after checking if there are reasonable duplicates before, and that means that if a question is so basic that people will be tired of answering it, then it is almost certainly a duplicate six years into this website.) – Asaf Karagila Jul 31 '16 at 17:08
  • In the whopping two lines that you wrote, I don't see how. Sorry. (Also because the answers in the linked questions do answer you question how does the empty set is a subset of every set, and how it is not an element of every set). – Asaf Karagila Jul 31 '16 at 17:46
  • "My question is different from what you are linked.you need to check" Actually, in this case, I don't see how your question is different. If Math had a frequently asked question, this exactly how you wrote it would be one of the top.... " If we all got tired of explaining this, this means that this should really be closed as a duplicate, rather than receive four answers." ... perhaps. But I see it more as a struggling student sincerely confused and not knowing how to find an answer. We should point out duplicate if we know where they are. .. Or try to explain if we can and have the patients – fleablood Jul 31 '16 at 20:50
  • I guess I understand the mechanics of this not being a tutoring site and the need for one definitive set of questions and answers... but sometimes the mechanics of it seem so.... rude. – fleablood Jul 31 '16 at 20:52

4 Answers4

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Imagine doing an checklist to determine if a set is a subset of another:

Case one: Is {fred, wilma, dino} a subset of {barney, dino, Mr. Slate, wilma, betty, fred}.

Okay, is fred on the list? Check. Is wilma on the list? Check. Is dino? Check. Okay it's a subset.

Case two: is {bananas, milk, peaches} a subset of {peaches, oranges, bananas, grapes}?

Okay, is bananas on the list? Check. Is milk? Nope. then it's not a subset.

Case three: is {} a subset of {green, red, yellow}?

Um... Well, go on... what's the first item on the list?... Um, there isn't any... Well, is everything on your list in my list? Well, there isn't anything on my list... Then is there anything on your list that isn't on my list?.. Um, there's nothing on my list... Well, then everything on your list is on my list, isn't it. After all, there's nothing on your list that isn't on my list, right? ... Um, I guess so.... So it's a subset--- a void, empty, trivial subset but a subset, none-the-less.

fleablood
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  • +1. I also like phrasing this in terms of looking for a counterexample: "Is ${}$ a subset of ${$green, red, yellow$}$?" "Well, in order to not be a subset, it would have to have some element which is not in ${$green, red, yellow$}$. Is there such an element of the emptyset?" "Well, there aren't any elements of the emptyset at all!" "Well there you go." – Noah Schweber Jul 31 '16 at 17:01
  • Thanks sirsir@flea blood. you answer looks awesome – Sathasivam K Jul 31 '16 at 17:24
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Definition: $A\subseteq B$ if $x\in A$ implies $x\in B$

$x\in \emptyset$ is false for any $x$, so the statement "$x\in \emptyset$ implies $x\in B$" is true for any set $B$.

Alternatively:

Definition: $A\subseteq B$ if every element of $A$ is in $B$

Since there is nothing in the empty set, every element of the empty set (i.e. nothing) is in every other set.

smcc
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By definition, $A\subseteq B$ if and only if $x\in A \Rightarrow x\in B$, which is equivalent to $x\notin B \Rightarrow x\notin A$. If $A=\emptyset$ then this last implication holds for any set $B$.

Lorenzo Stella
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Here is an intuitive way to think about it.

1) $\subseteq$ is a transitive relation

2) Throwing away an element from a set gives you a subset of the original set.

Start with $A = \{a,b,c\}$

Throw away $c$. By 2) you get a subset $\{a,b\}$ of $A$

Now throw away $b$. By 2) you get a subset $\{a\}$ of $\{a,b\}$ which by 1) is a subset of $A$.

Now throw away $a$. There is no reason to think that 2) all of a sudden breaks down. Hence you get a subset $\emptyset = \{\}$ of $\{a\}$ (hence of $A$). Thus the empty set is a subset of $A$.

The intuition is that subsets can be obtained by throwing away elements, and this is true even if you throw away all elements.

A related consideration involves intersections. It is easy to see that if $A$ and $B$ are any two sets then $A \cap B \subseteq A$, since any elements that are in both $A$ and $B$ are obviously in $A$. But this is true even when $A$ and $B$ are disjoint, in which case $A \cap B \subseteq A$ reduces to $\emptyset \subseteq A$.

John Coleman
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