Ever since day one of of my Mathematical Logic course, this fact has really bothered me. I cannot wrap my head around how an empty set is a subset of every possible set. Could someone kindly explain how this is true? Any help is appreciated!

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    $\begingroup$ To see that the null set is a subset of a set with some members in it is not difficult, I think. If you want to think of it in terms of plain English this could say that every set with some elements in is larger than the empty set. Hence the empty set is a subset. The fact that the empty set is a subset of the empty set is vacuously true $\endgroup$ – Keeran Brabazon Oct 16 '13 at 15:44
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    $\begingroup$ Think: can you come up with a set $A$ such that there exists $x\in \emptyset$ but $x\notin A$? That's what's necessary to show that $\emptyset\nsubseteq A$. $\endgroup$ – rschwieb Oct 16 '13 at 15:45
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    $\begingroup$ @Keeran: "Largeness" is misleading in this context. The set $\{1,2\}$ is "larger" than $\{0\},$ but that doesn't make $\{0\}$ a subset of $\{1,2\}$. $\endgroup$ – Cameron Buie Oct 16 '13 at 15:55
  • $\begingroup$ @CameronBuie I realise that this is the case. I was trying to put this into more accessible language. It may not be the best analogy, though $\endgroup$ – Keeran Brabazon Oct 16 '13 at 15:57
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    $\begingroup$ Be aware that there are two popular meanings for "null set", namely "an empty set" and "a set of measure zero". Using the second terminology the empty set is of measure zero, but there are sets of measure zero which are not empty. $\endgroup$ – dtldarek Oct 16 '13 at 16:09

If you're comfortable with proof by contrapositive, then you may prefer to prove that for any set $A,$ if $x\notin A,$ then $x\notin\emptyset$. But of course, $x\notin\emptyset$ is trivial since $\emptyset$ has no elements at all. Hence, $x\notin A\implies x\notin\emptyset,$ so by contrapositive, $x\in\emptyset\implies x\in A,$ meaning $\emptyset\subseteq A$.

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    $\begingroup$ Nice point. I'd never noticed before that accepting "vacuous truth" is essential if you want the law of the contrapositive to work correctly. $\endgroup$ – Nate Eldredge Oct 16 '13 at 17:24
  • $\begingroup$ I only noticed it recently, myself. $\endgroup$ – Cameron Buie Oct 16 '13 at 17:26

By definition, $A$ is a subset of $B$ if every element of $A$ is in $B$.

If we set $A=\emptyset$, then the above statement is vacuously true. Every element of $A$ is in fact an element of $B$ since the former has no elements.

  • $\begingroup$ I understand the conclusions of implication but can you clarify how a statement is vacuously true if the antecedent is False? $\endgroup$ – wonggr Oct 16 '13 at 17:35
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    $\begingroup$ "A conditional statement is called vacuously true if its antecedent is false". $\endgroup$ – The Chaz 2.0 Oct 16 '13 at 17:37
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    $\begingroup$ I might add that the conditional statement of import is, "for all $x$, if $ x \in A$ then $x \in B$. Since nothing is in $A$, ... $\endgroup$ – The Chaz 2.0 Oct 16 '13 at 17:38

A remarkably simple way of trying to grasp why that is the case is as follows:

We can consider any set and throw away all its elements, then we are left with the subset { }. This means that { } is a subset of any set.


I'm practicing my set proving skills, so shoot me down if I'm wrong.

I'm going to use proof contradiction as an alternative approach since that approach has not been mentioned as answer in this topic.

So we want to prove:

${ \emptyset \subseteq A }$

In other words

${\forall_x [(x \in \emptyset) \to (x \in A)]}$ def. subset

So since proof by contradiction is used, lets try to prove the opposite and hope to run into a contradiction.

${\lnot \forall_x [(x \in \emptyset) \to (x \in A)]}$

After moving the negation inside, we get

${\exists_x[(x \in \emptyset) \land \lnot(x \in A)]}$

So we want to demonstrate there exists an element of the empty set that isn't in A.

since ${x \in \emptyset}$ will always be false we get

${\exists_x[F \land \lnot(x \in A)]}$

${\exists_x[F]}$ domination law

So there doesn't exist such an element. And this contradicts with our assumption that such an element exists. Therefor we have proven ${ \emptyset \subseteq A }$


I am quite comfortable with the other two answers but there is a softer way to answer this question. I think the following can be made formal using the Axiom of Choice (perhaps in conjunction with a function $Y\rightarrow \{0,1\}$), but if you really want to be that formal just use one of the two perfectly good arguments above.

This soft approach is as follows. Where $X$ and $Y$ are sets, a subset of $X\subseteq Y$ corresponds to a choice of elements from $Y$.

For example, where $Y=\{1,2,3,4\}$, a subset $X$ is formed by answering the questions

$$1\in Y?,\,2\in Y?,\,3\in Y?,\,4\in Y?$$

For example, the subset $\{1,4\}\subset Y$ corresponds to choosing one and four but not two and three.

If we choose all of the elements of $Y$ we have the full subset and so, as it consists of a choice of elements of $Y$ --- namely all of them --- we have that

$$Y\subseteq Y.$$

Now what if we choose none of the elements of $Y$? This is a choice of elements and so is a subset of $Y$. It is of course the empty set and in this sense we have $$\emptyset\subset Y.$$ So a subset corresponds to a choice of elements from $Y$, choosing none is a choice, and therefore the empty set is always a subset.

You will need some of the more formal arguments from above to understand why $\emptyset\subseteq\emptyset$.

  • $\begingroup$ The downvoter care to explain the downvote? $\endgroup$ – JP McCarthy Nov 2 '17 at 16:15
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    $\begingroup$ As a side note, the Axiom of Choice has nothing whatever to do with this. AC says (roughly) that, given any infinite number of non-empty sets, there is a function choosing a single element from each of the sets. Your approach seems to have more to do with construction of indicator functions. $\endgroup$ – Cameron Buie Feb 3 '18 at 13:45

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