Take for example the set $X=\{a, b\}$. I don't see $\emptyset$ anywhere in $X$, so how can it be a subset?

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    $\begingroup$ "Subset of" means something different than "element of". Note $\{a\}$ is also a subset of $X$, despite $\{ a \}$ not appearing "in" $X$. $\endgroup$
    – user14972
    Jan 29, 2014 at 20:10
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    $\begingroup$ Why does this question get a downvote? It is not hard to see that someone can be asking this seriously. $\endgroup$
    – N. Owad
    Jan 29, 2014 at 22:02
  • $\begingroup$ Hint: Every element of the empty set is a pink elephant. Or an element of $X.$ (No joke) $\endgroup$ Jan 30, 2014 at 4:59
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    $\begingroup$ I personally like @HagenvonEitzen's question Or Can you name an element of ∅ that is not an element of X?. If you think it like this, given ∅ and X, you can't really find an element of ∅ (nothing) that you don't find in X, and as a subset A is a just a set whose elements (every element) are included in another set B, that is you can't find an element in A which is not in B, it makes more sense. $\endgroup$ Mar 15, 2018 at 20:24

4 Answers 4


Because every single element of $\emptyset$ is also an element of $X$. Or can you name an element of $\emptyset$ that is not an element of $X$?

  • $\begingroup$ But there are no elements in ∅, so how can every single element of ∅ be an element of X? You ask if we can name an element of ∅ that is not an element of X-- the answer is I can't, but conversely you can't name a single element of ∅ that IS an element of X. So how can we derive such a conclusion that the empty set is a subset of every set? It seems to me that the quality of being a sub-set cannot apply in this scenario because the definition assumes the existence of at least one element inside the set $\endgroup$
    – RandomUser
    Apr 27, 2022 at 18:51
  • $\begingroup$ @RandomUser vacuous truth. $\endgroup$
    – L. F.
    Jul 19, 2022 at 7:34

that's because there are statements that are vacuously true. $Y\subseteq X$ means for all $y\in Y$, we have $y\in X$. Now is it true that for all $y\in \emptyset $, we have $y\in X$? Yes, the statement is vacuously true, since you can't pick any $y\in\emptyset$.


You must start from the definition :

$Y \subseteq X$ iff $\forall x (x \in Y \rightarrow x \in X)$.

Then you "check" this definition with $\emptyset$ in place of $Y$ :

$\emptyset \subseteq X$ iff $\forall x (x \in \emptyset \rightarrow x \in X)$.

Now you must use the truth-table definition of $\rightarrow$ ; you have that :

"if $p$ is false, then $p \rightarrow q$ is true", for $q$ whatever;

so, due to the fact that :

$x \in \emptyset$

is not true, for every $x$, the above truth-definition of $\rightarrow$ gives us that :

"for all $x$, $x \in \emptyset \rightarrow x \in X$ is true", for $X$ whatever.

This is the reason why the emptyset ($\emptyset$) is a subset of every set $X$.

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    $\begingroup$ Shouldn't the last implication be "$\text{for all x, }x \in \emptyset \rightarrow x \in X$ is true" $\endgroup$
    – mauna
    Jun 30, 2014 at 13:20

Subsets are not necessarily elements. The elements of $\{a,b\}$ are $a$ and $b$. But $\in$ and $\subseteq$ are different things.


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