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

Anonymous - a group
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    "Subset of" means something different than "element of". Note ${a}$ is also a subset of $X$, despite ${ a }$ not appearing "in" $X$. –  Jan 29 '14 at 20:10
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    Why does this question get a downvote? It is not hard to see that someone can be asking this seriously. – N. Owad Jan 29 '14 at 22:02
  • Hint: Every element of the empty set is a pink elephant. Or an element of $X.$ (No joke) – Dan Christensen Jan 30 '14 at 04:59
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    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. – user3019105 Mar 15 '18 at 20:24

4 Answers4

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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 $X$?

Hagen von Eitzen
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  • 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 – RandomUser Apr 27 '22 at 18:51
  • @RandomUser vacuous truth. – L. F. Jul 19 '22 at 07:34
  • This is called proof by contradiction. If you can't prove it directly, then prove that is contradiction. – tbhaxor May 30 '25 at 05:25
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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 $y\in\emptyset$.

mathemagician
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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 $X$.

Mauro ALLEGRANZA
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    Shouldn't the last implication be "$\text{for all x, }x \in \emptyset \rightarrow x \in X$ is true" – mauna Jun 30 '14 at 13:20
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Subsets are not necessarily elements. The elements of $\{a,b\}$ are $a$ and $b$. But $\in$ and $\subseteq$ are different things.

Asaf Karagila
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