# Is $\{\varnothing\}$ a subset of every set $A$?

The following statement appears as a true or false question:

$\{ \varnothing \} \subseteq A$ for all sets $A$.

Because it is the set containing the null set, instead of just the null set itself, I'm hesitant to say whether this is true or false. Can anyone clarify for me?

• possible duplicate of Set notation confusion (Empty Sets) – M.K. Feb 8 '15 at 21:27
• @I.K. Not really a duplicate, even a conceptual duplicate. This question is about the concept of a subset and the other does not mention subsets. – 6005 Feb 8 '15 at 22:21
• @Goos, I don't agree. Even though the title 'Set notation confusion' is misleading, part of that post is dealing with the same concept as this post. – M.K. Feb 8 '15 at 22:48
• @I.K. Are you saying that you think that if part of one question is answered by a second question, the first should be closed as a duplicate of the second? I don't think anyone would agree with that, but if you were to insist I could bring it up on meta. – 6005 Feb 8 '15 at 22:51
• It's the closet I could get. I suppose a link in the comment would have been better. My main reason to alert the OP is that by reading that post he/she could have inferred the answer himself/herself. – M.K. Feb 8 '15 at 22:53

## 4 Answers

It is true that $\varnothing\subseteq A$ for all sets $A$, but as you suspected, $\{\varnothing\}$ and $\varnothing$ are two very different sets, so this says nothing about whether $\{\varnothing\}\subseteq A$ for all sets $A$; that has to be determined separately. Note that $\{\varnothing\}\subseteq A$ if and only if $\varnothing\in A$. This clearly isn’t true for all sets $A$. For example, we know that $x\notin\varnothing$ no matter what $x$ is, so in particular $\varnothing\notin\varnothing$, and therefore $\{\varnothing\}\nsubseteq\varnothing$.

You're rightfully reluctant.

We have $\varnothing\subseteq A$ for all sets $A$, but the set containing the null set as an element, namely, $\{\varnothing\}$ is not a subset of every set.

For example, given $$A = \{1, 4, 9, 16\},$$

• we have $1\in A,\;$ so $\;\{1\} \subseteq A$,
• while we do have $\varnothing \subseteq A,\,$ $\,\varnothing \notin A\,$ so $\,\{\varnothing\} \not\subseteq A$.

So it is not the case that for all sets $A$, $\,\{\varnothing\} \subseteq A$.

Remark: We can construct a set $A$ such that $\{\varnothing\} \subseteq A$. Put $A = \{\varnothing\}$. Then $\varnothing \subseteq A$ AND $\varnothing \in A$. So it follows that $\{\varnothing\} \subseteq A$.

• Thanks @Hugh, for the catch! – Namaste Feb 9 '15 at 14:28
• But if ${\bf A}_0 = \left\{ \emptyset \right\}$ and ${\bf A} = \left\{ {1,4,9,16} \right\}$, then what is ${\bf A}_0 \cup {\bf A}$ ? and does $\left\{ \emptyset \right\} \subseteq \left( {{\bf A}_0 \cup {\bf A}} \right)$ ? – G Cab Jun 19 '16 at 0:56

well if you set $A = \emptyset$ the statement is not true since cardinality of $A$ is zero, whereas the cardinality of $\{\emptyset\}$ is $1$

$\{\}$ is not the same as $\{\varnothing\}$.

$\{\}$ is the empty set; it contains no elements. Another way of writing an empty set is $\varnothing$.

However, $\{\varnothing\}$ is not empty; it contains the empty set. Therefore $\{\varnothing\}$ is not an empty set.