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?
 A: 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 $\{\varnothing\}\nsubseteq\varnothing$.
A: 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 $\{\varnothing\} \subseteq A$.
A: well if you set $A = \emptyset$ the statement is not true since cardinality of $A$ is zero, whereas the cardinality of $\{\emptyset\}$ is  $1$
A: $\{\}$ 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.
