Is $\varnothing\in\{\varnothing\}$ and $\varnothing\subseteq \{\varnothing\}?$ Is $\varnothing\in\{\varnothing\}$ and $\varnothing\subseteq \{\varnothing\}?$
As per me yes they are true because, in the first statement, the set $\{\varnothing\}$  contains one single element, $\varnothing$, the empty set.
In the second statement because of the fact that an empty set is a subset of every set
What about $\{\varnothing\}\subseteq \varnothing$ ?
 A: The only subset of $\emptyset$ is $\emptyset$. If $A\subseteq\emptyset$ and $A$ has an element $a$, then $a$ must also be an element of $\emptyset$.
A: The empty set is a subset of every set. Hence $\varnothing \subseteq \{\varnothing\}=:S$. The only element of $S$ is $\varnothing$ by inspection. Hence $\varnothing\in S$. Note that $S$ has one element whereas $\varnothing$ has none; hence $S\not\subseteq \varnothing$, since, otherwise, $S=\varnothing$.
A: Use $\{\}$ instead of $\emptyset$ and everything will become clear.

*

*$\{\} \in \{\{\}\}$ true

*$\{\} \in \{\}$ false

*$\{\} \subseteq \{\}$ true

*$\{\} \subseteq \{\{\}\}$ true

*$\{\{\}\} \subseteq \{\}$ false

A: Yes, by definition $\varnothing\in\{\varnothing\}$—the notation $\{\varnothing\}$ represents the set that contains the empty set (and no other element). And it is also true that $\varnothing\subseteq\{\varnothing\}$. This is because $A\subseteq B$ is just an abbreviation of
$$
\forall x:x\in A\implies x\in B \, .
$$
Therefore, the claim that $\varnothing\subseteq\{\varnothing\}$, by definition means
$$
\forall x:x\in\varnothing\implies x\in\{\varnothing\} \, .
$$
Recall that an implication $p\implies q$ is always satisfied when $p$ is false (see here). Since the empty set is, well, empty, there is no value of $x$ for which $x\in\varnothing$ is true. Hence, the claim$$\forall x:x\in\varnothing\implies x\in\{\varnothing\}$$ follows immediately. This is known as a vacuous truth. Note that exactly the same argument can be used to prove that the empty set is a subset of any given set.
As for $\{\varnothing\}\subseteq \varnothing$, this claim is not true. To prove this, consider that the negation of $$\forall x:x\in\{\varnothing\}\implies x\in\varnothing$$ is $$\exists x:x\in\{\varnothing\}\land x\notin\varnothing \, .$$The above statement is true because $\varnothing\in\{\varnothing\}$ and $\varnothing\not\in\varnothing$.
