Set theory. Is $\varnothing \in \{1,2,3\}$ true? Let $A = \{1, 2, 3\}$.
Is the following expression true then? 
$\varnothing \in A$.
I'm having issue understanding if the empty set is an element of $A$. Would love a brief explanation of why it is or isn't.
Thanks a lot.
 A: No it is not an element of $A$. It is a subset of $A$, which we denoted $\varnothing \subset A$.
A: I'm not positively certain someone proved $\emptyset \subseteq X$:  
Let there be a set $X$ such that $\emptyset \nsubseteq X$. So there exist $x \in \emptyset$ such that $x \notin X$. Wait, what?
QED ;-)
A: $A$ is a set with three elements, namely $1,2$, and $3$.  Since $\emptyset$ is none of these three, $\emptyset\notin A$.
It is indeed true that $\emptyset \subseteq A$.  This means that each element of $\emptyset$ is also an element of $A$.  Since $\emptyset$ has no elements, this statement is vacuously true, i.e. $\emptyset\subseteq S$ for all sets $S$.  In fact even $\emptyset\subseteq\emptyset$.
A: You must take care to the definitions.
For every set $A = \{ a_1, a_2, ...a_n, ... \}$ its elements are the "things" between the braces.
The emptyset is defined as $\forall x (x \notin \emptyset)$, and is denoted also as $\{ \}$, in order to show that it has no elements.
The definition of $A \subseteq B$ is 

$$\forall x (x \in A \rightarrow x \in B)$$

i.e.

"every object that is an element of $A$ is also an element of $B$".

Now, if you apply this condition to $\emptyset \subseteq B$, by properties of conditional ($\rightarrow$) you may check that it is always satisfied, for every $B$, because of the above definition of $\emptyset$ : no $x$ is in $\emptyset$ and so ($P \rightarrow Q$ is true when $P$ is false) $\emptyset$ is a subset of every set.

But $x \in B$ is a different relation from $A \subseteq B$.

In your example, $1$ and $2$ and $3$ are the elements of $A$; so we have that $B = \{ 1,2 \}$ is a subset of $A$.
But $\emptyset$ is not "listed" between the braces, so it is not an element of $A$.
Of course, being a subset of every set, we have $\emptyset \subseteq A$.
