For every set A, the empty set is a subset of A. The empty set is a set. Therefore, the empty set has a cardinality >= 1..

I have only recently been exposed to sets. According to wikipedia, as seen on the first bullet mark of the link, $\forall A: \emptyset \subseteq A$

Does this mean that $\emptyset$ is an element of all sets? (This is False, thank you those that answered)

Is the empty set also a set itself?

Assuming these statements are true, then the empty set therefore an element of the empty set. This does not sound right, please clarify for me. Thank you.

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Being a subset and being an element are different. For every set $A$, $\varnothing\subseteq A$ but not necessarily $\varnothing\in A$. –  Clayton Jan 8 '13 at 4:05
So an element is something without the {} curly brackets. And since $\emptyset$ has such brackets, it does not fit. –  Leonardo Jan 8 '13 at 4:07
Well, something inside of { } means we're thinking of it as a set, that does not mean that something without the brackets is an element. By the looks of it, the notation seems to be the barrier here. –  Andrew D Jan 8 '13 at 4:10
–  Asaf Karagila Jan 9 '13 at 0:55

The empty set is indeed a set (the set of no elements) and it is a subset of every set, including itself. $$\forall A: \emptyset \subseteq A,\;\text{ including if}\;\; A =\emptyset: \;\emptyset \subseteq \emptyset$$

$$\text{BUT:}\quad\emptyset \notin \emptyset \;\text{ (since the empty set, by definition, has no elements!)}$$

That is, being a subset of a set is NOT the same as being an element of a set: $$\quad\subseteq\;\, \neq \;\,\in: \;\; (\emptyset \subseteq \emptyset), \;\;(\emptyset \notin \emptyset).$$

$\emptyset \;\subseteq \;\{1, 2, 3, 4, 5\},\quad$ whereas $\;\;\emptyset \;\notin \;\{1, 2, 3, 4, 5\},\;$.

$\{3\} \subseteq \{1, 2, 3, 4, 5\},\quad$ whereas $\;\;3 \nsubseteq \{1, 2, 3, 4, 5\}, \text{... but}\; 3 \in \{1, 2, 3, 4, 5\}$.

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Yes that makes sense because $\emptyset = \emptyset$ –  Leonardo Jan 8 '13 at 4:12
Leonardo, Indeed, you are right; do you see the difference between being an element of and being a subset of? –  amWhy Jan 8 '13 at 4:17
I understand now, thank you. –  Leonardo Jan 8 '13 at 4:27
Your very welcome. It takes a little getting used to the distinction! –  amWhy Jan 8 '13 at 4:29

Being a subset and being an element are different. For every set $A$, $\varnothing\subseteq A$, but not necessarily $\varnothing\in A$. As an example, we can consider $$A=\left\{1,2,3,\{1,2\}\right\}\text{ and }B=\{1,2,3\}.$$ Then $\{1,2\}\in A$ AND $\{1,2\}\subseteq A$ whereas $\{1,2\}\subseteq B$ but $\{1,2\}\notin B$. Hopefully this clarifies that being an element and a being a subset are different things.

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That makes a bit more sense too.. according to your set definitions $\emptyset \notin A$, but, $\emptyset \subseteq A$, and also I think $\emptyset \subset A$ –  Leonardo Jan 8 '13 at 4:18
@Leonardo: If $\subset$ means proper containment, then $\emptyset\subset A$ is equivalent to $A\neq\emptyset$. I.e., $\emptyset\subset\emptyset$ is false with that convention, but that is the only exception. (You are probably talking about the explicit example $A$ given above, but I wanted to be explicit because $A$ is also quantified at the beginning of the answer.) –  Jonas Meyer Jan 8 '13 at 4:21
That is only because the empty set is the empty set. But A is not the empty set. Yes the empty set is not a proper subset of itself, I agree. For it would need to have an element of itself which itself does not have, which it has none to begin with. –  Leonardo Jan 8 '13 at 4:23
$\varnothing$ is a subset of $\varnothing$, but $\varnothing$ is a not an element of $\varnothing$, because $\varnothing$ has no elements by definition.
For every set $A$, there is no element of $\varnothing$ that is not in $A$, and therefore $\varnothing\subseteq A$. This is true whether or not $A$ is empty.
For an example of a nonempty set that doesn't have the empty set as an element, consider the set $A=\{\{1\}\}$. That is, $A$ is the set whose only element is the set $\{1\}$. Because $\{1\}$ contains the element $1$, it is not empty, $\{1\}\neq\varnothing$. Because the only element of $A$ is not $\varnothing$, $\varnothing\not\in A$. The set $\{\{\varnothing\}\}$ also does not have the empty set as an element for the same reason. On the other hand, the set $\{\varnothing\}$ does have the empty set as an element. By definition, $\varnothing$ is the only element of the set $\{\varnothing\}$.
Yes there is no element of $\emptyset$ that is not in $A$ because it has none. –  Leonardo Jan 8 '13 at 4:20
@Leonardo: That's right, and the only way for a set $B$ to not be a subset of a set $A$ is for $B$ to have some element that is not in $A$. –  Jonas Meyer Jan 8 '13 at 4:23