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If $\emptyset\neq\{\emptyset\}$, then the result of $\mathbf{A}\times\emptyset$ should differ from the result of $\mathbf{A}\times\{\emptyset\}$.

Does it? What is the result of the latter, if - let's say - $\mathbf{A}=\{1, 2\}$?

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    Not really sure why this was downvoted - it's a common conceptual stumbling block. – 2017-01-19
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    @NoahSchweber I've noticed that voting behaviors have gotten really bad on this site. Some people have started downvoting anything that they think is below them. We may need to address this issue. It makes the site completely toxic for new users. – 2017-01-19
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    Yes. For the first case it is the empty set. For the second case it is $\{(1, \varnothing), (2, \varnothing)\}$. – 2017-01-19

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Yup.

The Cartesian product of $X$ and $Y$ is the set of pairs $(a, b)$ with $a\in X$ and $b\in Y$.

The Cartesian product $A\times\emptyset$ is empty because there are no pairs of that form - $\emptyset$ has no elements!

But $\{\emptyset\}$ does have an element - namely, $\emptyset$. For example, if we take $A=\{1, 2\}$, then $$A\times\{\emptyset\}=\{(1, \emptyset), (2, \emptyset)\}.$$ Remember that the emptyset is not nothing! It just contains nothing. Think about it as an empty bag - it's still a thing that exists (namely, a bag).

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Recall that $X\times Y$ is the set of all pairs $(x,y)$ such that $x\in X$ and $y\in Y$. If $Y=\emptyset$ then no such pairs satisfy this condition, so $A\times\emptyset=\emptyset$. On the other hand, the set $\{\emptyset\}$ has precisely one element, and so $A\times\{\emptyset\}=\{(a,\emptyset)\,:\,a\in A\}$ which is nonempty if $A$ is nonempty, and in fact there is an obvious bijection $A\longrightarrow A\times\{\emptyset\}$.

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Yes, it does differ. Say, for your example, $\{1,2\}\times\{\varnothing\}=\{(1,\varnothing),(2,\varnothing)\}$.

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The number of elements in $\varnothing$ is $0$.

The number of elements in $\{\varnothing\}$ is $1$.

For the cartesian product, the number of elements is the product. So, if $A=\{1,2\}$ has two elements, we see:

$A \times \varnothing$ has $2 \times 0 = 0$ elements, but

$A \times \{\varnothing\}$ has $2 \times 1 = 2$ elements.