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Let $X_1$ be a set. What is $X_1 \times \{ \emptyset \}$? I know that a the product of a set and an empty set is an empty set, but what is the product of a set and an empty set WITHIN a set?

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    $\begingroup$ It is just the set $\{(a,\emptyset)\mid a\in X_1\}$. This set is of course empty if $X_1$ is. In any case, there is an obvious bijection between it and $X_1$. $\endgroup$ – Andrés E. Caicedo Mar 24 '14 at 5:43
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The empty set, considered as an element of another set, is nothing special! It is just an element. You could do the problem by finding $$X_1\times\{\,a\,\}$$ and then replacing $a$ by $\varnothing$.

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As a clarifying example, take $$ \{\emptyset\}\times\{\emptyset\} $$ It's the product of two sets with one element, and it's exactly $$ \{(\emptyset, \emptyset)\} $$ with the standard representation of pairs ($(x,y) = \{\{x\},\{x,y\}\}$), it becomes $$ \{ \{\{\emptyset\},\{\emptyset,\emptyset\}\} \} $$ which is the same as $$ \{ \{\{\emptyset\}\} \} $$

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Still empty. Cartesian product does not care about where your set lives. It's the set of all ordered pairs taken from the two sets, and when one of them is empty, there are no such pairs.

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    $\begingroup$ The second set is $\{\emptyset\}$, which is not empty. $\endgroup$ – Andrés E. Caicedo Mar 24 '14 at 5:45
  • $\begingroup$ I interpreted the question to be about the empty set as a subset of some other set, but the notation indicates otherwise, on second viewing. $\endgroup$ – Fred Byrd Mar 24 '14 at 5:52

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