Given the Cartesian product $X = A \times A$, is $A \subset X$?
$\endgroup$ 03 Answers
$\begingroup$An answer highly depends on the context and the meaning of the question.
The most probable one would be the context of axiomatic set theory, and the answer depends on your definition of an ordered pair, since, formally, $A\times A$ is the set of ordered pairs $(a_1, a_2)$ for all $a_1, a_2 \in A$. With Kuratowski definition of ordered pair, the statement $\forall A: A \not\subset A\times A$ is related to well-foundedness of set theory.
If some semantic context is the case and we think of sets as "types of objects", then $A\times A$ is the type of pairs, and we usually think that objects are not the same as pairs of objects, and the statement is clearly false.
If we consider "$A$ is a subset of $B$" as a somewhat broader notion, namely "$A$ can be embedded into $B$", then $A$ is definitelly a subset of $A\times A$ in a large number of ways: $x \mapsto (x,a)$ for a fixed $a\in A$, or $x \mapsto (a,x)$, or $x \mapsto (x,x)$, and, in general, many more.
I wish to emphasis that only the first answer makes sense if the question is about usual set theory, but I feel that thinking about it in some broader sense is quite useful.
$\endgroup$ 5 $\begingroup$If $A=\emptyset$, then indeed $A\subset X$. Otherwise, there is a copy of $A$ inside $X$ for every element of $A$, but the set $A$ itself is not necessarily exactly a subset of $X$. For every $a\in A$ we have $A\times\{a\}\subset X$ and $\{a\}\times A\subset X$. There is an obvious bijection between $A$ and $A\times\{a\}$.
A simple example for which it does not hold is $A=\{a\}$ with $a\neq\emptyset$. Then $X=\{(a,a)\}$ and $a\neq(a,a)$, so $A\not\subset X$.
$\endgroup$ 4 $\begingroup$Unfortunately its not.
Let $m \in A$. Then , $(m,m) \in A \times A=X$.
Now, What is the definition of $A \subset B$? --> Its $x \in A \Longrightarrow x \in B$.
Let your assumption $A \subset X$ is true, then $x \in A \Longrightarrow x \in X$
But we know $m \in A$ and $(m,m) \in X$. So this becomes $m \in A \Longrightarrow m \in X$. But we know $(m,m) \in X$ so this is clear contradiction. Hence your assumption is wrong.
Here I don't used Quantifiers, because it needs a bit higher knowledge of logic, and you are saying I am not very good at set theory, so...
