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I need help in proving this problem. The problem I had with the proof is that I've assumed that there is an element in $A$, such as:

$a \in A$.

Then $(a,a) \in A×A$

But from here I cannot just assume that $A×A$ is reflexive over $A$ or symmetric or transitive because I assumed that there is only 1 element in A, what happens if there are more?

I need a bit of help in the proof, thanks!

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    You want to prove that $A\times A $ is a relation with those properties? Not $R $? – 2017-01-11
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    What is this notion? $A(R\subseteq A\times A)$? What does it mean? And what do you mean by "$A\times A$ is reflexive over $A$"? – 2017-01-11
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    Sorry messed up, I meant A, not R. Now it should be ok. – 2017-01-11
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    $A \subseteq A \times A$ doesn't look right. – 2017-01-11
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    $A$ cannot be any sort of relation over itself? – 2017-01-11
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    Sorry, I meant $A×A \subseteq A$. – 2017-01-11
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    $A \times A \subseteq A$ doesn't look right either. – 2017-01-11
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    @NickS.That does not make sense. $A \times A$ is a set of ordered pairs and cannot be contained in the original set. – 2017-01-11
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    Why not? $A×A$ is a relation over $A$... – 2017-01-11
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    Yes, $A \times A$ is a relation over $A$, but $A \times A \subseteq A$ doesn't mean that, as pointed out by @rt6. – 2017-01-11
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    Oh ok I get it now, but still, Let's assume there is just a group A, it isn't over anything, can I prove it then? – 2017-01-11
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    Yes, apply the definitions of reflexivity, symmetry, and transitivity. All will work like a charm. $A \times A$ is the equivalence relation on $A$ with just one equivalence class: every member of $A$ is equivalent to all members of $A$. It also works for empty $A$. – 2017-01-11
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    @NickS. Yes $A\times A$ is indeed a relation over $A$, but that is very different from containment. Let's take $A=\{1,2\}$.Then $A\times A=\{(1,1), (1,2),(2,1),(2,2)\}$ Do you see how stating $A\times A\subseteq A$ does not make sense? – 2017-01-11
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    Ok, I see the difference now, thanks. But then comes the question, how come the statement $A×A$ is reflexive over $A$, symmetric and transitive for all group $A$? – 2017-01-11
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    Usually, we are considering some relation $R$ which is *contained* in $A\times A$ and we need to check that the conditions of reflexivity, symmetry and transitivity hold true. In the case of the relation $A\times A$, if you look at the definitions of reflexivity, etc., you will see $A\times A$ must satisfy these: since ALL possible ordered pairs made from elements in $A$ are included. – 2017-01-11

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After you edited your question:

$R = A \times A$;

Let us assume $a, b, c \in A$;

  • $a \in A \implies (a, a) \in A\times A \implies A\times A$ is reflexive;
  • $a, b \in A \implies (a, b) \in A\times A$, but swapping $a$ with $b$ yields $(b, a) \in A\times A$ and therefore $A \times A$ is symmetric;
  • $a, b, c \in A \implies (a, b), (b, c) \in A\times A$, but also $a, c \in A \implies (a, c) \in A\times A$ and therefore $A\times A$ is transitive.
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    Sorry, I meant A, not R. Check the edit now ;/ – 2017-01-11
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    @NickS. no problem, I edited my answer now. – 2017-01-11
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    Thanks a lot, after reading your proof, can I now tell at the end of the proof that the proof holds on for every size of $A$? I mean, can you do the same way with $a,b,c,d \in A$? – 2017-01-11
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    @NickS. you do not need the extra $d$ given that $a, b, c$ were **any** elements of $A$. For my proof to hold, I need only that $A \not= \emptyset$ – 2017-01-11
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$A \times A$ is simply all possible pairings $$ where $x \in A$ and $y \in A$, and hence it is reflexive, symmetrical, and transitive:

Reflexive: Take any $a \in A$. Since for any $x,y \in A: \in A \times A$, we can take $x = y = a$, and thus $ \in A \times A$, so yes, it's reflexive.

Symmetrical: Take any $a,b \in A$. Assume $ \in A \times A$. Since for any $x,y \in A: \in A \times A$, we can take $x = b$ and $y = a$, and so $ \in A \times A$, so yes, it's symmetrical.

Transitive: Take any $a,b,c \in A$. Assume $ \in A \times A$ and $ \in A \times A$. Since for any $x,y \in A, \in A \times A$, we can take $x = a$ and $y = c$, and so $ \in A \times A$, so yes, it's transitive.

This works for any size set $A$, even when $A$ is empty, since all universal statements (which is what the claims of refleixivity, symmetry, and transitivity are) are vacuously true over any empty domain.

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    Yeah I know, but is there any way I can write it mathematically? – 2017-01-11