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Prove or refute:

  • If R is reflexive than all subsets of R are reflexive.

This my solution:

Let {1,2,3) in R --> We know that R is reflexive -> R = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),*(3,3)}. We also know that R subset means: R is a set which all the elements are contained in another set, which means there are also reflexive.

Not sure if correct or totally wrong, please need help. Thank you very much.

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    $\begingroup$ It appears that you don't know what "reflexive" means. You might want to learn that first: it might be helpful in proving that other sets are reflexive :). $\endgroup$ – Erick Wong Sep 25 '15 at 16:31
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I am afraid you are wrong. The answer is no. Counterexample is:

set $X = \mathbb{N}$, and $R\subseteq P(X\times X)$

$R = \lbrace(1,1), (1,2), (2,1), (2,2)\rbrace$.

Of course $R$ is reflexive, because $xRx$ holds for every $x\in X$.

Now we can take a subset of $R$, for example $R_1 = \lbrace(1,2), (2,1)\rbrace$.

It is easy to see that $R_1$ is not reflexive.

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