# Proving reflexivity, symmetry and transitivity,…, on the subsets of $\mathbb{R}$

The relation R is the set of the subsets of $\mathbb{R}$ defined by $xRy$ iif $x \cap y \ne \emptyset$

I want to prove the following ...

1. Reflexive : Yes, because a set $x \cap x$ will have all the same elements.

2. Symmetric : Yes, because if $x \cap y$ then $y \cap x$

3. Transitive : Yes, We want to show that if xRy and yRw then xRw. if $x \cap y$ means y is a subset of x. if $y \cap w$ means w is a subset of y. Clearly, w is a subset of x.

4. Asymmetric : No, because xRx

5. Antisymmetric : No, because it's symetric

6. Irreflexive : No, because it's reflexive.

Can you show me my errors ?

And is it enough to say that 'No because its symetric/reflexive' ?

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$\emptyset \cap \emptyset = \emptyset$ – Lord Soth Apr 5 '13 at 2:29
$x\cap y\ne\emptyset$ does not imply $y\subseteq x$ like you claim. For instance $\mathbb Z\cap (\mathbb Q\setminus \{0\})\ne\emptyset$, but neither is a subset of the other. – Karl Kronenfeld Apr 5 '13 at 2:32
Transitivity is not there. – Shahab Apr 5 '13 at 2:35
+1 Nice effort Dave, by the way! Thanks for showing your thoughts and effort. – amWhy Apr 5 '13 at 2:40

$(a)$ Recall that reflexivity must hold FOR ALL subsets of a set, including $\varnothing$!

• $\varnothing \subset \mathbb R$.
• But $\varnothing \cap \varnothing = \varnothing$

Hence, since there exists a subset that doesn't satisfy reflexivity, the relation as a whole cannot be reflexive.

$(b)$ You are correct; the relation is symmetric: but it's symmetric because IF $x\cap y \neq \varnothing$, then $y\cap x = x \cap y \neq \varnothing.$

$(c)$ Recheck transitivity, too, to find a counterexample to the property: We want to show that if for any subsets $x, y, w \subseteq \mathbb R\,$ $\;x\,R\,y\;$ and $\;y\,R\,w,\;$ it follows that $\;x\,R\,w\;$ I think you got a little mixed up along the way as to *how the relation is defined. $\;x\,R\, y\;$ means $x\cap y\neq \varnothing$. (The intersection of subset $x$ and subset $y$ is non-empty.) And $\;y\, R \,w$ means $\;y\cap w \neq \varnothing.\;$ It does not necessarily follow that $x\,R\,w\;$, that is, there are counterexamples to $\;x \cap w \neq \varnothing.\;$

• E.g.: let $x, y, w$ be subsets of $\mathbb R$ defined by open intervals of reals: $x = (0, 2),\; y = (1, 3),\; w = (2, 4).\;$ Then $x \cap y = (1, 2) \neq \varnothing,\; y\cap w = (2, 3) \neq \varnothing,\;$ but $\;x\cap w = (0, 2) \cap (2, 4) = \varnothing$

$(d)$ In light of not being reflexive (1), refine your reason for not being irreflexive: There exist sets $x\subseteq \mathbb R$ such that $x \cap x\neq \varnothing$; and you need only provide an example of such a subset $x$ to show the relation is not irreflexive.

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You plowed the poor question. Nice Amy $+^{+^{+^+}}$ – Babak S. Apr 6 '13 at 7:33

One important thing is missing in your arguments, namely you are "rushing" things. Go slow, and make sure your every step is correct. For example,

1) Reflexive: We want to show $xRx$ for every subset $x$ of $\mathbb{R}$. Due to the "iff," this is equivalent to showing that $x\cap x \neq \emptyset,\,\forall x\subset\mathbb{R}$. On the other hand, if $x = \emptyset$, $\emptyset\cap \emptyset = \emptyset$, and thus $R$ is not reflexive.

2) Symmetric: We want to show if $xRy$ then $yRx$. This is equivalent to showing that $x\cap y\neq\emptyset \implies y\cap x\neq\emptyset$. The implication is obvious as $x\cap y = y \cap x$.

3) Transitivity: We want to show $xRy$ and $yRw$ implies $xRw$. Again, this is equivalent to showing that $(x\cap y\neq\emptyset ,\,y\cap w\neq\emptyset \implies x\cap w\neq\emptyset)$. Try to find a counterexample that this does not hold.

and so on...

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I think there's a problem with transitivity here. For example $\{1, 2\} \cap \{2, 3\} \neq \varnothing$, $\{2, 3\} \cap \{3, 4\} \neq \varnothing$, but $\{1, 2\} \cap \{3, 4\} = \varnothing$.

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