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

elementary-set-theoryrelationssolution-verification

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' ?

Best Answer

$(a)$ 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.

Related Question