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I have the relations $E_1$ and $E_2$.

A reflexive, symmetric and transitive property shows that two relations are equivalent to each other. I need to prove if this is true for the following

a) $E_1 ∩ E_2$

b) $E_1 ∪ E_2$

c) $E_1 ∘ E_2$

I found out that a) is true, but I am not sure about it. This is my proof for transitivity:

$E_1 ∩ E_2 ⇒ (a, b, c) ∈ E_1 ∧ (a, b, c) ∈ E_2 ⇒ \\((a, b) ∈ E_1 ∧ (b, c) ∈ E_1 → (a, c) ∈ E_1) ∧ ((a, b) ∈ E_2 ∧ (b, c) ∈ E_2 → (a, c) ∈ E_2)$

Also I found different arguments on the internet, so I'm pretty unsure about this.

Regarding b) $E_1 ∪ E_2$ a friend found out that this is not transitive and they wanted to prove it with an example. The example is the following:

$A = \{1,2,3\}$

$E_1 = \{(1,2), (2,1), (2,2)\}$

$E_2= \{(1,3), (3,1), (3,3)\}$

$E_1 ∪ E_2 = \{(1,2), (2,1), (2,2), (1,3), (3,1), (3,3)\}$

This is transitive, isn't it? If not, what is it what I did not understand?

And maybe you can help me with with c), as well.

Raphael
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LSR
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    Regarding your friend's example, the union relation indeed isn't transitive. – Yuval Filmus Nov 15 '15 at 16:09
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    A relation being reflexive, symmetric and transitive means it's called an equivalence relation. The claim is not that "$E_1$ and $E_2$ are equivalent to each other": it's that they are equivalence relations. Unfortunately, your attempt to prove transitivity makes no sense: $X\rightarrow Y$ means "If X is true, then Y must be true, too" but you've written "$E_1\cap E_2 \Rightarrow ..." and $E_1\cap $E_2$ isn't something that can be true or false: it's a set, not a Boolean value. $(a,b,c)$ cannot be an element of $E_1$, since $E_1$ is a binary relation, not ternary. – David Richerby Nov 15 '15 at 16:10
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    Your example to prove non-transitivity of the union doesn't work because your $E_1$ and $E_2$ aren't equivalence relations, they're not reflexive or transitive. However, given those relations $E_1$ and $E_2$, you've calculated $E_1\cup E_2$ correctly and you're correct that it's not transitive. I'm not sure why you're surprised that it's not transitive: you're looking for $E_1$ and $E_2$ such that $E_1\cup E_2$ isn't an equivalence relation, meaning you need $E_1\cup E_2$ to be non-reflexive and/or non-symmetric and/or non-transitive. – David Richerby Nov 15 '15 at 16:13
  • Why I don't understand or why it surprises me is because it is a simple fact and shouldn't be taken for granted: There are beginners and one of them is me who would like to understand. This is what ppl need to understand. I'm trying to learn. Question is and I thought I wrote it: Why is it not transitive? (E1 ∪ E2) And how do you prove that E1 ∩ E2 is equivalent? – LSR Nov 15 '15 at 16:24
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    Can you explain why this question belongs here on Computer Science? It looks like a pure math question to me (and thus more suitable to Math.SE). We are not 100% averse to pure math questions if the question can explain why it needs to be answered from a CS perspective -- do you have something like that in mind, or would this be reasonable to migrate to Math.SE? – D.W. Nov 15 '15 at 18:09
  • There's a legitimate question hiding here (though I agree with D.W. that there's nothing that indicates it mightn't be more appropriate for Mathematics.SE). The question, though, seriously lacks precision, for the reasons noted in the other comments. Both of your reasonings have problems that need to be addressed. – Rick Decker Nov 15 '15 at 20:45
  • This is part of theoretical computer science. If sb. is capable of, this person can migrate it to math exchange. What other questions in the comments are not addressed? And be precise. Since this is a topic I am not good at I need advice on how to do the things correctly. If I would be good at it I wouldn't ask here for help. – LSR Nov 15 '15 at 21:54
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    @Lintu I'm sorry -- I didn't mean to belittle you by saying I didn't know why you were surprised. You were trying to prove that $E_1\cup E_2$ isn't always an equivalence relation and you almost succeeded, so I was expecting you to be happy rather than surprised. ("Almost" because the relations you started with weren't reflexive but, if you make them reflexive, you're done). I hope my other comments were more helpful to you. I think the best thing you can do is talk to your TA or professor about this because I think you need a more interactive style of help than can happen on the internet. – David Richerby Nov 15 '15 at 21:55
  • This questions belongs in math.se. – Yuval Filmus Nov 15 '15 at 22:21
  • Thanks David. I'm attending the so called tutorial classes as much as I can. But the amount of what we have to learn in the short time is so much that I can not 'discover' the things on my own and I need a lot of help. I even search for books, tutorials and websites that only has the topic of proving things, since this seems to be a big issue for me. – LSR Nov 16 '15 at 20:46
  • @YuvalFilmus Please vote or flag the next time so that offtopic questions don't slip through the net. – Raphael Jan 15 '16 at 08:33

1 Answers1

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Here are some hints:

a) If $E_1,E_2$ are equivalence relations then so is $E_1 \cap E_2$. The proof is just by "unpacking" the definitions. (More generally, every class defined by universal statements – those of the form $\forall x_1,\ldots,x_n \phi(x_1,\ldots,x_n)$ – is closed under intersection.)

b) When trying to prove that $E_1 \cup E_2$ is transitive, we get stuck in the following case: if $a E_1 b$ and $b E_2 c$ then we have to show that $a E_1 c$ or $a E_2 c$, though there is no reason to believe that this must be the case. This helps you construct a counterexample.

c) When trying to prove that $E_1 \circ E_2$ is symmetric, we get stuck: if $a E_1 x E_2 b$ then there is no particular reason that $b E_1 y E_2 a$ for some $y$. This helps you construct a counterexample.

Yuval Filmus
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  • Thanks Yuval,

    I prove b and c by choosing one counterexample, because it exists at least one example that doesn't have one of the properties?

     – LSR Nov 16 '15 at 21:03
  • Well, to show that b) and c) aren't necessarily equivalence relations, by far the easiest way is to give explicit counterexamples. – Yuval Filmus Nov 16 '15 at 22:50