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I'm studying for a midterm and I just want to make sure that my understanding of these 2 problems that my teacher gave is logically sound. If you could take a look and give me some feedback I would really appreciate it.

Problem 1
In the case below, a relation on the set {1,2,3} is given. Of the three properties, reflexivity, symmetry, and transitivity, determine which ones the relation has. Give reasons.

a. R = {(1, 3), (3, 1), (2, 2)} I think it is not reflexive because for every x in the set {1,2,3} the (x,x) does not exisit in R. It is symmetric because for every element (x, y) there is a (y, x). Here is what I'm not so sure about. The no relation from x to y and from y to z exist. I think this case is similar to part C and therefore it is transitive. Though I'm not entirely sure if my reasoning is sound.

b. R = {(1, 1), (2, 2), (3, 3), (1, 2)}
I said reflexive because for every element x there was the pair (x,x). It is not symmetric because for the pair (1,2) the pair (2,1) does not exists. I said it is transitive but this goes off the logic I used in part A.

c. R = ∅

I already know this one is right from the post of previous person here. My reasoning was that it is not reflexive because for x there does not exist the pair (x,x). It is transitive based off of the logic in the previous post.

Problem 2 In each case below, say whether the given statement is true for the universe (0, 1) = {x ∈ R | 0 < x < 1}, and say whether it is true for the universe [0, 1] = {x ∈ R | 0 ≤ x ≤ 1}. For each of the four cases, you should therefore give two true-or-false answers. a. ∀x(∃y(x > y)) b. ∀x(∃y(x ≥ y)) c. ∃y(∀x(x > y)) d. ∃y(∀x(x ≥ y))

Honestly on this one I'm not sure where to start. My first guess would be that for A) Universe (0, 1) = false Universe [0, 1] = false

B) Universe (0, 1) = true Universe [0, 1] = true

C) Universe (0, 1) = false Universe [0, 1] = false

D) Universe (0, 1) = true Universe [0, 1] = true

Not entirely sure why

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thad
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    For 1a), transitivity, note that $(1,3)$ is in the relation, as is $(3,1)$. So if we had transitivity, then $(1,1)$ would have to be in the relation. But it isn't. – André Nicolas Mar 09 '14 at 17:42
  • Oh ok. That makes sense. Thanks @AndréNicolas. – thad Mar 09 '14 at 17:46

1 Answers1

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1.a. is not reflexive because it does not include $(1,1)$.

It is symmetric.

It is not transitive because, although it includes $(1,3)$ and $(3,1)$, it does not include $(1,1)$

1.b. Is reflexive and is not symmetric, for precisely the reasons you said.

It is also transitive because there is no counterexample to it being transitive. There are only two that you have to check.

1.c. is not reflexive. It is vacuously symmetric and transitive, because there is nothing to make a counterexample with.

2.a. says that every number in the universe has a number less than itself. That's true in the open interval, but not in the closed one.

2.b. says that every number in the universe has a number no bigger than itself. This is true in every universe.

Parts c and d mirror parts a and b exactly.

G Tony Jacobs
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