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Today, my teacher said that $R=\{(1,1),(2,2),(3,3)\}$ is reflexive, symmetric, antisymmetric and transitive on set $A=\{1,2,3\}$.

I can see that it is reflexive clearly. Moreover when I think $x=1,y=1$, then $yRx = (1,1)$ exists, and it is valid for other elements $2$ and $3$. So, it is symmetric, but I cannot understand why they are transitive or antisymmetric. For example, in transitivity, if xRy and yRz exist, then xRz must exist, but there is not such a pair here. My doubt is similar for anti symmetry. Can you please explain detailly and clearly? I am new in this math course and ashamed of asking in class.

Anne Bauval
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    It is transitive because for every couple of pair $(x,y), (y,z)$, also $(x,z)$ is there. Example: $(1,1), (1,1)$ and thus also $(1,1)$. – Mauro ALLEGRANZA Apr 15 '24 at 13:21
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    Btw, this just the identity relation on $A$, so certainly transitive. If it were not transitive you would manage to find a counterexample, i.e., $x,y,z\in A$ auch that $xRy$ and $yRz$ but not $xRz$. And for antisymmetric, just spell out the definition – Hagen von Eitzen Apr 15 '24 at 13:21
  • To emphasize, although we might refer to variables using different labels... that does not imply that they must refer to different values. Further, the non-existence of certain scenarios by itself does not always imply anything. The empty-relation or the relation containing only $(1,2)$ both qualify as transitive. Paraphrasing my linked answer, transitivity merely requires that any paths we can take (if any exist in the first place) must also be able to be taken using a single step. Although it is commonly worded talking about paths of length 2, it applies to any length. – JMoravitz Apr 15 '24 at 13:30
  • For real world example that you should be intimately familiar with... consider the $<$ relation talking about if one number is less than another. If we know that $a<b$ and we know that $b<c$ and $c<d$ and such... we should be able to conclude that $a<d$. Also, since $a<d$ we should also be able to conclude that it is not true that $d<a$ – JMoravitz Apr 15 '24 at 13:32
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    @AnneBauval although the question I linked to specifically asked about reflexivity... the answer that I wrote to that question deals with all of reflexivity, symmetry, antisymmetry and transitivity... where I relate the concepts to a graphical interpretation of these, and where I directly answer not only the question asked by the OP of the linked question, but I also answer the question asked by the OP here. I'm sure I've written similar or equivalent answers elsewhere as well, this was just the first I pulled up from a quick search through my post history. – JMoravitz Apr 15 '24 at 14:22

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Couple of things:

When transitivity says that if $xRy$ and $yRz$ implies $xRz$, please note that $x$, $y$, and $z$ do not need to be 3 different objects. For example, we could pick $x=1$, $y=2$, and $z=1$. Or we could pick $x=y=z=3$.

So, actually there are $x$, $y$, and $z$ such that $xRy$ and $yRz$: for example, when $x=y=z=3$, we have $xRy$ and $yRz$, and so for transitivity to hold, we do need to verify that we have $xRz$, but with $x=z=3$, that holds too, so we're good.

Also: even if there would not be any $x$, $y$, and $z$ such that $xRy$ and $yRz$, you can still have transitivity. In fact, you must have transitivity in such a case! Consider the 'empty' relation $R = \{ \}$. Well, this $R$ is trivially (or 'vacuously') transitive, because it is true that if we ever have $x$, $y$, and $z$ such that $xRy$ and $yRz$, then we have $xRz$ as well. That is: for all of the zero (!) cases where we have $xRy$ and $yRz$, we have $xRz$ as well ... and so we're good!

Put differently: for transitivity not to hold, we must have some $x$, $y$, and $z$ such that $xRy$ and $yRz$, but not $xRz$. But with this empty $R$, we never have $xRy$ and $yRz$, and so we certainly don't have $xRy$ and $yRz$, but not $xRz$. So again, transitivity holds.

Bram28
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I'll sum up all the comments you have under the post.

Starting with the definition of each of the properties, we see that:

  • A relation $R$ is reflexive if $\forall a \in A, (a,a) \in R$.
    Obviously, for all elements in your set $A$ it is true, since for $1 \in A$ there's $(1, 1)$, for $2 \in A$ there's $(2, 2)$, and for $3 \in A$ there's $(3, 3)$.

  • A relation $R$ is symmetric if $(a,b) \in R \Rightarrow (b,a) \in R$.
    Since the only pairs you have for a premise of the property are of the form $(a, a)$, it immediately follows that this is held in your case.

  • A relation $R$ is antisymmetric if $(a,b) \in R \, \cap \, (b,a) \in R \Rightarrow a=b$.
    Well, again, since all the pairs are of the form $(a, a)$, there's not a single pair which would go against the definition.

  • A relation $R$ is transitive if $(a,b) \in R \, \cap \, (b,c) \in R \Rightarrow (a,c) \in R$.
    Well, (here we go) again, since all the pairs are of the form $(a, a)$, the first condition of the premise implies that $b=a$, which in turn makes the second condition only true iff $c=b$, hence $a=b=c$, hence, $a=c$.

Egor Larionov
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