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Well... The three definitions I got for reflexive, symmetric, and transitive relations are:

R is said to be reflexive on A if $\forall x\in A : xRx$.

R is symmetric if $\forall x,y\in A : xRy \rightarrow yRx$.

R is transitive if $\forall x,y,z\in A : (xRy \wedge yRz) \rightarrow xRz$

So, if I understand it correctly, the empty relation {} is symmetric and transitive because the xRy will always be false, and so the implication is true. It's also not reflexive over A unless A is the null set.

What I'm wondering is: if I had a relation, say $R=\{(x,y)\in \mathbb{R}\times\mathbb{R} | x>0 \wedge x<10 \wedge x=y \}$, it's obviously not reflexive over $\mathbb{R}$, but would it be called "reflexive" in general?

I'm asking this because I have a homework problem that says R is a relationship over A, and R is reflexive. I'm not sure if that means R is reflexive over all of A, or in other words, if A is necessarily a subset (or equal to) the domain and range of R.

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When we say "$R$ is a relation on $A$", then the domain of discourse is understood to be $A$. That means that "$R$ is reflexive" will be understood to mean "reflexive on $A$". –  Arturo Magidin Nov 4 '11 at 3:41

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I would say that a relation should not be called "reflexive", without further qualification, unless the underlying set which the relation is on is specified (explicitly or implicitly).

For example your set of ordered pairs $R$ is a subset of $\mathbb{R} \times \mathbb{R}$ and hence is a relation on $\mathbb{R}$. It is not reflexive on $\mathbb{R}$ because there are real numbers $x$ with the property that $(x,x)$ is not in $\mathbb{R}$. But your set $R$ is also a subset of $(0, 10) \times (0,10)$ and hence is a relation on the open interval $I = (0,10)$, and it is is a reflexive relation on this set. To abstract this somewhat, the lesson is that whether a set of ordered pairs is reflexive or not, depends on more than just the set of ordered pairs. It depends on the set $A$ that you think of the set of ordered pairs as being a subset of, and this is not encoded in the set $R$ at all.

(This is a little bit like how the codomain of a function is not determined by the set of ordered pairs that represents the graph of a function.)

As you have defined them the notions of "symmetric" and "transitive" also depend on the underlying set, but this dependence is less sensitive, in the following sense: if $S$ is a set and $R$ is a subset of $S \times S$, and $R$ is symmetric (or transitive) as a relation on $S$, then for any set $T$ satisfying $T \supseteq S$ (so you have that $R$ is also a subset of $T \times T$, and hence a relation on $T$), $R$ will also be symmetric (or transitive) as a relation on $T$. As the above example shows, reflexivity does not behave like this. So symmetry and transitivity are far less sensitive to what set you consider a set of ordered pairs to be relation on, than reflexivity is. So you should not be casual about saying that a relation is "reflexive" (without further qualification) unless a reader knows from context what set you have in mind. (In actual mathematical practice this is never an issue because relations are not constructed arbitrarily for the purpose of testing knowledge of concepts like "reflexivity". But in classroom exercises it is a good practice to follow.)

If I may veer into metamathematics for a moment. Speaking very informally, reflexivity is fundamentally different in nature from the symmetry and transitivity because it is the only property that actually requires specific ordered pairs to be in a relation. (The other two only require that if some ordered pairs are in there, then other ordered pairs must be in there also.) In textbook exercises often people think of reflexivity as a silly and trivial property because it is easy to verify in many examples whether or not a relation is reflexive (on a given set). This is more an artifact of what happens in textbooks than it is an artifact of actual mathematical practice. In actual mathematical practice, it is often the case that reflexivity (when it occurs) is the most difficult property of these three to verify--- precisely because it is the only one that requires specific ordered pairs to be in a relation, it is the only one where in proving it you might be required to actually do something with the definition of the relation and not just use its formal properties.

As an illustration of this idea, suppose I let $A$ denote the set of simple closed curves in $\mathbb{R}^3$ and say that $C \sim D$ for $C, D \in A$ iff $C$ there is a surface $S$ in $\mathbb{R}^3$ satisfying [a long list of geometrical constraints] with the property that the boundary of $S$ is the disjoint union of a curve homeomorphic to $C$ and another curve homeomorphic to $D$. In checking that $\sim$ is symmetric and transitive, I only need to think about what happens to my long list of geometrical constraints if I permute the order of the data I feed into it, or if I piece two surfaces satisfying these constraints together. To prove that $\sim$ is reflexive I actually have to construct, for every curve $C$, a surface satisfying [long list of constraints] whose boundary is a disjoint union of two copies of $C$. A totally different kind of problem.

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Aha! Thanks. You explained my question and then some. –  NeuroFuzzy Nov 4 '11 at 5:15

I would interpret the homework problem to intend that $R$ is reflexive on $A$ in the sense of the definition you have given, but if you are uncertain, why not ask the person who assigned the homework?

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It's from a book. I could mail the author but... –  NeuroFuzzy Nov 4 '11 at 5:16
    
It's from a book, but it was assigned by someone with whom you have direct personal contact, no? So I repeat: why not ask the person who assigned the homework? –  Gerry Myerson Nov 4 '11 at 5:30
    
I'm teaching myself from a book, so I'm the one assigning the work. I have a tutor for these kinds of questions but I only meet with him once every few weeks, so this is the best way to get an answer. I guess I should have said that in my first comment, though. xD –  NeuroFuzzy Nov 4 '11 at 6:21
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When you write, "I have a homework problem," people will naturally assume that said problem has been assigned to you by some third party. If you don't want people to get that impression, it may be worth your while to express yourself in a different way. Anyway, you have an answer now, so we're all happy, right? –  Gerry Myerson Nov 4 '11 at 6:25

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