questions on finding if these relations are transitive or not

Question

1 X is the set of real-valued functions of a real variable whose domain contains the point 2 and R is the relation defined by fRg if and only if f(2) = g(2). Is this relation transitive? Explain.

For this one am I suppose to look at one side and fix a function (for example fix f(x)=x^2) to that side and then show if it is transitive or not?

2 Fix an $\epsilon > 0.$ $X$ is the set of real valued functions of a real variable whose domain is all of $R$ and $R$ is the relation defined by $fRg$ if and only if $\left |f(x)-g(x)\right |$ $< \epsilon$ for each $x \in R$. Is this relation transitive? Explain.

I think this one is not transitive because the fix $\epsilon > 0$ will cause some problems, but I don't know how I should explain that.

For the question number 1: It is transitive. Because if fRg and gRh then we have f(2)=g(2) and g(2)=h(2) which implies f(2)=h(2) .that is fRh. — Avinash N, Oct 31 '18 at 02:11
Answer for question 2: fix epsilon=1.5 and take f(x)=0, g(x)=1,h(x)=2. Then check it. Thus it is not transtive — Avinash N, Oct 31 '18 at 02:15

score 0 · Accepted Answer · answered Oct 31 '18 at 02:10

0

For the first question, let $f,g,h \in X$, if $f(2)=g(2)$ and $g(2)=h(2)$, can you conclude that $f(2)=h(2)$?

For the second question, let $f(x) = -\frac{\epsilon}2$, $g(x)=0$, $h(x)=\frac{\epsilon}2$, try to verify that this is a counterexample.

answered Oct 31 '18 at 02:10

Siong Thye Goh

149,520
20
88
149

1 Ahh I get it the relation is just at the point so then yes it would be transitive. – George Oct 31 '18 at 02:14
2 is it because since |f(x)−g(x)| <ϵ the fix a ϵ>0 will intersect which cannot happen? – George Oct 31 '18 at 02:18
part $1$ is right, perhaps compute $|f-g|$ and see what do you get, compute $|h-g|$ and also compute $|f-h|$? – Siong Thye Goh Oct 31 '18 at 02:21
let me reword my answer since at |f−h| you would get 0 but since there is a ϵ>0 thus resulting in the relation not being transitive – George Oct 31 '18 at 02:31
did you get $|f-h|=0$? I don't that's suppose to happen – Siong Thye Goh Oct 31 '18 at 02:33
is it suppose to be ϵ then? – George Oct 31 '18 at 02:41
That's right, and we have $|f(x)-h(x)| = \epsilon$ and it is not less than $\epsilon$. – Siong Thye Goh Oct 31 '18 at 02:42
ahh ok thanks for the help – George Oct 31 '18 at 02:43

questions on finding if these relations are transitive or not

1 Answers1