Question

Which of these relations on the set of all functions from Z to Z are equivalence relations?

(a) $\{(f,g) \mid f(1)=g(1)\}$

(b) $\{(f,g) \mid f(0)=g(0) \text{ or } f(1)=g(1)\}$

(c) $\{(f,g) \mid f(x)-g(x)=1 \text{ for all } x \in \textbf{ Z} \}$

In this question, what does it mean by f(1) = g(1) and f(1) - g(1)? And how it satisfies all the conditions (Symmetric, Transitive and Reflexive) of equivalence relation. Actually I cannot able to analyze when the relations defined on set of function. If anyone describe a bit it would helpful.

Answer 1

(a) equivalence relation

f has one element ,say i.e. 1 . Two function (f and g) merging in one element . Relation will be (1,1)

(b) not reflexive

f  has 2 elements , say i.e. 1,2 .But relation is (1,1) or (2,2) , but may not both

(c) not reflexive, not symmetric

distance between same element of f to g is 1 . But distance between g to f is -ve , which is not possible. So, it is not even symmetric

Comments

That is where my question belongs. f(1)=(g(1) means two functions f and g are mapped into same element, of course we can't say for all element it is true. But how do I check (1,1) i.e reflexivity from this.

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$f = g$, means for any $x, f(x) = g(x)$. $f(1) = g(1)$ means mapping is same for element 1, may or may not be for others.

Now, reflexivity requires $f$ related to $f$. $f=f$ is true here and this implies $f(1) = f(1)$ and hence $f$ related to $f$.

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(b) is reflexive and symmetric but not transitive