Is not only A⋂C=B but also A⋂C⊂B?

[round thunder]

A = {n | R(j(n)) is true}
B = {n | R(k(n)) is true}
C = {n | j(n) = k(n)}
R is a relation, j and k are sequences.

The textbook says A⋂C⊆B.
I thought A⋂C=B.
Is there a case that A⋂C⊂B?

Here are my effort notes, feel free to ignore if it doesn’t help your understanding.

j1, j2, j3, …
k1, k2, k3, …

{n | R(j(n)) is true and j(n) = k(n)}
{n | R(j(n)) is false and j(n) = k(n)}
{n | R(j(n)) is true and j(n) /= k(n)}

{n | R(k(n)) is true and j(n) = k(n)}
{n | R(k(n)) is false and j(n) = k(n)}
{n | R(k(n)) is true and j(n) /= k(n)}

I wanna be fluent in set theoryyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

[dim ether]

maybe show the part of the textbook you are talking about

I think there's information missing here

[round thunder]

Infinitesimal Calculus, page 29

[radiant fog]

Example: R is always true and the two series are disjoint. In that case, A=B, but C is the empty set.

This should be the kind of example you are looking for, no?

[round thunder]

Oh, I feel that A⋂C⊂B clearly in that case.

Hmm…

Thank you

[round thunder]

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