Transitivity in preferences seems as a flawed concept because there might be a situation where A>B, B>C but A<C. Going by this analogy it seems that it does not qualify as a reason for Indifference curves intersecting. I need to understand how can actually two indifference curves not intersect.

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    $\begingroup$ I don't understand what you are trying to ask. If two indifference curves intersect, then transitivity is violated. $\endgroup$
    – Herr K.
    Nov 3 '20 at 15:53

Just because there might be instances where transitivity is not satisfied does not mean that the entire concept is flawed. The concept of the color blue is not flawed just because things can exist that have a different color.

Generally, preferences that are both complete and transitive are called rational, and I believe such preferences make a lot of sense. Take your A(pple juice), B(eer), and C(offee). If your preferences are A>B, B>C but A<C, you will never be able to decide on an optimal drink. Suppose you grab A because A>B, then you should put it down and grab C as A<C, but then you should take B because B>C, but then you should grab A because A>B, and so on. Because of such circles you would never decide for an option.

Two indifference curves representing rational preferences do not intersect, and rational implies transitive. Take two indifference curves (IC) of two different utility levels $U_1>U_2$. If point $P$ is on $IC_1$ it gives utility $U_1$. If it were also on $IC_2$, it would also give utility $U_2\neq U_1$, a contradiction.

  • $\begingroup$ To be clear on what this has to do with transitivity. Suppose A is only on IC1, B is only on IC2 and C is on both IC1 and IC2. All bundles on the same IC give the same utility. So, U(A)=U(C) and U(B)=U(C), but how can U(A)>U(B)=U(C)=U(A)? $\endgroup$
    – Bayesian
    Nov 3 '20 at 16:58

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