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From Perloff, Microecomics, 6th edition (Pg. 75):

A consumer chooses between bundles of goods by ranking them as to the pleasure the consumer gets from consuming each. We summarize a consumer’s ranking using a preference relation, such as the consumer weakly prefers Bundle a to Bundle b, if the consumer likes Bundle a at least as much as Bundle b.

Given this weak preference relation, we can derive two other relations. If the consumer weakly prefers Bundle a to b, but the consumer does not weakly prefer b to a, then we say that the consumer strictly prefers a to b—would definitely choose a rather than b if given a choice—which we write as a b.

If the consumer weakly prefers a to b and weakly prefers b to a —— then we say that the consumer is indifferent between the bundles, or likes the two bundles equally.

My question is: When you have declared that a consumer weakly prefers A to B, how could you then argue that a consumer could also prefer (weakly) B to A? Aren't they mutually exclusive?

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    – BCLC
    Mar 5, 2016 at 15:28
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    – Kitsune Cavalry ♦
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No, weak orderings are not mutually exclusive, but strict orderings are, exactly as for the usual comparison of real numbers: you can have $x \geq y$ and $y \geq x$ (if $x=y$), but $x>y$ and $y>x$ are incompatible with each other.

The weak orderings $A \succeq B$ and $B \succeq A$ mean that the decision-maker is indifferent between $A$ and $B$: he might therefore choose any of these options indifferently.

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  • $\begingroup$ Thank you. That clears up a few things. If A ⪰ B mean that the decision-maker is indifferent between A and B, then why does the author raise a third possibility that if A ⪰ B and B ⪰ A, then A ∼ B ( The consumer is indifferent between the two). Precisely, What I'm asking is: What is the difference between A ⪰ B and A ∼ B if both notations mean that the consumer is unable to make a decision? $\endgroup$
    – WorldGov
    Mar 6, 2016 at 4:06
  • $\begingroup$ @AkalankPrakash $A \succeq B$ does not mean that the consumer is indifferent: it means that he weakly prefers $A$ to $B$ (but not strictly). It is the opposite of $B \succ A$. It is only the combination of $A \succeq B$ and $B \succeq A$ that is equivalent to $A \sim B$. Thinking of $A$ and $B$ as real numbers instead of bundles.might help you. $\endgroup$
    – Oliv
    Mar 6, 2016 at 8:15

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