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We say $\succsim$ represents weak monotonic preferences if $$x,y \in X, \,\, y >> x \implies y \succ x $$

where $y >> x$ means that every element of $y$ is greater than every element of $x$.

And we say $\succsim$ represents strong monotonic preferences if $$x,y \in X, \,\, y \geq x, y \neq x \implies y \succ x$$

where $y \geq x$ means that at least one element of $y$ is greater than an element of $x$ and all others are equal.

My question is: do strong monotonic preferences imply weak monotonic preferences?

My answer: yes. The reasoning is as follows:

Since the set $A = \{ x,y \in X: y >> x \}$ is a subset of $B = \{x,y \in X: y \geq x, y \neq x \}$, then if $B \implies y \succ x$ it must also be true that $A \implies y \succ x$ and thus strong monotonic preferences imply weak monotonic preferences.

In plain English, if a bundle with more of one commodity and the same of all others is preferred, then a bundle with more of every commodity must also be preferred.

Is my reasoning correct?

Thanks!

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  • $\begingroup$ Thanks for the input, @1muflon1. I don't think I made myself very clear with regards to the math. I think we can agree that the "plain English" explanation is correct, right? What I meant to say with the math is exactly the same. If $B \implies y \succ x$, that is, if having more of one commodity and the same of others implies that $y \succ x$, then it must also be true that having more of all commodities, that is, $A$, also implies $y \succ x$. I'm not saying $B \implies A$, I'm saying that if $B \implies y \succ x$ than it must also be true that $A \implies y \succ x$, by their design. $\endgroup$ May 13 '20 at 11:34
  • $\begingroup$ When you say "if you say that strong monotonic preferences imply weak then all results for weak monotonic preferences should apply to strong ". That's not true, because weak monotonicity requires a stronger condition than strong monotonicity, that is for all elements of a bundle to be greater than the elements of another bundle. Thus, if strong monotonicity happens, there's no guarantee weak monotonicity will also happen. Take the sets $A = (2,3)$ and $B = (3,3)$ and $C=(3,4)$. Then if s.m is valid, $B \succ A, C \succ A, C \succ B$. However, if only w.m is valid $C \succ A$ is all you can say $\endgroup$ May 13 '20 at 11:48
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    $\begingroup$ The proof seems fine, albeit could be cleaner. $\endgroup$ May 13 '20 at 17:41
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    $\begingroup$ @1muflon1: "Does it make sense to say that a natural number implies real number? I don’t think so in my mind those are two different concepts, but I might be wrong." Yes, if $x$ is a natural number, then $x$ is a real number. Yes you are wrong. $\endgroup$
    – user18
    May 14 '20 at 2:57
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    $\begingroup$ @PedroCunha Kenny's answer was the cleaner more direct proof I would have written! $\endgroup$ May 14 '20 at 4:24
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Let $x,y\in X$. Suppose $y \gg x$ (so that in particular, $y\geq x$ and $y\neq x$). By assumption, $\succsim$ satisfies strong monotonicity; and so, $y\succ x$.

(We've just shown that for any $x,y\in X$ such that $y \gg x$, we have $y\succ x$. Hence, $\succsim$ satisfies weak monotonicity.)


The argument you've given above is correct and roughly the same as the proof I've just given. The only problem is that your argument is somewhat indirect, convoluted, and unclear.

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    $\begingroup$ I'm not sure what your math/econ background is, but it is probably worthwhile reviewing the basics of proof techniques. $\endgroup$
    – user18
    May 14 '20 at 2:51
  • $\begingroup$ I'm working on it. I'm taking my first 'proof-based' course now. And also it doesn't help that I'm not a native english speaker. Thank you very much for the proof. $\endgroup$ May 14 '20 at 3:56

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