# Strict preference relation implies weak preference relation

Condition A:

Given x, y in X such that $$yRx$$ then it follows that

$$\lambda y +(1-\lambda)xRx$$ for all $$0< \lambda<1$$

Condition B:

Given x, y in X such that $$yPx$$ then it follows that

$$\lambda y +(1-\lambda)xPx$$ for all $$0< \lambda<1$$

Show that the condition B implies the condition A.

R refers a weak preference relation and P is a strict preference relation.

I don’t understand how to show this implication.

What do you think? How can I show this? I am very confused.

This question is duplicated. I also asked on math-stack exchange website. But there, I could not get any proper answer. What do you think about my question? Thank you.

Condition B does not imply condition A. Consider an example of the preference relation defined on $$\mathbb{R}$$ that is represented by the following utility function: $$\begin{eqnarray*} u(x)= \begin{cases} 0 &\text{if } x = 0 \\ 1 & \text{if } x \neq 0\end{cases}\end{eqnarray*}$$ The preference relation $$R$$ represented by $$u$$ satisfies condition B but not condition A.

• Thank you so much professor :) Commented Sep 12, 2022 at 2:16
• The above statement, in my humble view, is not true. The above utility function also satisfies condition A. Indeed, If $u(x'') > u(x')$, which can hold if and only if $x' = 0$, you necessarily have that $u(\lambda x'' +(1-\lambda)x') > u(x')$, which implies condition A as well. Since in any example where you have a strict preference in the above example you need $x=0$, I don't think that the above statement is correct. A strict preference relation always implies a weak preference relation, if preferences are rational. Commented Oct 15, 2022 at 8:58
• It does not satisfy condition A because $1 R (-1)$, but for $\lambda = \frac{1}{2}$ we get $\neg 0 R (-1)$.
– Amit
Commented Oct 15, 2022 at 9:32
• EDIT: a strict preference relation always implies a weak preference relation, by definition. Commented Oct 15, 2022 at 9:32
• Ah, ok, you're considering also the negatives. Note, however, that you just assumed that you can represent this preference relation with an utility function, which is not in the text. Commented Oct 15, 2022 at 11:41

By definition $$y \succ x$$ means that $$y \succeq x$$ but not $$x \succeq y$$

Thus if $$y \succ x$$, then $$y \succeq x$$, which is the first part of the two conditions. If $$y \succ x$$ implies (for some random reason) that for any $$\lambda\in(0,1)$$, you have that $$\lambda y + (1-\lambda)x \succ x$$, then, by definition of $$\succ$$, we have that $$\lambda y + (1-\lambda)x \succeq x$$ but not $$x \succeq \lambda y + (1-\lambda)x$$. Read the proof again: I just showed that any time condition B is true, thus any time that $$y \succ x$$ and, for some random reason, for any $$\lambda \in (0,1)$$, $$\lambda y + (1-\lambda)x \succ x$$, then condition A is also true.

Note that: 1) the problem doesn't state that $$\succeq$$ is rational (necessary but not sufficient condition for $$\succeq$$ to be represented by an utility function), 2) the problem doesn't state that $$\succeq$$ can be represented by an utility function.

• Dear Matreo I have a question on PBE. Someone draw the extensive form game tree but I could not solve it. Can you please help me to do part (b) (c) ? economics.stackexchange.com/questions/52557/… Commented Oct 18, 2022 at 14:01
• I'll try tonight, I need to work, sorry, but I promise I'll have a look. Commented Oct 18, 2022 at 14:09
• I see, I will be very happy. Many thanks:) Commented Oct 18, 2022 at 14:40