<p>I think a proof by contradiction would be easier (if your preferences are complete). If there is no such $\alpha$ then each $\alpha x + (1-\alpha)z$ is in the (weak) lower contour set of $y$, which is closed. Now construct a sequence (or net).</p>

CC BY-SA 3.0
2017-10-08T15:40:26.157Z
2017-10-08T15:40:26.157Z
James Hunt
<p>HINT: it is easier if you go through the intermediate step that the set $\{\alpha \in[0,1]: \alpha x+(1-\alpha) z \succsim y\} $ is closed for every $x,y,z$</p>

CC BY-SA 3.0
2017-07-06T14:55:06.333Z
2017-07-06T14:55:06.333Z
Enrico
<p>One of the definitions of continuity is that the Upper Contour Set and the Lower Contour Set are closed.</p>

<p>I am trying to show that if preference is continuous and $x&gt;y&gt;z$, then there is some $\alpha \in [0,1]$ such that $\alpha x+(1-\alpha)z&gt;y$</p>

<p>Can you help me to prove second one by using closeness?</p>

CC BY-SA 3.0
2017-06-20T16:18:55.900Z
Or you could do your own homework.
Giskard
CC BY-SA 3.0
2017-06-20T16:20:18.333Z
This is not my homework but I am also trying to find that. (my math is not good enough) But I can not find any kind of resource even similar to that idea.
ozozzz
CC BY-SA 3.0
2017-06-20T16:27:20.370Z
I am also open to any kind of hints.
ozozzz
CC BY-SA 3.0
2017-06-20T16:48:46.740Z
How come you are trying to find this proof? What motivates you to do it?
Giskard
CC BY-SA 3.0
2017-06-20T16:54:09.970Z
Hint: You can start by reminding yourself of the definition and characterizations of the closedness of a set.
Herr K.
CC BY-SA 3.0
2017-06-20T18:58:56.880Z
I am studying temptation for my thesis. The paper that I am reading includes 3 types of continuity assumptions:(paper deals with menus, not singletons) USC, Lower vNM cont. and lower singleton continuity. I am trying to show that we may not need lower vNM continuity in some cases. This idea is not directly related to my thesis, but you can think that I am looking for ways to simplify my assumptions.
But in order to do that, first I want to learn about continuity with different definitions so that I can improve my understanding of continuity.
ozozzz
CC BY-SA 3.0
2017-06-21T05:54:56.570Z
Best of luck to you then. IMO learning is thinking about unfamiliar concepts. As this question seems relatively simple it is a good exercise to solve by yourself.
Giskard
CC BY-SA 3.0
2017-06-21T07:52:14.297Z
You might consider if you can restate this such that it's purely mathematical.  Math.SE might be more open to such a question.
Brythan
CC BY-SA 3.0
2017-07-07T20:17:41.890Z
I am confused. Can you not set $\alpha = 1$?
201p
<utility><preferences>
Equivalence of definitions of continuity
ozozzz
# Equivalence of definitions of continuity

(Asked by: username_0 on 2017-06-20)

One of the definitions of continuity is that the Upper Contour Set and the Lower Contour Set are closed.

I am trying to show that if preference is continuous and $x>y>z$, then there is some $\alpha \in [0,1]$ such that $\alpha x+(1-\alpha)z>y$

Can you help me to prove second one by using closeness?

**Tags:** utility, preferences

> **Comments:**
> **username_1**: Or you could do your own homework.  
> **username_0**: This is not my homework but I am also trying to find that. (my math is not good enough) But I can not find any kind of resource even similar to that idea.  
> **username_0**: I am also open to any kind of hints.  
> **username_1**: How come you are trying to find this proof? What motivates you to do it?  
> **username_2**: Hint: You can start by reminding yourself of the definition and characterizations of the closedness of a set.  
> **username_0**: I am studying temptation for my thesis. The paper that I am reading includes 3 types of continuity assumptions:(paper deals with menus, not singletons) USC, Lower vNM cont. and lower singleton continuity. I am trying to show that we may not need lower vNM continuity in some cases. This idea is not directly related to my thesis, but you can think that I am looking for ways to simplify my assumptions. But in order to do that, first I want to learn about continuity with different definitions so that I can improve my understanding of continuity.  
> **username_1**: Best of luck to you then. IMO learning is thinking about unfamiliar concepts. As this question seems relatively simple it is a good exercise to solve by yourself.  
> **username_3**: You might consider if you can restate this such that it's purely mathematical.  Math.SE might be more open to such a question.  
> **username_4**: I am confused. Can you not set $\alpha = 1$?  

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## 2 answers

### Answer by username_5 on 2017-10-08. Score: 0

I think a proof by contradiction would be easier (if your preferences are complete). If there is no such $\alpha$ then each $\alpha x + (1-\alpha)z$ is in the (weak) lower contour set of $y$, which is closed. Now construct a sequence (or net).


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### Answer by username_6 on 2017-07-06. Score: 0

HINT: it is easier if you go through the intermediate step that the set $\{\alpha \in[0,1]: \alpha x+(1-\alpha) z \succsim y\} $ is closed for every $x,y,z$


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