<p>There are two situations with us when we know that preferences are locally non satiated. 
1) Preferences are monotonic - In this case we know that we have "good" goods which necessarily means that more of every good is better, and thus the consumer would want to exhaust his income 
2) Preferences are non monotonic - This permits the existence of "bad" goods but that doesn't mean that all goods are bad because that would mean that there is no bundle in the neighbourhood of origin (no good is consumed because all are bad) that is preferred to origin, thus we reach at a bliss/satiation point, violating LNS assumption. So, we need atleast one good which is not bad. In this case, the consumer would not spend anything on all the "bad" goods and exhaust his/her entire income on that one good which gives him/her positive utility.</p>

CC BY-SA 3.0
2017-10-21T18:53:00.277Z
Why do you make two separate cases? Everything you write for case 2) also goes for case 1). More importantly, the proof is lacking rigor. In case of $U(x,y) = \min(x,y)$ consider bundle (1,1). Neither good $x$ nor good $y$ is "good", yet this is clearly not a bliss point.
Giskard
CC BY-SA 3.0
2017-10-21T11:36:10.937Z
2017-10-21T11:36:10.937Z
LUCIFER
<p>The OP is correct in pointing out that "Local Non Satiation (LNS) only says there's a (utility) increasing direction but doesn't say which direction it is increasing". Namely, we entertain the possibility on dealing with "bads" also, not only with "goods". MWG Microeconomic Theory book page 43 Figure 3.B.1 depicts exactly such a situation.</p>

<p>But it <em>is</em> the case that, when the bundle set is $\mathbb R_+$, <strong>under LNS not all items can be bads</strong>. Because then, <em>the zero vector will be a point of satiation</em> (and so it would violate the LNS assumption).   </p>

<p>So using non-negative quantities of items and imposing LNS forces us to consider only the cases where at least one item in the bundle is a good and not a bad, in which case "more is better" for this item.</p>

<p>Then, we can prove that local non-satiation implies exhaustion of the available budget.</p>

<p>Ad absurdum, assume that $px^* &lt; m$. Under LNS <em>for every $\epsilon &gt;0$</em>, there exists a $y(\epsilon)$ that is more preferred to $x^*$. If some $y(\epsilon)$ is feasible, $py(\epsilon) \leq m$, then $x^*$ cannot be the optimal choice in the first place.</p>

<p>So the question is : Is it possible that <em>all</em> $y(\epsilon)$ that are preferred to $x^*$ under LNS, are infeasible, $py(\epsilon)&gt;m,\;\; \forall \epsilon&gt;0$?</p>

<p>I guess the OP can take it from here.</p>

CC BY-SA 3.0
2017-10-14T22:09:49.660Z
This really helps! Before I missed the underlying assumption that not all items can be bads.
Neucoder
CC BY-SA 3.0
2017-10-14T14:17:18.927Z
2017-10-14T14:17:18.927Z
Alecos Papadopoulos
<p>Local non satiation says that for any $x \in X$ and $\epsilon &gt; 0$, there exists $y \in X$ such that $d(x,y) &lt; \epsilon$ and $U(x) &lt; U(y)$.</p>

<p>I don't understand why this implies that $px^* = m$ if $x^*$ sovles consumer problem. If we think of $x \in R^2$, it implies that you can find a $y$ that is strictly preferred in the small neighborhood of $x$. In that case, even $x$ is on $px = m$, LNS seems to imply that there is a $y$ that is strictly preferred than $x$, and that $y$ may not be on the boundary since LNS only says there's a increasing direction but doesn't say which direction it is increasing. </p>

CC BY-SA 3.0
2017-10-13T20:15:38.293Z
You want to disprove $px^* < m$. Why are you examining the case where $px = m$? There is no need to study that at all.
Giskard
CC BY-SA 3.0
2017-10-14T04:07:28.543Z
Suppose you are at some point where the budget binds, and you can move into the interior of the budget set to find a more preferred bundle, as you are postulating. But then that bundle must have a more preferred bundle around it due to LNS. So you keep moving in a direction and either go to where...
Kitsune Cavalry
CC BY-SA 3.0
2017-10-14T04:33:15.047Z
https://economics.stackexchange.com/questions/8449/local-non-satiation-proof this question may be of assistance
Kitsune Cavalry
CC BY-SA 3.0
2017-10-14T06:37:54.627Z
LNS doesn't say anything about budget constraints. So yes, pick a point $x$ on the budget line, and there will be a $y$ in its neighborhood that is strictly preferred to $x$, only that it may not be feasible given the budget.
Herr K.
<microeconomics><utility><consumer-theory><preferences>
Why does local non satiation imply the constraint is binding?
Neucoder
# Why does local non satiation imply the constraint is binding?

(Asked by: username_0 on 2017-10-13)

Local non satiation says that for any $x \in X$ and $\epsilon > 0$, there exists $y \in X$ such that $d(x,y) < \epsilon$ and $U(x) < U(y)$.

I don't understand why this implies that $px^\* = m$ if $x^\*$ sovles consumer problem. If we think of $x \in R^2$, it implies that you can find a $y$ that is strictly preferred in the small neighborhood of $x$. In that case, even $x$ is on $px = m$, LNS seems to imply that there is a $y$ that is strictly preferred than $x$, and that $y$ may not be on the boundary since LNS only says there's a increasing direction but doesn't say which direction it is increasing.

**Tags:** microeconomics, utility, consumer-theory, preferences

> **Comments:**
> **username_1**: You want to disprove $px^* < m$. Why are you examining the case where $px = m$? There is no need to study that at all.  
> **username_2**: Suppose you are at some point where the budget binds, and you can move into the interior of the budget set to find a more preferred bundle, as you are postulating. But then that bundle must have a more preferred bundle around it due to LNS. So you keep moving in a direction and either go to where...  
> **username_2**: https://economics.stackexchange.com/questions/8449/local-non-satiation-proof this question may be of assistance  
> **username_3**: LNS doesn't say anything about budget constraints. So yes, pick a point $x$ on the budget line, and there will be a $y$ in its neighborhood that is strictly preferred to $x$, only that it may not be feasible given the budget.  

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## 2 answers

### Answer by username_4 on 2017-10-21. Score: -1 | Accepted answer

There are two situations with us when we know that preferences are locally non satiated.
1) Preferences are monotonic - In this case we know that we have "good" goods which necessarily means that more of every good is better, and thus the consumer would want to exhaust his income
2) Preferences are non monotonic - This permits the existence of "bad" goods but that doesn't mean that all goods are bad because that would mean that there is no bundle in the neighbourhood of origin (no good is consumed because all are bad) that is preferred to origin, thus we reach at a bliss/satiation point, violating LNS assumption. So, we need atleast one good which is not bad. In this case, the consumer would not spend anything on all the "bad" goods and exhaust his/her entire income on that one good which gives him/her positive utility.

> **Comments:**
> **username_1**: Why do you make two separate cases? Everything you write for case 2) also goes for case 1). More importantly, the proof is lacking rigor. In case of $U(x,y) = \min(x,y)$ consider bundle (1,1). Neither good $x$ nor good $y$ is "good", yet this is clearly not a bliss point.  

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### Answer by username_5 on 2017-10-14. Score: 2

The OP is correct in pointing out that "Local Non Satiation (LNS) only says there's a (utility) increasing direction but doesn't say which direction it is increasing". Namely, we entertain the possibility on dealing with "bads" also, not only with "goods". MWG Microeconomic Theory book page 43 Figure 3.B.1 depicts exactly such a situation.

But it *is* the case that, when the bundle set is $\mathbb R\_+$, **under LNS not all items can be bads**. Because then, *the zero vector will be a point of satiation* (and so it would violate the LNS assumption).

So using non-negative quantities of items and imposing LNS forces us to consider only the cases where at least one item in the bundle is a good and not a bad, in which case "more is better" for this item.

Then, we can prove that local non-satiation implies exhaustion of the available budget.

Ad absurdum, assume that $px^\* < m$. Under LNS *for every $\epsilon >0$*, there exists a $y(\epsilon)$ that is more preferred to $x^\*$. If some $y(\epsilon)$ is feasible, $py(\epsilon) \leq m$, then $x^\*$ cannot be the optimal choice in the first place.

So the question is : Is it possible that *all* $y(\epsilon)$ that are preferred to $x^\*$ under LNS, are infeasible, $py(\epsilon)>m,\;\; \forall \epsilon>0$?

I guess the OP can take it from here.

> **Comments:**
> **username_0**: This really helps! Before I missed the underlying assumption that not all items can be bads.  

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