<p>One can prove the existence of such optimal plans using the extreme value theorem of Weierstrass, but it requires some advanced math.</p>
<p>Here is a toy version of the model without energy and emissions. Both the instantaneous utility functions <span class="math-container">$u:\mathbb{R}_+\to\mathbb{R}$</span> and the production function <span class="math-container">$f:\mathbb{R}_+\to\mathbb{R}$</span> are assumed to be continuous and nondecreasing. Moreover, <span class="math-container">$u$</span> is assumed to be bounded (!). There is given initial capital stock <span class="math-container">$k_1\geq 0$</span>. The space of feasible consumption and production plans is defined as <span class="math-container">$$F=\big\{(c_1,k_1,c_2,k_2,\ldots)\mid 0\leq k_{t+1}\leq f(k_t-c_t)\}, k_t\geq c_t\geq 0\big\}.$$</span>
<span class="math-container">$F$</span> describes all feasible paths of consumption and capital.
This set is a nonempty compact subset of <span class="math-container">$\mathbb{R}^\infty$</span> endowed with the <a href="https://en.wikipedia.org/wiki/Product_topology" rel="nofollow noreferrer">product topology</a>. Here is why: By <a href="https://en.wikipedia.org/wiki/Tychonoff%27s_theorem" rel="nofollow noreferrer">Tychonoff's theorem</a>, the set <span class="math-container">$$\prod_{t=1}^\infty [0,f^t(k_1)]^2$$</span> with <span class="math-container">$f^0$</span> the identity function <span class="math-container">$f^{t+1}=f\circ f^t$</span> is compact and <span class="math-container">$F$</span> is a subset of this compact set.  By definition, all coordinate functions are continuous. Each of the relevant inequalities defines a closed set and <span class="math-container">$F$</span> is, therefore, a closed subset of a compact set and, therefore, compact itself. Also, the plan that never invests and consumes everything is in <span class="math-container">$F$</span>, so <span class="math-container">$F$</span> is nonempty.</p>
<p>The utility function <span class="math-container">$U:F\to\mathbb{R}$</span> given by
<span class="math-container">$$U(c_1,k_1,c_2,k_2,\ldots)=\sum_{t=1}^\infty \beta^t u(c_t)$$</span>
is well-defined and continuous in the product topology. Since the utility function is bounded, <span class="math-container">$U$</span> will always be finite and, therefore, well-defined. To see that it is continuous, note that since <span class="math-container">$u$</span> is bounded, there exists for each <span class="math-container">$\epsilon&gt;0$</span> some <span class="math-container">$T$</span> such that the utility of any two paths that only differ at times later than <span class="math-container">$T$</span> can differ by at most <span class="math-container">$\epsilon/2$</span>. Since all coordinate functions are continuous, if the two consumption plans are close enough in the first <span class="math-container">$T$</span> coordinates, they will differ at most by <span class="math-container">$\epsilon/2$</span>. So if the paths are close enough at finitely many coordinates, the corresponding utility will be close enough. So <span class="math-container">$U$</span> is continuous in the product topology.</p>
<p>So <span class="math-container">$U$</span> is a continuous function on the nonempty compact set <span class="math-container">$F$</span>. By the extreme value theorem of Weierstrass, <span class="math-container">$U$</span> takes on a maximum at some point in <span class="math-container">$F$</span> and such a point is an optimal plan.</p>
<p>You can find a more general proof of the existence of optimal plans along these lines in the book &quot;Dynamic Programming in Economics&quot; by Le Van and Dana. The assumption that <span class="math-container">$F$</span> is increasing is not needed, you can replace the product of intervals that includes <span class="math-container">$F$</span> by bounding it with the largest values that can be produced instead of the one that comes from always reinvesting capital. That <span class="math-container">$u$</span> is bounded was assumed to guarantee that <span class="math-container">$U$</span> is finite. One can replace this by an assumption that guarantees that the instantaneous utility cannot increase to
fast along a feasible path.</p>

CC BY-SA 4.0
2020-10-30T20:16:33.340Z
Thank you Michael, very informative! Alongside that, thank you for the book recomendation!
EB3112
CC BY-SA 4.0
2020-10-31T00:41:02.873Z
Why is $F$ compact?
Michael
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2020-10-31T01:11:41.910Z
@Michael The set $\prod_{t=1}^\infty [0,f^{t-1}(k)]^2$ is compact by [Tychonoff's theorem](https://en.wikipedia.org/wiki/Tychonoff%27s_theorem) and the continuity of $f$ implies that $F$ is a closed subset.
Michael Greinecker
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2020-10-31T13:48:32.947Z
Hi Michael, could I just confirm with you that I am understanding the point above? Namely, the utility function is well-defined and continuous; and the production function is compact (by Tychonoff's theorem). Therefore the EVT is confirmed/an optimal plan exists.
EB3112
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2020-11-01T14:16:02.177Z
@AndrewCoyle It is not the production function that is compact, it is the set of feasible production ad consumption paths.
Michael Greinecker
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2020-11-01T22:05:59.917Z
Seems that additional assumption is needed to ensure that $f(k) \leq k$, which then implies $F$ is bounded in $l^{\infty}$. E.g. take $f(k) = \sqrt{k} + y$ where $y > 0$ is a constant, which is concave. Take $c_t = 0$ and $k_{t+1} = f(k_t)$, then the feasible consumption/production plan $\{ (0, k_t) \}_{t\geq 0}$ would not lie in a compact set of the Tychonoff type. (That $f(0) = y >0$ is admittedly uneconomical, and the resulting problem is not well-posed. Requiring $f(0) = 0$ would address the both issues.)
Michael
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2020-11-02T00:28:02.917Z
@Michael Compactness in the product topology is very different than compactness in the norm topology on $\ell^\infty$. The product topology is the topology of pointwise convergence in contrast to the norm topology on $\ell^\infty$, which is the topology of uniform convergence. The production function in your example causes no problems.
Michael Greinecker
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2020-11-02T22:52:46.873Z
@MichaelGreinecker Right, my bad. Compact sets can be of the form $\Pi_i [-a_i, a_i]$ where $a_i \rightarrow \infty$. (The example then seems a bit funny economically. The production function gives "free lunch" but since $u$ is bounded the problem is well-posed.)
Michael
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2020-11-02T23:15:46.020Z
@Michael Exactly. If $u$ was unbounded, one would need to make some assumption that guarantees that utility cannot grow too fast along a feasible path.
Michael Greinecker
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2020-11-03T18:54:14.197Z
Can I ask what difference would be made to this problem was made a two-sector problem (with a final goods sector, and say, some intermediate sector), or the fact that the problem is in finite-horizon? I see for example a lot of reference to infinity above.
EB3112
CC BY-SA 4.0
2020-11-03T19:44:12.643Z
@AndrewCoyle None of that should cause a problem. Things are actually easier with a finite time horizon and you would not need the assumption that $u$ is bounded.
Michael Greinecker
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2020-11-03T20:43:03.573Z
Thanks Michael. With two sectors, would C and K still remain the only concerns in the feasible set?
EB3112
CC BY-SA 4.0
2020-11-03T20:54:21.017Z
@AndrewCoyle What ultimately matters is the possible consumption paths and in defining them, you need to take into account which sector one invests in influences the resulting paths. But the set of feasible consumption paths will still be compact.
Michael Greinecker
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2020-12-12T18:54:06.883Z
the talk about optimal plans might suggest looking into optimal mass transportation theory using the Wasserstein distance via linear programming, which I have not seen applied to EVT problems yet
develarist
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2020-10-30T15:49:14.350Z
2021-03-07T10:59:02.417Z
2021-03-07T10:59:02.417Z
Michael Greinecker
<p>At present, I would need more information to answer this question. (Missing details might be obvious to an economist, my background is in applied math.)</p>
<p>The extreme value theorem (e.g. Theorem 4.16 of Rudin’s <em>Principles of Mathematical Analysis</em>) says that if <span class="math-container">$f$</span> is a continuous real function on a compact metric space, then for a compact subset <span class="math-container">$M$</span>, then the supremum and infimum of <span class="math-container">$f$</span> are achieved at some point(S) within <span class="math-container">$M$</span>.</p>
<p>Examples to keep in mind.</p>
<ul>
<li>Compact sets are not infinite. The function <span class="math-container">$f(x) = 1 - \frac{1}{x}$</span> does not achieve its supremum on the set <span class="math-container">$x \geq 1$</span>.</li>
<li>It is not an if and only if condition. E.g. <span class="math-container">$f(x) = x^2$</span>, it achieves its infimum on <span class="math-container">$(-1,1)$</span>, even though the set is not compact.</li>
</ul>
<p>The question does not specify all the constraints for the mathematical system. All I can offer is the following points if one wants to apply the theorem.</p>
<ul>
<li>The time horizon has to be finite, as otherwise the feasible solution set would  not be compact (if it is non-empty).</li>
<li>At each time point, it is necessary to show that the feasible values at that time point is compact. If we have linear constraints (budget constraints), a non-negative condition on variables may be enough for this.</li>
<li>Constraints <span class="math-container">$F_F$</span> have to either hold with equality, or be non-strict inequalities. A strict inequality could be disqualifying (it would be OK if it was not binding.)</li>
<li>It might only be necessary to show that the feasible set lies with a bounded set, and is non-empty (e.g. find one solution).</li>
<li>Continuity of the utility function should be straightforward.</li>
</ul>

CC BY-SA 4.0
2020-10-29T20:15:18.473Z
Thank you Brian, your answer is very informative.
EB3112
CC BY-SA 4.0
2020-10-30T00:56:15.290Z
As an economics point, it doesn't matter if the maximum utility is negative. The maximum is still the maximum. Behaviorally, people should still act the same, at least up until prospect theory.
RegressForward
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2020-10-30T01:31:43.783Z
My understanding is that the non-negative condition referred to is on variables like consumption, investment, etc. Otherwise, the feasible set would be unbounded.
Brian Romanchuk
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2021-03-09T12:24:12.240Z
I am a bit confused by your statement "Compact sets are not infinite". Clearly there are compact, infinite sets, did you mean under the discrete topology? Or are you referring to the "length" of the set?
Alba
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2021-03-09T12:40:03.843Z
@Alba - I am using the definition of compact as in Rudin. On the real line, a set that is infinite is not compact. As my example  f(x) = 1 - 1/x (x=>1) shows, the supremum is not achieved (since “infinity” is not normally considered a point on the real line), thus showing the IVT does not apply to such a set.
Brian Romanchuk
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2021-03-09T20:41:05.140Z
Alright, thanks for explaining. I thought you were referring to cardinality, but I see how that was a misunderstanding.
Alba
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2020-10-29T20:06:43.927Z
2020-10-29T20:06:43.927Z
Brian Romanchuk
<p>Would appreciate some thoughts on proving the existence of a solution (a maximum) in an applied context.</p>
<p>Suppose the objective function is one of maximising utility:</p>
<p><span class="math-container">$$\max\Sigma\beta U(C_t)$$</span></p>
<p>Subject to some constraints on resources, say:</p>
<p><span class="math-container">$$\\C_t+K_{t+1} = F_F(K_F,E_F,S_t)$$</span></p>
<p>Where the constraint is the economy's resource constraint, inclusive of current consumption <span class="math-container">$C_t$</span> and investment possibilities <span class="math-container">$K_{t+1}$</span>, set equal to the production function <span class="math-container">$F_F$</span>, which includes arguments on capital <span class="math-container">$K_t$</span>, energy, <span class="math-container">$E_t$</span> and the emissions stock <span class="math-container">$S_t$</span> (i.e. emissions affect production costs).</p>
<p>The extreme value theorem states 'that a continuous function throughout a non-empty compact set will have a max (min. respectively)'.</p>
<p>In this respect, my questions is: what are the requirements for the feasible set of solutions to be on a compact set?</p>
<ul>
<li>non-negativity of variables?</li>
<li>terminal conditions on the variables (i.e. they are all exhausted in the future)?</li>
</ul>
<p>Would appreciated some thoughts, for sure.</p>

CC BY-SA 4.0
2020-10-29T18:41:28.967Z
I’m a bit confused. You are attempting to apply the theorem? The theorem says nothing about “non-negativity.” To apply it, you need to prove that the set of feasible solutions is compact (and the function is continuous).
Brian Romanchuk
CC BY-SA 4.0
2020-10-29T18:52:57.857Z
Thank you Brian. You put your finger on it better than I could've. How would you prove/state that the set of feasible solutions is compact? What would be the economic analog required for this to be upheld?
EB3112
CC BY-SA 4.0
2020-10-29T19:11:03.897Z
I realise after I wrote that what you meant. You probably need to add the non-negative constraints to ensure compactness. I’ll make a stab at an answer. I’m an applied mathematician, and not an economist, and it might help if you specified the meaning of the variables, as well any missing dynamics. That’s why I had a hard time parsing the question.
Brian Romanchuk
CC BY-SA 4.0
2020-10-29T19:15:13.283Z
Thank you Brian, your time is very appreciated. The variables in the constraint are capital, energy, and the subsequent emissions. (i.e. it is assumed that a producer, uses the two resources, but emissions impose a cost; so they're included with a constraint). The overall objective function however, is maximising welfare/utility.
EB3112
CC BY-SA 4.0
2020-10-29T19:19:48.810Z
At the minimum, it needs to be a finite horizon problem, as the solution set would not be compact. After that, you would need to show that the set of feasible solutions is compact at each finite time point, which is not going to be possible with an arbitrary nonlinear constraint function $F_F$.
Brian Romanchuk
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2020-10-29T19:25:48.690Z
Yeah, I think that's a sound observation. Is that to say that the there are some terminal conditions on the variables?
EB3112
CC BY-SA 4.0
2020-10-29T19:39:54.740Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/115662/discussion-between-brian-romanchuk-and-andrew-coyle).
Brian Romanchuk
CC BY-SA 4.0
2020-10-30T07:50:42.923Z
Can you please [edit your post](https://economics.stackexchange.com/posts/40533/edit) to make the setup of the problem a little clearer? What is(are) the choice variable(s) in the objective function? Are the $u_t$'s functions or numbers? Which set does $t$ belong, $\mathbb N$ or some finite subset of it? Are you sure there is no discounting in the objective? What are $F_F$, $K_F$, $E_F$, and $S_t$? Are these exogenous or endogenous? If endogenous, how?
Herr K.
CC BY-SA 4.0
2020-10-30T11:43:47.357Z
Hi Herr, I tried to add some more content to the question.
EB3112
<mathematical-economics>
Extreme Value Theorem in Economics
EB3112
# Extreme Value Theorem in Economics

(Asked by: username_0 on 2020-10-29)

Would appreciate some thoughts on proving the existence of a solution (a maximum) in an applied context.

Suppose the objective function is one of maximising utility:

$$\max\Sigma\beta U(C\_t)$$

Subject to some constraints on resources, say:

$$\\C\_t+K\_{t+1} = F\_F(K\_F,E\_F,S\_t)$$

Where the constraint is the economy's resource constraint, inclusive of current consumption $C\_t$ and investment possibilities $K\_{t+1}$, set equal to the production function $F\_F$, which includes arguments on capital $K\_t$, energy, $E\_t$ and the emissions stock $S\_t$ (i.e. emissions affect production costs).

The extreme value theorem states 'that a continuous function throughout a non-empty compact set will have a max (min. respectively)'.

In this respect, my questions is: what are the requirements for the feasible set of solutions to be on a compact set?

* non-negativity of variables?
* terminal conditions on the variables (i.e. they are all exhausted in the future)?

Would appreciated some thoughts, for sure.

**Tags:** mathematical-economics

> **Comments:**
> **username_1**: I’m a bit confused. You are attempting to apply the theorem? The theorem says nothing about “non-negativity.” To apply it, you need to prove that the set of feasible solutions is compact (and the function is continuous).  
> **username_0**: Thank you username_1. You put your finger on it better than I could've. How would you prove/state that the set of feasible solutions is compact? What would be the economic analog required for this to be upheld?  
> **username_1**: I realise after I wrote that what you meant. You probably need to add the non-negative constraints to ensure compactness. I’ll make a stab at an answer. I’m an applied mathematician, and not an economist, and it might help if you specified the meaning of the variables, as well any missing dynamics. That’s why I had a hard time parsing the question.  
> **username_0**: Thank you username_1, your time is very appreciated. The variables in the constraint are capital, energy, and the subsequent emissions. (i.e. it is assumed that a producer, uses the two resources, but emissions impose a cost; so they're included with a constraint). The overall objective function however, is maximising welfare/utility.  
> **username_1**: At the minimum, it needs to be a finite horizon problem, as the solution set would not be compact. After that, you would need to show that the set of feasible solutions is compact at each finite time point, which is not going to be possible with an arbitrary nonlinear constraint function $F_F$.  
> **username_0**: Yeah, I think that's a sound observation. Is that to say that the there are some terminal conditions on the variables?  
> **username_1**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/115662/discussion-between-username_1-username_1-and-andrew-coyle).  
> **username_2**: Can you please [edit your post](https://economics.stackexchange.com/posts/40533/edit) to make the setup of the problem a little clearer? What is(are) the choice variable(s) in the objective function? Are the $u_t$'s functions or numbers? Which set does $t$ belong, $\mathbb N$ or some finite subset of it? Are you sure there is no discounting in the objective? What are $F_F$, $K_F$, $E_F$, and $S_t$? Are these exogenous or endogenous? If endogenous, how?  
> **username_0**: Hi username_2, I tried to add some more content to the question.  

---

## 2 answers

### Answer by username_3 on 2020-10-30. Score: 4

One can prove the existence of such optimal plans using the extreme value theorem of Weierstrass, but it requires some advanced math.

Here is a toy version of the model without energy and emissions. Both the instantaneous utility functions $u:\mathbb{R}\_+\to\mathbb{R}$ and the production function $f:\mathbb{R}\_+\to\mathbb{R}$ are assumed to be continuous and nondecreasing. Moreover, $u$ is assumed to be bounded (!). There is given initial capital stock $k\_1\geq 0$. The space of feasible consumption and production plans is defined as $$F=\big\{(c\_1,k\_1,c\_2,k\_2,\ldots)\mid 0\leq k\_{t+1}\leq f(k\_t-c\_t)\}, k\_t\geq c\_t\geq 0\big\}.$$
$F$ describes all feasible paths of consumption and capital.
This set is a nonempty compact subset of $\mathbb{R}^\infty$ endowed with the [product topology](https://en.wikipedia.org/wiki/Product_topology). Here is why: By [Tychonoff's theorem](https://en.wikipedia.org/wiki/Tychonoff%27s_theorem), the set $$\prod\_{t=1}^\infty [0,f^t(k\_1)]^2$$ with $f^0$ the identity function $f^{t+1}=f\circ f^t$ is compact and $F$ is a subset of this compact set. By definition, all coordinate functions are continuous. Each of the relevant inequalities defines a closed set and $F$ is, therefore, a closed subset of a compact set and, therefore, compact itself. Also, the plan that never invests and consumes everything is in $F$, so $F$ is nonempty.

The utility function $U:F\to\mathbb{R}$ given by
$$U(c\_1,k\_1,c\_2,k\_2,\ldots)=\sum\_{t=1}^\infty \beta^t u(c\_t)$$
is well-defined and continuous in the product topology. Since the utility function is bounded, $U$ will always be finite and, therefore, well-defined. To see that it is continuous, note that since $u$ is bounded, there exists for each $\epsilon>0$ some $T$ such that the utility of any two paths that only differ at times later than $T$ can differ by at most $\epsilon/2$. Since all coordinate functions are continuous, if the two consumption plans are close enough in the first $T$ coordinates, they will differ at most by $\epsilon/2$. So if the paths are close enough at finitely many coordinates, the corresponding utility will be close enough. So $U$ is continuous in the product topology.

So $U$ is a continuous function on the nonempty compact set $F$. By the extreme value theorem of Weierstrass, $U$ takes on a maximum at some point in $F$ and such a point is an optimal plan.

You can find a more general proof of the existence of optimal plans along these lines in the book "Dynamic Programming in Economics" by Le Van and Dana. The assumption that $F$ is increasing is not needed, you can replace the product of intervals that includes $F$ by bounding it with the largest values that can be produced instead of the one that comes from always reinvesting capital. That $u$ is bounded was assumed to guarantee that $U$ is finite. One can replace this by an assumption that guarantees that the instantaneous utility cannot increase to
fast along a feasible path.

> **Comments:**
> **username_0**: Thank you username_3, very informative! Alongside that, thank you for the book recomendation!  
> **username_4**: Why is $F$ compact?  
> **username_3**: @username_4 The set $\prod_{t=1}^\infty [0,f^{t-1}(k)]^2$ is compact by [Tychonoff's theorem](https://en.wikipedia.org/wiki/Tychonoff%27s_theorem) and the continuity of $f$ implies that $F$ is a closed subset.  
> **username_0**: Hi username_3, could I just confirm with you that I am understanding the point above? Namely, the utility function is well-defined and continuous; and the production function is compact (by Tychonoff's theorem). Therefore the EVT is confirmed/an optimal plan exists.  
> **username_3**: @AndrewCoyle It is not the production function that is compact, it is the set of feasible production ad consumption paths.  
> **username_4**: Seems that additional assumption is needed to ensure that $f(k) \leq k$, which then implies $F$ is bounded in $l^{\infty}$. E.g. take $f(k) = \sqrt{k} + y$ where $y > 0$ is a constant, which is concave. Take $c_t = 0$ and $k_{t+1} = f(k_t)$, then the feasible consumption/production plan $\{ (0, k_t) \}_{t\geq 0}$ would not lie in a compact set of the Tychonoff type. (That $f(0) = y >0$ is admittedly uneconomical, and the resulting problem is not well-posed. Requiring $f(0) = 0$ would address the both issues.)  
> **username_3**: @username_4 Compactness in the product topology is very different than compactness in the norm topology on $\ell^\infty$. The product topology is the topology of pointwise convergence in contrast to the norm topology on $\ell^\infty$, which is the topology of uniform convergence. The production function in your example causes no problems.  
> **username_4**: @username_3 Right, my bad. Compact sets can be of the form $\Pi_i [-a_i, a_i]$ where $a_i \rightarrow \infty$. (The example then seems a bit funny economically. The production function gives "free lunch" but since $u$ is bounded the problem is well-posed.)  
> **username_3**: @username_4 Exactly. If $u$ was unbounded, one would need to make some assumption that guarantees that utility cannot grow too fast along a feasible path.  
> **username_0**: Can I ask what difference would be made to this problem was made a two-sector problem (with a final goods sector, and say, some intermediate sector), or the fact that the problem is in finite-horizon? I see for example a lot of reference to infinity above.  
> **username_3**: @AndrewCoyle None of that should cause a problem. Things are actually easier with a finite time horizon and you would not need the assumption that $u$ is bounded.  
> **username_0**: Thanks username_3. With two sectors, would C and K still remain the only concerns in the feasible set?  
> **username_3**: @AndrewCoyle What ultimately matters is the possible consumption paths and in defining them, you need to take into account which sector one invests in influences the resulting paths. But the set of feasible consumption paths will still be compact.  
> **username_5**: the talk about optimal plans might suggest looking into optimal mass transportation theory using the Wasserstein distance via linear programming, which I have not seen applied to EVT problems yet  

---

### Answer by username_1 on 2020-10-29. Score: 1

At present, I would need more information to answer this question. (Missing details might be obvious to an economist, my background is in applied math.)

The extreme value theorem (e.g. Theorem 4.16 of Rudin’s *Principles of Mathematical Analysis*) says that if $f$ is a continuous real function on a compact metric space, then for a compact subset $M$, then the supremum and infimum of $f$ are achieved at some point(S) within $M$.

Examples to keep in mind.

* Compact sets are not infinite. The function $f(x) = 1 - \frac{1}{x}$ does not achieve its supremum on the set $x \geq 1$.
* It is not an if and only if condition. E.g. $f(x) = x^2$, it achieves its infimum on $(-1,1)$, even though the set is not compact.

The question does not specify all the constraints for the mathematical system. All I can offer is the following points if one wants to apply the theorem.

* The time horizon has to be finite, as otherwise the feasible solution set would not be compact (if it is non-empty).
* At each time point, it is necessary to show that the feasible values at that time point is compact. If we have linear constraints (budget constraints), a non-negative condition on variables may be enough for this.
* Constraints $F\_F$ have to either hold with equality, or be non-strict inequalities. A strict inequality could be disqualifying (it would be OK if it was not binding.)
* It might only be necessary to show that the feasible set lies with a bounded set, and is non-empty (e.g. find one solution).
* Continuity of the utility function should be straightforward.

> **Comments:**
> **username_0**: Thank you username_1, your answer is very informative.  
> **username_6**: As an economics point, it doesn't matter if the maximum utility is negative. The maximum is still the maximum. Behaviorally, people should still act the same, at least up until prospect theory.  
> **username_1**: My understanding is that the non-negative condition referred to is on variables like consumption, investment, etc. Otherwise, the feasible set would be unbounded.  
> **username_7**: I am a bit confused by your statement "Compact sets are not infinite". Clearly there are compact, infinite sets, did you mean under the discrete topology? Or are you referring to the "length" of the set?  
> **username_1**: @username_7 - I am using the definition of compact as in Rudin. On the real line, a set that is infinite is not compact. As my example  f(x) = 1 - 1/x (x=>1) shows, the supremum is not achieved (since “infinity” is not normally considered a point on the real line), thus showing the IVT does not apply to such a set.  
> **username_7**: Alright, thanks for explaining. I thought you were referring to cardinality, but I see how that was a misunderstanding.  

---

