<p>What the book calls the "Weierstrass Theorem" is more commonly known as the <a href="https://en.wikipedia.org/wiki/Extreme_value_theorem" rel="nofollow noreferrer"><strong>Extreme Value Theorem</strong></a>, which states that a continuous function defined on a compact domain attains a maximum and a minimum.</p>

<p><a href="https://en.wikipedia.org/wiki/Extreme_value_theorem#Proof_of_the_extreme_value_theorem" rel="nofollow noreferrer">A common proof</a> of this theorem involves the use of the <a href="https://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem" rel="nofollow noreferrer"><strong>Bolzano–Weierstrass theorem</strong></a>, which you learned in your math course, and which says every bounded sequence has a convergent subsequence. </p>

<p>So the two theorems are actually different, albeit related, statements.</p>

CC BY-SA 4.0
2018-08-22T18:57:55.337Z
2018-08-22T18:57:55.337Z
Herr K.
<p>Weierstrass Theorem states that any bounded sequence has a convergent subsequence.</p>

<p>I did that in my maths course and understood it completely. But when I was learning optimization techniques in economics, the definition by the book Sundaram was tweaked slightly to make it fit into the understanding of optimization. It goes like this - (see Figure 1)</p>

<p>If you read it, it is saying just what original theorem said. The only difference is that, instead of starting off with a bounded sequence and reaching the conclusion that it will have convergent subsequence, it started off with a convergent subsequence (by assuming Compact set) and reached the conclusion that it should be bounded. It makes sense right? It did to me too until I found a result on compact set, stated on the same book. (See figure 2)</p>

<p>This redefining of the theorem made it a circular definition - if a set can only be compact if n only if it is bounded then by assuming Compactness in redefinition we implicitly assumed boundedness. </p>

<p>So, my question is please tell me where am I going wrong. Have I understood the redefinition incorrectly or missed so point? </p>

<p><img src="https://i.sstatic.net/GTa4F.jpg" alt="Figure 1">
<img src="https://i.sstatic.net/s72gs.jpg" alt="Figure 2"></p>

CC BY-SA 4.0
2018-08-22T19:03:48.043Z
As a matter of writing style, may I suggest that you should perhaps avoid the excessive use of the pronoun "it", whose referent is extremely unclear, especially in the second and third paragraphs.
Herr K.
CC BY-SA 4.0
2018-08-23T04:04:42.040Z
I don't see what you consider circular here.
Michael Greinecker
CC BY-SA 4.0
2018-08-23T08:15:47.150Z
Compactness needs boundedness and closed bounded sequence is compact. If the 'let us consider the compact set' took care of the closed and bounded set (already a theorem [1.21] which established their relationship), then concluding it with the fact that the set will have a maximum and minimum is obvious. This redefinition took away the energy of the original Weierstrass Theorem. @Michael Greinecker
Elina Gilbert
<optimization>
Weierstrass Theorem in Optimization
Elina Gilbert
# Weierstrass Theorem in Optimization

(Asked by: username_0 on 2018-08-22)

Weierstrass Theorem states that any bounded sequence has a convergent subsequence.

I did that in my maths course and understood it completely. But when I was learning optimization techniques in economics, the definition by the book Sundaram was tweaked slightly to make it fit into the understanding of optimization. It goes like this - (see Figure 1)

If you read it, it is saying just what original theorem said. The only difference is that, instead of starting off with a bounded sequence and reaching the conclusion that it will have convergent subsequence, it started off with a convergent subsequence (by assuming Compact set) and reached the conclusion that it should be bounded. It makes sense right? It did to me too until I found a result on compact set, stated on the same book. (See figure 2)

This redefining of the theorem made it a circular definition - if a set can only be compact if n only if it is bounded then by assuming Compactness in redefinition we implicitly assumed boundedness.

So, my question is please tell me where am I going wrong. Have I understood the redefinition incorrectly or missed so point?

![Figure 1](https://i.sstatic.net/GTa4F.jpg)
![Figure 2](https://i.sstatic.net/s72gs.jpg)

**Tags:** optimization
> **Comments:**
> **username_1**: As a matter of writing style, may I suggest that you should perhaps avoid the excessive use of the pronoun "it", whose referent is extremely unclear, especially in the second and third paragraphs.  
> **username_2**: I don't see what you consider circular here.  
> **username_0**: Compactness needs boundedness and closed bounded sequence is compact. If the 'let us consider the compact set' took care of the closed and bounded set (already a theorem [1.21] which established their relationship), then concluding it with the fact that the set will have a maximum and minimum is obvious. This redefinition took away the energy of the original Weierstrass Theorem. @Michael Greinecker  

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## 1 answers

### Answer by username_1 on 2018-08-22. Score: 4

What the book calls the "Weierstrass Theorem" is more commonly known as the [**Extreme Value Theorem**](https://en.wikipedia.org/wiki/Extreme_value_theorem), which states that a continuous function defined on a compact domain attains a maximum and a minimum.

[A common proof](https://en.wikipedia.org/wiki/Extreme_value_theorem#Proof_of_the_extreme_value_theorem) of this theorem involves the use of the [**Bolzano–Weierstrass theorem**](https://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem), which you learned in your math course, and which says every bounded sequence has a convergent subsequence.

So the two theorems are actually different, albeit related, statements.


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