<p>The sequence you proposed <span class="math-container">$0,1,0,2,0,3,\ldots$</span> i.e. <span class="math-container">$x_n=\begin{cases}0 &amp;\text{if $n$ is odd} \\ \frac{n}{2} &amp; \text{if $n$ is even} \end{cases}$</span> is not a bounded sequence and is therefore, not from a compact set. The following result holds for sequences from compact sets:</p>
<p>In a compact metric space <span class="math-container">$(X,d)$</span>, if every convergent subsequence of a sequence converges to the same limit, say <span class="math-container">$l$</span>, then the original sequence also converges to <span class="math-container">$l$</span>. [ Ref: See this result- <a href="https://math.stackexchange.com/q/461610/378131">https://math.stackexchange.com/q/461610/378131</a> ]</p>

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2024-05-14T05:17:12.453Z
2024-05-14T05:17:12.453Z
Amit
<p><em><strong>See Proposition 3.C.1 from MWG</strong></em></p>
<p>Continue from this <a href="https://economics.stackexchange.com/questions/58392/proof-let-epsilon0-and-x-in-mathbbrl-be-such-that-x-x-geq-eps">post</a>, <em>Microeconomic Theory</em> by Mas-Colell et al. book said the following:</p>
<blockquote>
<p>What remains is to establish that all convergent subsequences of <span class="math-container">$\{\alpha(x^n)\}_{n=1}^{\infty}$</span> converge to <span class="math-container">$\alpha(x)$</span>.</p>
</blockquote>
<p>The book then proceed to prove this and conclude:</p>
<blockquote>
<p>Since all convergent subsequences of <span class="math-container">$\{\alpha(x^n)\}_{n=1}^{\infty}$</span> must converge to <span class="math-container">$\alpha(x)$</span>, we have <span class="math-container">$\lim_{n\to\infty}\alpha(x^n)=\alpha(x)$</span>, and we are done.</p>
</blockquote>
<p>I seriously think this is wrong. Consider the sequence <span class="math-container">$0,1,0,2,0,3,0,4,0,5,\dots$</span>. Every convergent subsequence of this sequence converges to 0, but this sequence diverges.</p>
<p><em><strong>We need to prove every subsequence (not just the convergent ones) of <span class="math-container">$\alpha(x^n)$</span> converges to <span class="math-container">$\alpha(x)$</span></strong></em> (although this idea is not helpful because the original sequence is a subsequence of itself).</p>
<p>What am I missing here? How should we fix this? Thank you very much!</p>
<hr />
<p>Here is the full proof of the final step of Proposition 3.C.1 from MWG:
<a href="https://i.sstatic.net/82X77a8T.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/82X77a8T.png" alt="enter image description here" /></a>
<a href="https://i.sstatic.net/AJO5K4W8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/AJO5K4W8.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2024-05-13T19:12:07.280Z
Please keep including explicit references (as edited by me in your last two posts and now this one) in subsequent posts instead of  referring to "the book". Thank you!
Richard Hardy
CC BY-SA 4.0
2024-05-13T19:49:58.713Z
@RichardHardy Actually, the "MWG" stands for the book *Microeconomic Theory*.
Champa
CC BY-SA 4.0
2024-05-14T05:37:43.870Z
Acronyms are helpful when they are introduced properly. That ensures the reader is on the same page as the writer and hardly costs anything to include. Failing to introduce an acronym may often be a sign of sloppiness, though some may perceive it as disrespect for the reader, too.
Richard Hardy
<microeconomics><mathematical-economics><utility><preferences><theory>
Question About Proving $\alpha(\cdot)$ is Continuous in the Proof of Proposition 3.C.1 from MWG
Champa
# Question About Proving $\alpha(\cdot)$ is Continuous in the Proof of Proposition 3.C.1 from MWG

(Asked by: username_0 on 2024-05-13)

***See Proposition 3.C.1 from MWG***

Continue from this [post](https://economics.stackexchange.com/questions/58392/proof-let-epsilon0-and-x-in-mathbbrl-be-such-that-x-x-geq-eps), *Microeconomic Theory* by Mas-Colell et al. book said the following:

> What remains is to establish that all convergent subsequences of $\{\alpha(x^n)\}\_{n=1}^{\infty}$ converge to $\alpha(x)$.

The book then proceed to prove this and conclude:

> Since all convergent subsequences of $\{\alpha(x^n)\}\_{n=1}^{\infty}$ must converge to $\alpha(x)$, we have $\lim\_{n\to\infty}\alpha(x^n)=\alpha(x)$, and we are done.

I seriously think this is wrong. Consider the sequence $0,1,0,2,0,3,0,4,0,5,\dots$. Every convergent subsequence of this sequence converges to 0, but this sequence diverges.

***We need to prove every subsequence (not just the convergent ones) of $\alpha(x^n)$ converges to $\alpha(x)$*** (although this idea is not helpful because the original sequence is a subsequence of itself).

What am I missing here? How should we fix this? Thank you very much!

---

Here is the full proof of the final step of Proposition 3.C.1 from MWG:
[![enter image description here](https://i.sstatic.net/82X77a8T.png)](https://i.sstatic.net/82X77a8T.png)
[![enter image description here](https://i.sstatic.net/AJO5K4W8.png)](https://i.sstatic.net/AJO5K4W8.png)

**Tags:** microeconomics, mathematical-economics, utility, preferences, theory
> **Comments:**
> **username_1**: Please keep including explicit references (as edited by me in your last two posts and now this one) in subsequent posts instead of  referring to "the book". Thank you!  
> **username_0**: @RichardHardy Actually, the "MWG" stands for the book *Microeconomic Theory*.  
> **username_1**: Acronyms are helpful when they are introduced properly. That ensures the reader is on the same page as the writer and hardly costs anything to include. Failing to introduce an acronym may often be a sign of sloppiness, though some may perceive it as disrespect for the reader, too.  

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## 1 answers

### Answer by username_2 on 2024-05-14. Score: 3 | Accepted answer

The sequence you proposed $0,1,0,2,0,3,\ldots$ i.e. $x\_n=\begin{cases}0 &\text{if $n$ is odd} \\ \frac{n}{2} & \text{if $n$ is even} \end{cases}$ is not a bounded sequence and is therefore, not from a compact set. The following result holds for sequences from compact sets:

In a compact metric space $(X,d)$, if every convergent subsequence of a sequence converges to the same limit, say $l$, then the original sequence also converges to $l$. [ Ref: See this result- <https://math.stackexchange.com/q/461610/378131> ]


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