<p>We don't need <span class="math-container">$X$</span> (domain of the utility) to be equal to the set of its limit points to use the sequential characterisation of continuity. In other words, the sequential characterisation of continuity also holds when <span class="math-container">$X$</span> is not equal to the set of its limit points.</p>
<p>Given <span class="math-container">$x\in\mathbb{R}^L_+$</span>, to get <span class="math-container">$\alpha_0,\alpha_1$</span>,
consider
<span class="math-container">$\alpha_0=0$</span> and
<span class="math-container">$\alpha_1=\max_{1\leq i \leq L} x_i+\epsilon$</span></p>
<p>Check that for any <span class="math-container">$x'\in\mathbb{R}^L_+$</span> such that <span class="math-container">$||x'-x||\leq\epsilon$</span>, <span class="math-container">$\alpha(x')\in [\alpha_0,\alpha_1]$</span>.</p>

CC BY-SA 4.0
2024-05-14T11:18:11.417Z
Indeed, $X$ need not be closed or equal to the set of its limit points. But every point of $X$ have to be a limit point of $X$, in order for the sequential characterization of continuity to hold for all $x\in X$. In other words, let $E$ be the set of limit points of $X$, we need $X\subseteq E$.
Champa
CC BY-SA 4.0
2024-05-14T11:28:59.603Z
I don't think this condition is required. Please provide the definition of limit point of the set you are using. I'll try and provide a proof of characterisation.
Amit
CC BY-SA 4.0
2024-05-14T14:19:47.870Z
Consider, for example, $X=[0,2]\bigcup\{4\}\bigcup(8,10]$ and the function $f:X\to\mathbb{R}$.  Then $f$ is always continuous at $x=4$ (according to the $\epsilon-\delta$ definition of continuity). However, the sequential characterization of continuity does not hold at $x=4$ because it is not a limit point.
Champa
CC BY-SA 4.0
2024-05-14T14:35:33.790Z
Sequential characterisation holds because any sequence $(x_n)$ that converges to 4 consists of constant 4 in all but finitely many terms of the sequence. Consequently, $\lim f(x_n)=  \lim f(4) = f(4)$ for all such sequences.
Amit
CC BY-SA 4.0
2024-05-14T14:41:59.470Z
Not exactly. By definition (see Baby Rudin), a point $p$ is a *limit point* of the set $E$ if *every* neighborhood of $p$ contains a point $q\neq p$ such that $q\in E$. As a result, "a finite point set has no limit points". Moreover, the sequential characterization of continuity says that a function $f$ is continuous at a *limit point* $p$ if and only if $\lim_{x\top}f(x)=f(p)$.
Champa
CC BY-SA 4.0
2024-05-14T14:54:06.153Z
@Champa It is not true that  the sequential charaterization says necessarily that the point must be a limit point. Some texts may say that because they consider obviuous what Amit said in the comment above, that at an isolated point the condition is obvious, and they focus on the more interesting case of a limit point.
BakerStreet
CC BY-SA 4.0
2024-05-14T15:06:23.367Z
@BakerStreet I see. Yes, that does makes sense. Thanks a lot!
Champa
CC BY-SA 4.0
2024-05-14T15:07:00.383Z
@Amit Thank you very much! Now I get your point.
Champa
CC BY-SA 4.0
2024-05-14T15:10:13.603Z
@Champa You are welcome. You are right that this matter in some text can be confusing, but many text don't make this confusion. The point is exactly what Amit said.
BakerStreet
CC BY-SA 4.0
2024-05-14T17:09:30.097Z
@Amit May I ask how would you prove $\alpha(x')\leq\alpha_1$, where $\alpha_1=\max\{x_1,\dots,x_L\}+\epsilon$. I wanted to prove it by contradiction, and I think the contradiction should be that if $\alpha(x')>\alpha_1$ then $\|x'-x\|$ would be bigger than $\epsilon$. But I couldn't formalize a proof. Can you please help me with it?
Champa
CC BY-SA 4.0
2024-05-14T17:38:31.300Z
@Amit Does this work: Assume to the contrary that $\alpha(x')>\alpha_1$. Let $x''=(x_1+\epsilon,\dots,x_L+\epsilon)$. Then $\alpha(x')e\gg x''$ and the monotonicity implies $x'\sim\alpha(x')e\succ x''$. However, $x_i''=x_i+\epsilon\geq x_i'$, because otherwise $\|x'-x\|=\sqrt{\sum_{i=1}^L(x_i'-x_i)^2}>\epsilon$. Thus $x''\geq x'$ and the lemma in this post (https://economics.stackexchange.com/questions/58383/question-about-the-step-proving-alphax-represents-preferences-in-the-proof) implies $x''\succsim x'$. We have a contradiction.
Champa
CC BY-SA 4.0
2024-05-15T05:05:36.770Z
Consider any $x'\in\mathbb{R}^L_+$ such that $||x'-x||\leq \epsilon$. This implies that $|x_i'-x_i|\leq \epsilon$ for all $i\in\{1,2,\ldots,L\}$. This further implies that $x_i'\leq x_i+\epsilon\leq \alpha_1$ for all $i\in\{1,2,\ldots,L\}$. Therefore, by monotonicity, $\alpha(x')\leq \alpha_1$.
Amit
CC BY-SA 4.0
2024-05-14T11:10:39.647Z
2024-05-14T11:15:56.333Z
2024-05-14T11:15:56.333Z
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/58383/question-about-the-step-proving-alphax-represents-preferences-in-the-proof">post</a>, the book (MWG) then started the proof that <span class="math-container">$\alpha(x)$</span> is a continuous function:</p>
<blockquote>
<p>We now argue that <span class="math-container">$\alpha(x)$</span> is a continuous function at all <span class="math-container">$x$</span>; that is, for any sequence <span class="math-container">$\{x^n\}_{n=1}^{\infty}$</span> with <span class="math-container">$x=\lim_{n\to\infty}x^n$</span>, we have <span class="math-container">$\lim_{n\to\infty}\alpha(x^n)=\alpha(x)$</span>. Hence, consider a sequence <span class="math-container">$\{x^n\}_{n=1}^{\infty}$</span> such that <span class="math-container">$x=\lim_{n\to\infty}x^n$</span>.</p>
</blockquote>
<p><em><strong>Quick question here:</strong></em> The reason that this paragraph is true is based on the fact that every point of <span class="math-container">$\mathbb{R}^L_+$</span> is a limit point of <span class="math-container">$\mathbb{R}^L_+$</span>, am I right? Because otherwise the sequential characterization of continuity will not be sufficient to prove <span class="math-container">$\alpha(x)$</span> is continuous at every <span class="math-container">$x\in\mathbb{R}^L_+$</span>.</p>
<p>Then the book put:</p>
<blockquote>
<p>We note that the sequence <span class="math-container">$\{\alpha(x^n)\}_{n=1}^{\infty}$</span> must have a convergent subsequence. <strong>By monotonicity, for any <span class="math-container">$\epsilon&gt;0$</span>, <span class="math-container">$\alpha(x')$</span> lies in a compact subset of <span class="math-container">$\mathbb{R}_+$</span>, <span class="math-container">$[\alpha_0,\alpha_1]$</span>, for all <span class="math-container">$x'$</span> such that <span class="math-container">$\|x'-x\|\leq\epsilon$</span> (see the figure below).</strong> Since <span class="math-container">$\{x^n\}_{n=1}^{\infty}$</span> converges to <span class="math-container">$x$</span>, there exists an <span class="math-container">$N$</span> such that <span class="math-container">$\alpha(x^n)$</span> lies in this compact set for all <span class="math-container">$n&gt;N$</span>. But any infinite sequence that lies in a compact set must have a convergent subsequence.
<a href="https://i.sstatic.net/bOrQtQUr.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/bOrQtQUr.png" alt="enter image description here" /></a></p>
</blockquote>
<p><em><strong>My question about the bolded part in this paragraph:</strong></em> Intuitively, this makes sense to me, but I had a difficult time proving this rigorously. I rephrased the bolded part as follows:</p>
<blockquote>
<p>Let <span class="math-container">$\epsilon&gt;0$</span>. Let <span class="math-container">$x'\in\mathbb{R}^L_+$</span> be such that <span class="math-container">$\|x'-x\|\leq\epsilon$</span>. Then there are <span class="math-container">$\alpha_0$</span> and <span class="math-container">$\alpha_1$</span> in <span class="math-container">$\mathbb{R}_+$</span> such that <span class="math-container">$\alpha(x')\in[\alpha_0,\alpha_1]$</span>.</p>
</blockquote>
<p>Here is my attempt to prove it:</p>
<blockquote>
<p>Assume to the contrary that there does not exist <span class="math-container">$\alpha_0$</span> and <span class="math-container">$\alpha_1$</span> in <span class="math-container">$\mathbb{R}_+$</span> such that <span class="math-container">$\alpha(x')\in[\alpha_0,\alpha_1]$</span>. Then this means either <span class="math-container">$\alpha(x')$</span> is negative or <span class="math-container">$\alpha(x')=+\infty$</span>. If <span class="math-container">$\alpha(x')&lt;0$</span>, then <span class="math-container">$0\gg\alpha(x')e$</span>, and monotonicity would imply <span class="math-container">$0\sim0\succ\alpha(x')e\sim x'$</span>, contradicts the fact that <span class="math-container">$x\succsim0$</span> for all <span class="math-container">$x\in\mathbb{R}^L_+$</span>. Now suppose <span class="math-container">$\alpha(x')=+\infty$</span>. Denote <span class="math-container">$x'=(x'_1,\dots,x'_L) \in \mathbb{R}^L_+$</span> and let <span class="math-container">$x''=(x'_1+1,\dots,x'_L+1)$</span>. Then <span class="math-container">$x''\gg x'$</span> and so <span class="math-container">$x''\succ x'$</span>. However, <span class="math-container">$\alpha(x'')\leq\alpha(x')=+\infty$</span>, which contradicts the fact that <span class="math-container">$\alpha(\cdot)$</span> represents <span class="math-container">$\succsim$</span>. Therefore, the above statement holds.</p>
</blockquote>
<p>Is my attempt correct?</p>
<p>Thanks a lot for any help!</p>

<microeconomics><mathematical-economics><utility><preferences><theory>
Proof: Let $\epsilon>0$ and $x'\in\mathbb{R}^L_+$ be such that $\|x'-x\|\geq\epsilon$. Then $\alpha(x')$ belongs to some $[\alpha_0,\alpha_1]$
Champa
# Proof: Let $\epsilon>0$ and $x'\in\mathbb{R}^L_+$ be such that $\|x'-x\|\geq\epsilon$. Then $\alpha(x')$ belongs to some $[\alpha_0,\alpha_1]$

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

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

Continue from this [post](https://economics.stackexchange.com/questions/58383/question-about-the-step-proving-alphax-represents-preferences-in-the-proof), the book (MWG) then started the proof that $\alpha(x)$ is a continuous function:

> We now argue that $\alpha(x)$ is a continuous function at all $x$; that is, for any sequence $\{x^n\}\_{n=1}^{\infty}$ with $x=\lim\_{n\to\infty}x^n$, we have $\lim\_{n\to\infty}\alpha(x^n)=\alpha(x)$. Hence, consider a sequence $\{x^n\}\_{n=1}^{\infty}$ such that $x=\lim\_{n\to\infty}x^n$.

***Quick question here:*** The reason that this paragraph is true is based on the fact that every point of $\mathbb{R}^L\_+$ is a limit point of $\mathbb{R}^L\_+$, am I right? Because otherwise the sequential characterization of continuity will not be sufficient to prove $\alpha(x)$ is continuous at every $x\in\mathbb{R}^L\_+$.

Then the book put:

> We note that the sequence $\{\alpha(x^n)\}\_{n=1}^{\infty}$ must have a convergent subsequence. **By monotonicity, for any $\epsilon>0$, $\alpha(x')$ lies in a compact subset of $\mathbb{R}\_+$, $[\alpha\_0,\alpha\_1]$, for all $x'$ such that $\|x'-x\|\leq\epsilon$ (see the figure below).** Since $\{x^n\}\_{n=1}^{\infty}$ converges to $x$, there exists an $N$ such that $\alpha(x^n)$ lies in this compact set for all $n>N$. But any infinite sequence that lies in a compact set must have a convergent subsequence.
> [![enter image description here](https://i.sstatic.net/bOrQtQUr.png)](https://i.sstatic.net/bOrQtQUr.png)

***My question about the bolded part in this paragraph:*** Intuitively, this makes sense to me, but I had a difficult time proving this rigorously. I rephrased the bolded part as follows:

> Let $\epsilon>0$. Let $x'\in\mathbb{R}^L\_+$ be such that $\|x'-x\|\leq\epsilon$. Then there are $\alpha\_0$ and $\alpha\_1$ in $\mathbb{R}\_+$ such that $\alpha(x')\in[\alpha\_0,\alpha\_1]$.

Here is my attempt to prove it:

> Assume to the contrary that there does not exist $\alpha\_0$ and $\alpha\_1$ in $\mathbb{R}\_+$ such that $\alpha(x')\in[\alpha\_0,\alpha\_1]$. Then this means either $\alpha(x')$ is negative or $\alpha(x')=+\infty$. If $\alpha(x')<0$, then $0\gg\alpha(x')e$, and monotonicity would imply $0\sim0\succ\alpha(x')e\sim x'$, contradicts the fact that $x\succsim0$ for all $x\in\mathbb{R}^L\_+$. Now suppose $\alpha(x')=+\infty$. Denote $x'=(x'\_1,\dots,x'\_L) \in \mathbb{R}^L\_+$ and let $x''=(x'\_1+1,\dots,x'\_L+1)$. Then $x''\gg x'$ and so $x''\succ x'$. However, $\alpha(x'')\leq\alpha(x')=+\infty$, which contradicts the fact that $\alpha(\cdot)$ represents $\succsim$. Therefore, the above statement holds.

Is my attempt correct?

Thanks a lot for any help!

**Tags:** microeconomics, mathematical-economics, utility, preferences, theory

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## 1 answers

### Answer by username_1 on 2024-05-14. Score: 3 | Accepted answer

We don't need $X$ (domain of the utility) to be equal to the set of its limit points to use the sequential characterisation of continuity. In other words, the sequential characterisation of continuity also holds when $X$ is not equal to the set of its limit points.

Given $x\in\mathbb{R}^L\_+$, to get $\alpha\_0,\alpha\_1$,
consider
$\alpha\_0=0$ and
$\alpha\_1=\max\_{1\leq i \leq L} x\_i+\epsilon$

Check that for any $x'\in\mathbb{R}^L\_+$ such that $||x'-x||\leq\epsilon$, $\alpha(x')\in [\alpha\_0,\alpha\_1]$.

> **Comments:**
> **username_0**: Indeed, $X$ need not be closed or equal to the set of its limit points. But every point of $X$ have to be a limit point of $X$, in order for the sequential characterization of continuity to hold for all $x\in X$. In other words, let $E$ be the set of limit points of $X$, we need $X\subseteq E$.  
> **username_1**: I don't think this condition is required. Please provide the definition of limit point of the set you are using. I'll try and provide a proof of characterisation.  
> **username_0**: Consider, for example, $X=[0,2]\bigcup\{4\}\bigcup(8,10]$ and the function $f:X\to\mathbb{R}$.  Then $f$ is always continuous at $x=4$ (according to the $\epsilon-\delta$ definition of continuity). However, the sequential characterization of continuity does not hold at $x=4$ because it is not a limit point.  
> **username_1**: Sequential characterisation holds because any sequence $(x_n)$ that converges to 4 consists of constant 4 in all but finitely many terms of the sequence. Consequently, $\lim f(x_n)=  \lim f(4) = f(4)$ for all such sequences.  
> **username_0**: Not exactly. By definition (see Baby Rudin), a point $p$ is a *limit point* of the set $E$ if *every* neighborhood of $p$ contains a point $q\neq p$ such that $q\in E$. As a result, "a finite point set has no limit points". Moreover, the sequential characterization of continuity says that a function $f$ is continuous at a *limit point* $p$ if and only if $\lim_{x\top}f(x)=f(p)$.  
> **username_2**: @Champa It is not true that  the sequential charaterization says necessarily that the point must be a limit point. Some texts may say that because they consider obviuous what Amit said in the comment above, that at an isolated point the condition is obvious, and they focus on the more interesting case of a limit point.  
> **username_0**: @BakerStreet I see. Yes, that does makes sense. Thanks a lot!  
> **username_0**: @Amit Thank you very much! Now I get your point.  
> **username_2**: @Champa You are welcome. You are right that this matter in some text can be confusing, but many text don't make this confusion. The point is exactly what Amit said.  
> **username_0**: @Amit May I ask how would you prove $\alpha(x')\leq\alpha_1$, where $\alpha_1=\max\{x_1,\dots,x_L\}+\epsilon$. I wanted to prove it by contradiction, and I think the contradiction should be that if $\alpha(x')>\alpha_1$ then $\|x'-x\|$ would be bigger than $\epsilon$. But I couldn't formalize a proof. Can you please help me with it?  
> **username_0**: @Amit Does this work: Assume to the contrary that $\alpha(x')>\alpha_1$. Let $x''=(x_1+\epsilon,\dots,x_L+\epsilon)$. Then $\alpha(x')e\gg x''$ and the monotonicity implies $x'\sim\alpha(x')e\succ x''$. However, $x_i''=x_i+\epsilon\geq x_i'$, because otherwise $\|x'-x\|=\sqrt{\sum_{i=1}^L(x_i'-x_i)^2}>\epsilon$. Thus $x''\geq x'$ and the lemma in this post (https://economics.stackexchange.com/questions/58383/question-about-the-step-proving-alphax-represents-preferences-in-the-proof) implies $x''\succsim x'$. We have a contradiction.  
> **username_1**: Consider any $x'\in\mathbb{R}^L_+$ such that $||x'-x||\leq \epsilon$. This implies that $|x_i'-x_i|\leq \epsilon$ for all $i\in\{1,2,\ldots,L\}$. This further implies that $x_i'\leq x_i+\epsilon\leq \alpha_1$ for all $i\in\{1,2,\ldots,L\}$. Therefore, by monotonicity, $\alpha(x')\leq \alpha_1$.  

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