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See Proposition 3.C.1 from MWG

Continue from this post, 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

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!

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1 Answer 1

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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 $||x'-x||\leq\epsilon$, $\alpha(x')\in [\alpha_0,\alpha_1]$.

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  • $\begingroup$ 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 $X\subseteq E$. $\endgroup$
    – Champa
    Commented May 14 at 11:18
  • $\begingroup$ 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. $\endgroup$
    – Amit
    Commented May 14 at 11:28
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    $\begingroup$ 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. $\endgroup$
    – Amit
    Commented May 14 at 14:35
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    $\begingroup$ @BakerStreet I see. Yes, that does makes sense. Thanks a lot! $\endgroup$
    – Champa
    Commented May 14 at 15:06
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    $\begingroup$ @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. $\endgroup$ Commented May 14 at 15:10

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