<p>I am assuming that the following facts do not require proofs for the purposes of this question.</p>

<p><strong>Fact 1:</strong> Let $h_n$ be a sequence in $\mathbb{R}^K$ such that $\lim_{n\rightarrow \infty} h_n =h\in \mathbb{R}^K$. Then, for each $i\in \{1,2,\ldots,K\}$, we have $h^i_n\rightarrow h^i$.</p>

<p><strong>Fact 2:</strong> Let $z_n$ and $q_n$ be sequence in $\mathbb{R}$ such that $\lim_{n\rightarrow \infty} z_n =z\in \mathbb{R}$ and $\lim_{n\rightarrow \infty} q_n =q\in \mathbb{R}$. Then, </p>

<p><strong>i)</strong> $\lim_{n\rightarrow \infty}(z_n\pm q_n)=z\pm q$</p>

<p><strong>ii)</strong> $\lim_{n\rightarrow \infty}(z_n \times q_n)=z\times q$</p>

<p><strong>Fact 3:</strong> Let $z_n$ in $\mathbb{R}$ such that for each $n$, $z_n\leq a\in \mathbb{R}$ and $\lim_{n\rightarrow \infty} z_n =z$. Then, $z\leq a$.</p>

<p><strong>Definition:</strong> A correspondence $G:X\rightrightarrows Y $ is upper hemicontinuous at $x\in X$ if for any open neighborhood of $V$ of $G(x)$. There exists a neighborhood $U$ of $x$ such that for all $x'$ in $U$, $G(x')\subseteq V$.</p>

<p>Unlike the previous remarks, the following fact requires proof. </p>

<p><strong>Fact 4:</strong> Let $G:X\rightrightarrows Y $ be a correspondence. If for every $x_n\rightarrow x\in X$ and $y_n\in G(x_n)$ there exists a subsequence $y_{n_k}$ of $y_n$ with $y_{n_k}\rightarrow y$ and $y\in G(x)$, then $G$ is upper hemicontinuous. Moreover, if $G$ is compact valued, then the converse is also true.</p>

<p>Now, let's go back to your question. Let $(p_n,w_n)$ be a sequence in $\mathbb{R}^{L+1}_{++}$ such that $\lim_{n\rightarrow \infty} (p_n,w_n) =(p,w)\in \mathbb{R}^{L+1}_{++}$. Moreover, let $x_n\in B(p_n,w_n)$ for all $n$. </p>

<p>Let $w^\ast = \sup_n w_n$. Since $w_n\rightarrow w\in \mathbb{R}_{++}$, it must be true that $w^\ast \in \mathbb{R}_{++}$ because otherwise we could have created a subsequence $w_{n_k}$ of $w_n$ for which $w_{n_k}\rightarrow\infty$ which would contradict with $w_n\rightarrow w$. Similar arguments would imply that $p^{\ast} = \min_i( \inf_n p^i_n)&gt;0$ since $p^n\rightarrow p\in\mathbb{R}_{++}^L$.</p>

<p>The previous observations imply that for all $n$ and for all $i$ we have $x_n^i\leq w^\ast/p^\ast$. Therefore, the sequence $x_n$ is bounded. By Bolzano-Weierstrass Theorem, $x_n$ has a convergent subsequence $x_{n_k}$ with $x_{n_k}\rightarrow x$. From here onwards, we will suppress the subscript $k$.</p>

<p>Since $x_n\in B(p_n,w_n)$, we have $\sum_i p_n^i\times x_n^i-w_n\leq 0$. Let $c_n = \sum_i p_n^i\times x_n^i$. By fact 1, we have $p_n^i\rightarrow p_i$ and $x_n^i\rightarrow x_i$ for each $i$, and by fact 2, we have $c_n=\sum_i p_n^i\times x_n^i\rightarrow p\cdot  x$. </p>

<p>By fact 2, $c_n-w_n\rightarrow (p\cdot  x-w)$. Fact 3 implies that $(p\cdot  x-w)\leq 0$, which in turn implies that $x\in B(p,w)$. This concludes the proof that the budget correspondence is upper hemicontinuous.</p>

<p><strong>Original (incorrect) Answer</strong></p>

<p>The following answer has been posted earlier. However, I realized that there is a mistake with this one. I am keeping this here since the original question had been wondering specifically about how Bolzano-Weierstrass theorem is relevant to the question and comparison of this incorrect answer and the correct version shows why we would need this theorem.</p>

<p><strong>Definition:</strong> A correspondence $G:X\rightrightarrows Y $ is upper hemicontinuous at $x\in X$ if for all sequences $x_n\in X$ with $x_n\rightarrow x$, for all $y_n\in G(x_n)$ with $y_n \rightarrow y \in Y$, we have $y\in G(x)$.</p>

<p>Now, let's go back to your question. Let $(p_n,w_n)$ be a sequence in $\mathbb{R}^{L+1}$ such that $\lim_{n\rightarrow \infty} (p_n,w_n) =(p,w)$. Moreover, let $x_n\in B(p_n,w_n)$ for all $n$, with $x_n\rightarrow x$. </p>

<p>Since $x_n\in B(p_n,w_n)$, we have $\sum_i p_n^i\times x_n^i-w_n\leq 0$. Let $c_n = \sum_i p_n^i\times x_n^i$. By fact 1, we have $p_n^i\rightarrow p_i$ and $x_n^i\rightarrow x_i$ for each $i$, and by fact 2, we have $c_n=\sum_i p_n^i\times x_n^i\rightarrow p\cdot  x$. </p>

<p>By fact 2, $c_n-w_n\rightarrow (p\cdot  x-w)$. Fact 3 implies that $(p\cdot  x-w)\leq 0$, which in turn implies that $x\in B(p,w)$. This concludes the proof that the budget correspondence is upper hemicontinuous.</p>

CC BY-SA 3.0
2015-10-07T03:55:51.997Z
Gosh, I hope I can as smooth as you in a few years. Thanks a lot for your help!
Kenneth Chen
CC BY-SA 3.0
2015-10-07T06:15:44.330Z
One follow-up question. Therefore Bolzano-Weierstrass theorem in this case is used simply to prove that such a sequence of $x_n$ is convergent?
Kenneth Chen
CC BY-SA 3.0
2015-10-07T06:42:24.273Z
wikipedia seems to have lied to me in terms of the definition of hemicontinuity. can you describe in your question what exactly is the definition you are using.
ramazan
CC BY-SA 3.0
2015-10-07T06:50:37.930Z
Exactly yours. That is, a correspondence $\Gamma:  R_+^{L+1} \rightarrow R_+^L$ is upper hemi-continuous if,   $(p_n, w_n) \to (p,w)$ and $x_n \to x$ where for each $n$ it's true $x_n \in \Gamma(p_n,w_n)$ will  ensure $x \in \Gamma(p,w)$.
Kenneth Chen
CC BY-SA 3.0
2015-10-07T07:18:55.710Z
The definition does not seem to be correct. I am making some changes in the proof. Should be ready soon.
ramazan
CC BY-SA 3.0
2015-10-07T07:39:14.653Z
Thanks in advance. That definition is what I have from my class, could you tell me what could goes wrong with that definition? (If you care to)
Kenneth Chen
CC BY-SA 3.0
2015-10-07T03:30:59.470Z
2015-10-07T07:56:32.950Z
2015-10-07T07:56:32.950Z
ramazan
<p>Let $p \in \mathbb{R}_+^L$ be price vector and let $w \in \mathbb{R}_+$ be wealth of the consumer. Define the Budget correspondence $B(p,w) =\{x \in \mathbb{R}_+^L : p\cdot x\le w  \}$ . How to prove that this budget correspondence is upper hemi-continuous?</p>

<p>A correspondence $Γ:R_+^{L+1}+→R^L_+$ is upper hemi-continuous if, $(p_n,w_n)→(p,w) $ and $x_n \to x$ where for each $n$ it's true $x_n \in 
\Gamma(p_n,w_n)$, will ensure $x \in \Gamma(p,w)$.</p>

<p>I've seen a possible solution using Bolzano–Weierstrass theorem by Mark Dean at the following link <a href="http://www.econ.brown.edu/fac/mark_dean/Maths_HW4_13.pdf" rel="noreferrer">http://www.econ.brown.edu/fac/mark_dean/Maths_HW4_13.pdf</a>. But I failed to understand the last step in his approach where he claims that if every sequence $x_m \in B(p^m, w_m)$ has convergent sub-sequence converging to a point in set $B(p^*, w^*)$, then we have hemi-continuity for budget set, aren't we supposed to show that $x_m$ converge to $B(p,w)$? Am I missing any theorem?</p>

<microeconomics><consumer-theory>
Prove the budget correspondence is upper hemi-continuous
Kenneth Chen
# Prove the budget correspondence is upper hemi-continuous

(Asked by: username_0 on 2015-10-07)

Let $p \in \mathbb{R}\_+^L$ be price vector and let $w \in \mathbb{R}\_+$ be wealth of the consumer. Define the Budget correspondence $B(p,w) =\{x \in \mathbb{R}\_+^L : p\cdot x\le w \}$ . How to prove that this budget correspondence is upper hemi-continuous?

A correspondence $Γ:R\_+^{L+1}+→R^L\_+$ is upper hemi-continuous if, $(p\_n,w\_n)→(p,w) $ and $x\_n \to x$ where for each $n$ it's true $x\_n \in
\Gamma(p\_n,w\_n)$, will ensure $x \in \Gamma(p,w)$.

I've seen a possible solution using Bolzano–Weierstrass theorem by Mark Dean at the following link <http://www.econ.brown.edu/fac/mark_dean/Maths_HW4_13.pdf>. But I failed to understand the last step in his approach where he claims that if every sequence $x\_m \in B(p^m, w\_m)$ has convergent sub-sequence converging to a point in set $B(p^\*, w^\*)$, then we have hemi-continuity for budget set, aren't we supposed to show that $x\_m$ converge to $B(p,w)$? Am I missing any theorem?

**Tags:** microeconomics, consumer-theory


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## 1 answers

### Answer by username_1 on 2015-10-07. Score: 6 | Accepted answer

I am assuming that the following facts do not require proofs for the purposes of this question.

**Fact 1:** Let $h\_n$ be a sequence in $\mathbb{R}^K$ such that $\lim\_{n\rightarrow \infty} h\_n =h\in \mathbb{R}^K$. Then, for each $i\in \{1,2,\ldots,K\}$, we have $h^i\_n\rightarrow h^i$.

**Fact 2:** Let $z\_n$ and $q\_n$ be sequence in $\mathbb{R}$ such that $\lim\_{n\rightarrow \infty} z\_n =z\in \mathbb{R}$ and $\lim\_{n\rightarrow \infty} q\_n =q\in \mathbb{R}$. Then,

**i)** $\lim\_{n\rightarrow \infty}(z\_n\pm q\_n)=z\pm q$

**ii)** $\lim\_{n\rightarrow \infty}(z\_n \times q\_n)=z\times q$

**Fact 3:** Let $z\_n$ in $\mathbb{R}$ such that for each $n$, $z\_n\leq a\in \mathbb{R}$ and $\lim\_{n\rightarrow \infty} z\_n =z$. Then, $z\leq a$.

**Definition:** A correspondence $G:X\rightrightarrows Y $ is upper hemicontinuous at $x\in X$ if for any open neighborhood of $V$ of $G(x)$. There exists a neighborhood $U$ of $x$ such that for all $x'$ in $U$, $G(x')\subseteq V$.

Unlike the previous remarks, the following fact requires proof.

**Fact 4:** Let $G:X\rightrightarrows Y $ be a correspondence. If for every $x\_n\rightarrow x\in X$ and $y\_n\in G(x\_n)$ there exists a subsequence $y\_{n\_k}$ of $y\_n$ with $y\_{n\_k}\rightarrow y$ and $y\in G(x)$, then $G$ is upper hemicontinuous. Moreover, if $G$ is compact valued, then the converse is also true.

Now, let's go back to your question. Let $(p\_n,w\_n)$ be a sequence in $\mathbb{R}^{L+1}\_{++}$ such that $\lim\_{n\rightarrow \infty} (p\_n,w\_n) =(p,w)\in \mathbb{R}^{L+1}\_{++}$. Moreover, let $x\_n\in B(p\_n,w\_n)$ for all $n$.

Let $w^\ast = \sup\_n w\_n$. Since $w\_n\rightarrow w\in \mathbb{R}\_{++}$, it must be true that $w^\ast \in \mathbb{R}\_{++}$ because otherwise we could have created a subsequence $w\_{n\_k}$ of $w\_n$ for which $w\_{n\_k}\rightarrow\infty$ which would contradict with $w\_n\rightarrow w$. Similar arguments would imply that $p^{\ast} = \min\_i( \inf\_n p^i\_n)>0$ since $p^n\rightarrow p\in\mathbb{R}\_{++}^L$.

The previous observations imply that for all $n$ and for all $i$ we have $x\_n^i\leq w^\ast/p^\ast$. Therefore, the sequence $x\_n$ is bounded. By Bolzano-Weierstrass Theorem, $x\_n$ has a convergent subsequence $x\_{n\_k}$ with $x\_{n\_k}\rightarrow x$. From here onwards, we will suppress the subscript $k$.

Since $x\_n\in B(p\_n,w\_n)$, we have $\sum\_i p\_n^i\times x\_n^i-w\_n\leq 0$. Let $c\_n = \sum\_i p\_n^i\times x\_n^i$. By fact 1, we have $p\_n^i\rightarrow p\_i$ and $x\_n^i\rightarrow x\_i$ for each $i$, and by fact 2, we have $c\_n=\sum\_i p\_n^i\times x\_n^i\rightarrow p\cdot x$.

By fact 2, $c\_n-w\_n\rightarrow (p\cdot x-w)$. Fact 3 implies that $(p\cdot x-w)\leq 0$, which in turn implies that $x\in B(p,w)$. This concludes the proof that the budget correspondence is upper hemicontinuous.

**Original (incorrect) Answer**

The following answer has been posted earlier. However, I realized that there is a mistake with this one. I am keeping this here since the original question had been wondering specifically about how Bolzano-Weierstrass theorem is relevant to the question and comparison of this incorrect answer and the correct version shows why we would need this theorem.

**Definition:** A correspondence $G:X\rightrightarrows Y $ is upper hemicontinuous at $x\in X$ if for all sequences $x\_n\in X$ with $x\_n\rightarrow x$, for all $y\_n\in G(x\_n)$ with $y\_n \rightarrow y \in Y$, we have $y\in G(x)$.

Now, let's go back to your question. Let $(p\_n,w\_n)$ be a sequence in $\mathbb{R}^{L+1}$ such that $\lim\_{n\rightarrow \infty} (p\_n,w\_n) =(p,w)$. Moreover, let $x\_n\in B(p\_n,w\_n)$ for all $n$, with $x\_n\rightarrow x$.

Since $x\_n\in B(p\_n,w\_n)$, we have $\sum\_i p\_n^i\times x\_n^i-w\_n\leq 0$. Let $c\_n = \sum\_i p\_n^i\times x\_n^i$. By fact 1, we have $p\_n^i\rightarrow p\_i$ and $x\_n^i\rightarrow x\_i$ for each $i$, and by fact 2, we have $c\_n=\sum\_i p\_n^i\times x\_n^i\rightarrow p\cdot x$.

By fact 2, $c\_n-w\_n\rightarrow (p\cdot x-w)$. Fact 3 implies that $(p\cdot x-w)\leq 0$, which in turn implies that $x\in B(p,w)$. This concludes the proof that the budget correspondence is upper hemicontinuous.

> **Comments:**
> **username_0**: Gosh, I hope I can as smooth as you in a few years. Thanks a lot for your help!  
> **username_0**: One follow-up question. Therefore Bolzano-Weierstrass theorem in this case is used simply to prove that such a sequence of $x_n$ is convergent?  
> **username_1**: wikipedia seems to have lied to me in terms of the definition of hemicontinuity. can you describe in your question what exactly is the definition you are using.  
> **username_0**: Exactly yours. That is, a correspondence $\Gamma:  R_+^{L+1} \rightarrow R_+^L$ is upper hemi-continuous if,   $(p_n, w_n) \to (p,w)$ and $x_n \to x$ where for each $n$ it's true $x_n \in \Gamma(p_n,w_n)$ will  ensure $x \in \Gamma(p,w)$.  
> **username_1**: The definition does not seem to be correct. I am making some changes in the proof. Should be ready soon.  
> **username_0**: Thanks in advance. That definition is what I have from my class, could you tell me what could goes wrong with that definition? (If you care to)  

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