# Prove the budget correspondence is upper hemi-continuous

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?

• Can I use closed graph theorem for this problem? – user7584 Mar 9 '16 at 13:45

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.

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.

• Gosh, I hope I can as smooth as you in a few years. Thanks a lot for your help! – Kenneth Chen Oct 7 '15 at 3:55
• 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 Oct 7 '15 at 6:15
• 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 Oct 7 '15 at 6:42
• 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 Oct 7 '15 at 6:50
• The definition does not seem to be correct. I am making some changes in the proof. Should be ready soon. – ramazan Oct 7 '15 at 7:18