# 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
Commented Mar 9, 2016 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! Commented Oct 7, 2015 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? Commented Oct 7, 2015 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. Commented Oct 7, 2015 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 $x \in \Gamma(p,w)$. Commented Oct 7, 2015 at 6:50
• The definition does not seem to be correct. I am making some changes in the proof. Should be ready soon. Commented Oct 7, 2015 at 7:18