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In order to prove the existence of a solution to the equation $f_1(x)-c_1(x) = 0$, I have shown that: \begin{align*} \lim_{x \rightarrow 0} f_1(x) &> \lim_{x \rightarrow \bar{x}} f_1(x) \\ \lim_{x \rightarrow 0} c_1(x) &< \lim_{x \rightarrow \bar{x}} c_1(x) \\ \frac{d f_1(x)}{d x}&<0 \\ \frac{d c_1(x)}{d x}&>0 \end{align*} That is, some version of the intermediate value theorem.

Now, I have two endogenous variables (x,y) and I want to prove that a solution to \begin{align*} f_1(x,y)-c_1(x,y) = 0 \\ f_2(x,y)-c_2(x,y) = 0 \end{align*}

Is there a similar approach I can use? I have also heard about Kakutani's fixed point theorem, but I do not now how to use it.

Thanks for your help!!

Thomas Andrews
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Karl Smith
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  • If $g=(f_1-c_1,f_2-c_2):\mathbb R^2\to \mathbb R^2,$ the equivalent of the intermediate value theorem is, if $g:\mathbb R^2\to\mathbb R^2$ is a continuous function and some circle $S$ and $y=(y_1,y_2)\in\mathbb R^2\setminus g(S),$ is such that $g_{|S}$ is not contractible in $\mathbb R^2\setminus {y},$ then there is some $x=(x_1,x_2)$ such that $g(x)=y.$ But that condition is more complicated - you have to know what contractible means, and how to characterize contractible curves in the punctured plane. – Thomas Andrews Jun 21 '23 at 17:55
  • Intuitively, the function is not contractible if the image of $S$ winds around $y$ zero or more times, total, in one direction. So a function which goes clockwise once and counterclockwise once is contractible, but an image that goes around $2$ times clockwise and one time counter-clockwise is counted as going $1$ time clockwise, total, and is not contractible. But there is a fair amount of definitions to make this rigorous - search for "fundamental group" in the topic of topology. – Thomas Andrews Jun 21 '23 at 18:01

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