Answer
(a) We show that the IVT is false if ${\cal D}$ is not connected.
(b) We state the Intermediate Value Theorem for two variables and prove it.
Work Step by Step
(a) Suppose that the domain ${\cal D}$ is not connected and let $P,Q \in {\cal D}$. Then, there exist a curve that does not lie entirely in ${\cal D}$ but joins $P$ and $Q$. Thus, there is some point along the curve that is not in ${\cal D}$. Therefore, $f\left( {x,y} \right)$ does not take on every value between $f\left( P \right)$ and $f\left( Q \right)$. Hence, the IVT is false. This implies that the IVT is false if ${\cal D}$ is not connected.
(b) We state the Intermediate Value Theorem for two variables and prove it.
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Theorem: Intermediate Value Theorem for two variables.
Let ${\cal D}$ be a closed connected domain and $P,Q \in {\cal D}$. If $f\left( {x,y} \right)$ is continuous on ${\cal D}$, then $f\left( {x,y} \right)$ takes on every value between $f\left( P \right)$ and $f\left( Q \right)$ at some point in ${\cal D}$.
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Proof.
Since ${\cal D}$ is connected, then there exists a path $c\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right)$ connecting $P$ and $Q$ such that $c\left( 0 \right) = P$ and $c\left( 1 \right) = Q$.
By the Intermediate Value Theorem (IVT) in one variable (Theorem 1 of Section 2.8), for every value $M$ between $f\left( {c\left( 0 \right)} \right)$ and $f\left( {c\left( 1 \right)} \right)$, there exists at least one value ${t_0} \in \left( {0,1} \right)$ such that $f\left( {c\left( {{t_0}} \right)} \right) = M$. Since ${\cal D}$ is connected, $f\left( {x,y} \right)$ lies entirely in ${\cal D}$. Hence, $f\left( {x,y} \right)$ takes on every value between $f\left( P \right)$ and $f\left( Q \right)$ at some point in ${\cal D}$.
