Continuum set and connected graph For topological spaces $X$, a set $A\subset X$ is continuum if $A$ is both compact and connected. Notice that if $Y\subset X$ be an open subspace of $X$ and $B\subset Y $ is continuum then $B$ is continuum in $X.$ There is well know is says : For a function $g\colon \Bbb R \to \Bbb R$, a graph of $g$ is connected iff  whenever $D$ is continuum in $R^2$ which contains points above and below the graph of $g$, then D must meet the graph of $g.$
Now, Let $f\colon \Bbb (1,2)\to \Bbb R$ be a function  that has a disconnected graph. By using that fast above there exists a continuum $D\subset (1,2)\times \Bbb R^2$ that has pints above and below the graph such that $$D\cap Gr  f=\emptyset,$$ where $Gr f$ denotes to the graph of $f.$  Since $(1,2)\times \Bbb R$  is an open set in $\Bbb R^2$, then $D$ is continuum in $R^2.$ Define a function $h\colon \Bbb R\to \Bbb R$ as follows
$$ h(x) = \begin{cases} f(x) , \ \ \ x\in(1,2) \\
      0  ,  \ \ \ \ \ \text{Otherwise}\end{cases}$$
Then $h$ still has disconnected graph. Since $D$ is continuum of $\Bbb R^2$ and has point below and above  the graph and $D\cap Gr h=\emptyset.$  Does that correct? if not how I can show that $h$ has disconnected graph. Is there any easy way to show $h$ has disconnected graph.  Any help will appreciated greatly.
 A: Assume $f: (1, 2) \to \Bbb{R}$ has a disconnected graph. Then the graph $\mathrm{Gr}(f)$ is contained in the union of two disjoint non-empty closed sets, $A$ and $B$ say, such that $A \cap \mathrm{Gr}(f) \neq \emptyset$, $B \cap \mathrm{Gr}(f) \neq \emptyset$ and $A \cap B \cap \mathrm{Gr}(f) = \emptyset$. We may assume that $A$ and $B$ are both subsets of $[1, 2] \times \Bbb{R}$, (since if not, we may replace $A$ and $B$ by $A \cap ([1, 2] \times \Bbb{R})$ and $B \cap ([1, 2] \times \Bbb{R})$, respectively).
If we extend $f$ to a function $h : \Bbb{R} \to \Bbb{R}$, by setting $h(x) = 0$ for $x \not\in (1, 2)$, then $\mathrm{Gr}(h) = I \cup \mathrm{Gr}(f) \cup J$, where $I = (-\infty, 1] \times \{0\}$ and $J =[2, \infty) \times \{0\}$. But then, by considering the various cases as to whether $(1, 0) \in A$, $(2, 0) \in A$, $(1, 0) \in B$ and $(2, 0) \in B$, we can find non-empty closed sets $A'$ and $B'$ such that $A' \cap \mathrm{Gr}(h) \neq \emptyset$, $B' \cap \mathrm{Gr}(h) \neq \emptyset$ and $A' \cap B' \cap \mathrm{Gr}(h) = \emptyset$. E.g., if $(1, 0) \in A$ and $(2, 0) \in B$, we take $A' = I \cup A$ and $B' = B \cup J$, while if $(1,0)$ and $(2, 0)$ are both in $A$ we take $A' = I \cup A \cup J$ and $B'= B$.
