Let $X$ be a metric space and $f\colon X \rightarrow \mathbb R$ be a continuous function. Let $G = \{ (x , f(x) ) : x \in X \}$ be the graph of $f$. Then which one is true?
$G$ is homeomorphic to $X$
$G$ is homeomorphic to $\mathbb R$
$G$ is homeomorphic to $X \times \mathbb R$
$G$ is homeomorphic to $\mathbb R \times X$
My attempt : My answer None of the option is correct
For option $1)$ Consider $X=(0,1)$ and $f(x)=x$ then $G=\{(x,x):x\in (0,1)\}$ which is a closed set but $(0,1)$ is not
option $2)$ $X=[0,1]$ which is compact and $f(x)=x$.Its graph is compact but $\mathbb R$ is not.
option $3)$ $f(x)=0$ .Then graph of $f=\{(x,0):x\in \mathbb R\}$ i.e. the $x$ axis .Now remove the point $(0,0)$ from the graph of $f$ ,it becomes disconnected but $\mathbb R^2\setminus \{(0,0)\}$ is not.
option $4)$.Consider $f(x)=x$ .Then graph of $f=\{(0,y):x\in \mathbb R\}$ i.e. the $y$ axis .Now remove the point $(0,0)$ from the graph of $f$ ,it becomes disconnected but $\mathbb R^2\setminus \{(0,0)\}$ is not.
Is my answer is correct or not correct ?
Pliz tell me
Your counter example for (1) doesn't work for two reasons: because $\{(x,x)\mid x \in (0,1)\} $ is not closed in the plane (it does not contain the limit points $(0,0)$ and $(1,1)$), and moreover, being open or closed are weak conditions to look at, since every space is both open and closed with respect to itself.
Item (1) is actually true, with homeomorphism $$X\ni x\mapsto (x,f(x))\in G $$and inverse $$G\ni (x,y)\mapsto x \in X. $$