Prove that $X=\{(x,y): y=mx+c\}$ is homeomorphic to $\mathbb{R}$
I've taken the hint from Prove that $X= \{ (x,y): y = mx + c \}$ is homeomorphic to $\mathbb R$.
So, my function is $f: \mathbb{R} \rightarrow X, f(x)=(x,mx+c)$
I have proved this is one-one and onto.
Now I need to prove that $f$ and $f^{-1}$ maps open sets into open sets.
Here is where I'm stuck: what do open sets look like in the relative topology on $X$?
EDIT: Expanding on the hint given:
Let $H$ be an open set in the topology on $X$.
Then $H=X\cap G$ where $G$ is an open set on the standard topology on $\mathbb{R^2}$
So, $H=X\cap [\cup \{(a,b) \times (c',d) \}]$, using the base of standard topology on $\mathbb{R^2}$
So, $f^{-1}(H)=f^{-1}(X) \cap f^{-1} [\cup \{(a,b) \times (c',d) \}]$, using the bijection of $f$,
i.e. $f^{-1}(H)=f^{-1}(X) \cap [\cup f^{-1} \{(a,b) \times (c',d) \}]$
i.e. $f^{-1}(H)=\mathbb{R} \cap [\cup f^{-1} \{(a,b) \times (c',d) \}]$
Now, I think $f^{-1} (a,b) \times (c,d)$ will be $(a,c')$ such that$ b=ax+c,d=c'x+c$. Am I correct?
It might be easier to give a specific inverse, here we have $f^{-1} ((x,y)) = x$, where $(x,y) \in X$.
This function is continuous of all of $\mathbb{R}^2$ hence it is continuous on the restriction to $X$.