Prove that there is an embedding of the line as a closed subset of the plane, and there is an embedding of the line as a bounded subset of the plane, but there is no embedding of the line as a closed and bounded subset of the plane.
I'm not sure what the book means by "line," but I intuit an continuous subset of the real line to be a "line," for example the subsets $(0,1), [0,1] \subset \mathbb{R}.$ Now, I know that $[0,1]$ is open in $\mathbb{R}^2.$ Indeed, is it not true that every line is open in $\mathbb{R^2}?$
I think I may be incorrect in my interpretation of what "line" means in the problem. The problem is not more specific than "line."
For a line to be embedded as a subset of the plane, does that mean a homeomorphism $h: \mathbb{R} \rightarrow \mathbb{R},$ considering the graph $\mathbb{R} \times h(\mathbb{R}) \subseteq \mathbb{R} \times \mathbb{R}$ to be the "plane," or $h: \mathbb{R} \rightarrow \mathbb{R}^2?$
By line they mean $\mathbb{R}$. Embedding is a homeomorphism onto image, i.e. $f:\mathbb{R}\to\mathbb{R}^2$ such that $f$ as a function from $\mathbb{R}$ to $f(\mathbb{R})$ is a homeomorphism.
First of all $[0,1]$ is not a subset of $\mathbb{R}^2$. And there's no way to embed it as an open subset because nonempty open subsets of $\mathbb{R}^2$ are not compact.
So $\mathbb{R}$ can be embedded into $\mathbb{R}^2$ via $x\mapsto (x,0)$. The image is closed but not bounded.
It can also be embedded via $x\mapsto (arctan(x),0)$. The image is not closed but it is bounded this time.
It cannot be embedded as a closed and bounded subset because closed and bounded subsets of $\mathbb{R}^2$ are compact while $\mathbb{R}$ is not.