Consider the map $f\colon(\mathbb{R}^2,\tau)\to (\mathbb{R},J)$ given by $f(x,y)=x+y$, where $\tau$ is the standard topology on $\mathbb{R}^2$ and $J$ is the order topology on $\mathbb{R}$. Find and sketch on the same plane:
- (a) The image of $\{0\}$ under inverse of $f$.
- (b) The image of $[0,1)$ under inverse of $f$.
- (c) Is $f$ continuous? Why?
In $\mathbb{R}^2$, define a map $d\colon\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}$ by $d(x,y)=|x_1-y_1|+|x_2-y_2|$ where $x=(x_1,x_2)$, $y=(y_1,y_2)$.
- (a) Show that $d$ is a metric on $\mathbb{R}^2$.
- (b) Find the basis for the topology on $\mathbb{R}^2$ induced by $d$.
- (c) How is the topology induced by $d$ related to the standard topology on $\mathbb{R}^2$?
Let $S$ denote the unit circle $S=\{(x,y)\in\mathbb{R}^2\mid x^2+y^2=1\}$, considered as a subspace of the plane $\mathbb{R}^2$, and let $f\colon [0,1)\to S$ be the map defined $by f(t)=(\cos(2\pi t),\sin(2\pi t))$. Show that $f$ is continuous but not a homeomorphism.
Show that the function $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=3x+1$ is a homeomorphism.
My attempt.
Question 1. (a) $f^{-1}(\{0\}) = \{ (x,y)\in\mathbb{R}^2\mid f(x,y)=0\} = \{\text{all points in the line } x=-y\}$.
Question 2. (a) $d(x,y)=|x_1-y_1|+|x_2-y_2|=|y_1-x_1|+|y_2-x_2|=d(y,x)$ $$\begin{align*} d(x,y)&=|x_1-y_1|+|x_2-y_2|=|x_1-z_1 + z_1-y_1|+|x_2-z_2 +z_2-y_2|\\ &\leq |x_1-z_1|+|x_2-z_2|+|z_1-y_1|+|z_2-y_2|\\ &\leq d(x,z)+d(z,y) \end{align*}$$ therefore $d$ is a metric on $\mathbb{R}^2$.
Question 4. $f(1)=4$, $f(2)=7$ so $f$ is one to one hence bijection, I think $f$ carries a basis element from $\mathbb{R}$ to $\mathbb{R}$ and vice versa, hence $f$ is automatically homeomorphism
I wish to know the correctness of my attempt , and how I can attempt Question 1(b),(c); Question 2(b),(c); and Question 3.