Why is $\mathbb{R}^2$ endowed with the taxicab metric isomorphic to the infinity distance model of the cartesian real plane? (Hartshorne exercise 8.9)

My question deals with the following exercise from Hartshorne’s Euclid: Geometry and Beyond:

Following our general principles, we say that two models $ M,M’$ of our geometry are isomorphic if there exists a 1-to-1 mapping $ \phi:M\to M’$ of the set of points of $ M$ onto the set of points $ M’$ , written $ \phi(A) = A’$ , that sends lines to lines, preserves betweenness, i.e., $ A\ast B\ast C$ in M $ \iff$ $ A’\ast B’\ast C’$ in M’, and preserves congruence of line segments, i.e., $ AB\cong CD$ in M$ \iff$ $ A’B’\cong C’D’$ in $ M’$

If $ A=(a_1,a_2),B=(b_1,b_2)\in\mathbb{R}^2$ , the exercise asks you to first show that $ \mathbb{R}^2$ endowed with the taxicab metric: $ $ d(A,B)=|a_1-b_1|+|a_2-b_2|$ $ Is isomorphic to the model of $ \mathbb{R}^2$ endowed with the infinity distance: $ $ d(A,B)=sup\{|a_1-b_1|,|a_2-b_2|\}$ $ How can we proceed in order to prove these models are isomorphic? How should we describe our isomorphism $ \phi:\mathbb{R}^2\to \mathbb{R}^2$ ? I’m definitely having trouble with sending congruent lines to congruent lines.

The second part of the exercise claims that the taxicab metric model is not isomorphic to that of the standard model, i.e., the cartesian plane $ \mathbb{R}^2$ equipped with the usual Euclidean distance. Hartshorne leaves a hint to this exercise: in order to prove two models are not isomorphic, one just has to find a statement which is true in one model but not in the other. I’m thinking, for example, that the line: $ $ y=x+1$ $ Cuts the circle of center the origin $ (0,0)$ , and radius $ 1$ in infinitely many points in the taxicab metric model, a property that never holds in the standard model, which should make it clear that both models are not isomorphic. However, the first part still puzzles me, and I can’t imagine how to proceed.

Thanks in advance for your time, any comments will be appreciated.

Constructing tangent to any curve in $\mathbb{R}^2$

As I mentioned in my previous question. I’ve been studying Basic Mathematics for Physics courses.

While teaching about derivatives my prof. said that there are actually two points the tangent at a point passes through (and those points are almost coincident). Symbolically $ x_0$ and $ x_0+\mathrm dx$ . Clearly this sounds absurd because a tangent by definition touches any curve only at a single point. But also, this puts forth the discrepancy that at least two points are required to construct a line.

I thought that if we somehow know the curvature of the curve we would indeed be able to construct a tangent using only a single point by using definition of curvature, $ \kappa=1/r$ and follow as we do in the case of a circle.

But how to actually measure the curvature of curve using only its derivative? Feel free to present a model of constructing a tangent this way or any other that you may find relevant and fitting into intermediate-to-advanced calculus courses.

How to determine subsets of $\mathbb{R}^2$ open, closed, both, neither or compact?

I need help with this problem:

What of the following subsets of $ \mathbb{R}^2$ are (i) open, (ii) closed, (iii) both open and clossed, (iv) neither open or closed, (v) compact?

  1. $ 1<x_1<2$ and $ -1<x_2<2$
  2. $ x_1^2+x_2^2>0$
  3. $ \Vert \mathbf{x}-(1,3) \Vert <1$
  4. $ x_1>x_2$

I now the definitions of closed, open and compact, but I’m cofused with both and neither. How do I show using the definitios if those subsets open, closed, etc? Hope you can help me.

Why is $\mathbb{R}^2$ without the x-axis simply connected?

So in an Exam Question I found on the Internet, i saw that $ \mathbb{R}^2$ without the x-axis is simply connected. But in my opinion, this shouldn’t even be connected and thus not simply connected. I think of it as $ \mathbb{R}\setminus \{0\}$ which is also not connected. So why should $ \mathbb{R}^2$ without the x-axis be connected?

If $A \times B$ is Lebesgue measurable in $\mathbb{R}^2$ and $B$ is Lebesgue measurable in $\mathbb{R}$ then $A$ is so?

Though it seems simple but I am struggling to find a proof for it being a starter to measure theoty. I think this statement is true and I am trying to prove it. Here is my trial: Write $ A= A \times \{0\}= \cup_i(A_{i} \times B_{i})$ ,

$ m_1(A)=m_2(A \times \{0\})=\cup_i m_1(A_i)m_1(B_i)$

I don’t know what to do next! Is my approach correct?

where $ m_1,m_2$ are 1 and 2 dimensional measures in $ \mathbb{R},\mathbb{R^2}$ respectively.

Finding the interior and closure of set $A \subset \mathbb{R}^2$

I am trying to determine the interior and closure of the following set $ $ A := \{ (x,y) \in \mathbb{R}^2 : |x| \geq |y|\} $ $ and would like to vertify my answer

For every $ (x,y) \in A$ we can find an $ \epsilon$ -neighbourhood that is completely contained in $ A$ unless $ |x| = |y|$ . If we let $ a \in A$ such that $ a = (x,x)$ and $ \epsilon > 0$ then $ b = (x-\epsilon, x+\epsilon) \in B_{\epsilon}(a)$ and $ b\notin A$ which means $ A^{°} := \{ (x,y) \in A : |x| \neq |y| \}$

The set $ A$ is closed, which means $ A = \overline{A}$ . I tried to show that $ \mathbb{R}^2 \setminus A$ is open, but I am having trouble finding an $ \epsilon$ such that $ $ c \in \mathbb{R}^2 \setminus A \ \text{and} \ d = (x,y) \in B_{\epsilon}(c) \Longrightarrow |x| < |y| \ \text{and} \ d \in \mathbb{R}^2 \setminus A $ $

Note: we are using the Euclidean metric on $ \mathbb{R}^2$