$\mathbb{R}^2\setminus\{0\}$ is path connected

Question

I am trying to construct an explicit path which connects two arbitrary points $a,b \in \mathbb{R}^2\setminus\{0\}$. My idea is to take a ball of radius $\epsilon>0$ around zero, $B := \{(x,y) : |x|,|y| = \epsilon\}$, then construct a line from $a$ to $b$ parameterised by $ta + (1-t)b$, next if the line intersects $B$ at any time, the path will then proceed to follow the boundary of the ball around by the required amount, and then continue to follow the line.

So I tried to define $f:[0,1]\rightarrow \mathbb{R}^2\setminus\{0\}$ by a piecewise definition $$f(t) = \begin{cases} ta + (1-t)b, &\text {if } f(ta+(1-t)b) \cap B = \emptyset\\ \epsilon (\cos(?), \sin(?)), &\text{if ? } \end{cases}$$

But can't figure out what conditions I should place so that I have what I need to connect my points.

The picture looks something like this, thanks.

Answer 1

You're making it harder for yourself than it needs to be by choosing such a relatively complex path. Instead I would do

• If the two points $a$ and $b$ are not collinear with $0$ (that is, if their vectors are linearly independent), just interpolate linearly between them.

• If they are collinear, then choose a point $c$ outside the line they form with $0$, and go in a straight line from $a$ to $c$ on $t\in[0,\frac12]$ and then down from $c$ to $b$ on $t\in[\frac12,1]$.

Either $(0,1)$ or $(1,0)$ will work as $c$ -- or you can construct it systematically by rotating $a$ by $90^\circ$ around the origin.

Or you could even always use the three-point strategy, and choose $c$ as the first of $(0,1)$, $(1,1)$ and $(1,0)$ that is not collinear with $0$ and one of $a$ and $b$.

Answer 2

To simplify the notations, consider we work in $\Bbb C$.

Use the polar form of $a$ and $b$: $a=r_a \exp (i\theta_a)$ and $b=r_b\exp (i \theta_b)$. Then the path can be defined on $[0,1]$ by

• If $t\in[0,1/3]$, $f(t)=[r_a(1-3t)+3t]\exp (i\theta_a)$
• If $t\in[1/3,2/3]$, $f(t)=\exp \left(i[(2-3t)\theta_a+(3t-1)\theta_b]\right)$
• If $t\in[2/3,1]$, $f(t)=[(3-3t)+(3t-2)r_b]\exp (i\theta_b)$

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An even simpler path:

For $t\in[0,1]$, $f(t)=((1-t)r_a+tr_b)\exp\left(i[(1-t)\theta_a+t\theta_b)]\right)$

Answer 3

You can do that I personally would use two different types of paths.

Let $a(x_0,y_0)$ and $b(x_1,y_1)$

Case I: If the origin is not on segment from $a$ to $b$

$f(t)=ta+(1-t)b$ for $0 \le t \le 1$

Case II: If the origin is on segment from $a$ to $b$

Measure the angle from the positive $x$-axis to $a$ and $b$ in the standard way and call the $\theta_a$, $\theta_b$ respectively. Let's insist that they're both on interval $(-\pi,\pi]$ so we don't accidentallywind around the circle more than once. Let $(x_m,y_m)$ be the midpoint between $a$ and $b$. Then the following circular path should work:

$f(t)=\left<x_m+r\cos t, y_m+r\sin t\right>$ for $\theta_a \le t \le \theta_b$

Answer 4

Viewing $\mathbb{R}^2$ as the complex plane $\mathbb{C}$, a path from $a$ to $b$ in $\mathbb{C} \smallsetminus \{0\}$ is given by $$f(t) = \exp((1-t) \log a + t \log b).$$

Answer 5

Here we formulate explicit paths connecting any two different points from the set

$V = \{(1,1), (1,-1), (-1,-1), (-1,1)\}$

We leave it to the reader to write down the 'inverse' paths.

To begin, we have the four 'easy' ones:

$\tag 1 [(1,1) \text{ to } (1,-1)]: t \mapsto (1, 1 - 2 t)$ $\tag 2 [(1,1) \text{ to } (-1,1)]: t \mapsto (1 - 2 t, 1)$ $\tag 3 [(1,-1) \text{ to } (-1,-1)]: t \mapsto (1 - 2 t, -1)$ $\tag 4 [(-1,-1) \text{ to } (-1,1)]: t \mapsto (-1, -1 + 2 t)$

Here is the mapping for $[(1,1) \text{ to } (-1,-1)]$:

$\tag 5 t \mapsto (1, 1 - 4 t) \text{ for } t \le 1/2 \text{ else } t \mapsto ( 3 - 4 t, -1)$

Here is the mapping for $[(1,-1) \text{ to } (-1,1)]$:

$\tag 6 t \mapsto (1 - 4 t, -1) \text{ for } t \le 1/2 \text{ else } t \mapsto (-1, -3 + 4 t)$

We can now get explicit mappings for any two points. The following is a very simple assignment:

Exercise: If $a \in \mathbb{R}^2\setminus\{0\}$ formulate an explicit path connecting $a$ to a point in $V$.

To really 'pin down' this elementary piece-wise linear approach to logically specifying a path connecting any two points, you have to carefully organize your work. One idea when working out the exercise is to specify a set theoretic partition of $\mathbb{R}^2\setminus\{0\}$ into four quadrants thereby getting some logical focus (cut down the 'choice tree').