Artin, Chapter 2, Misc.6

Question

I am trying to solve miscellaneous exercise 6 in Chapter 2 of Artin's book, Algebra. Below is the statement of the problem.

Let $a = (a_1, \ldots, a_k)$ and $b = (b_1, \ldots, b_k)$ be points in $k$-dimensional space $\mathbb{R}^k$. A path from $a$ to $b$ is a continuous function on the unit interval $[0,1]$ with values in $\mathbb{R}^k$, a function $X: [0,1] \to \mathbb{R}^k$, sending $t \rightsquigarrow X(t) = (x_1 (t), \ldots, x_k (t))$, such that $X(0) = a$ and $X(1) = b$. If $S$ is a subset of $\mathbb{R}^k$ and if $a$ and $b$ are in $S$, define $a \sim b$ if and $a$ and $b$ can be joined by a path lying entirely in $S$.

(a) Show that $\sim$ is an equivalence relation on $S$. Be careful to check that any paths you construct stay within the set $S$.

(b) A subset $S$ is path connected if $a \sim b$ for any two points $a$ and $b$ in $S$. Show that every subset $S$ is partitioned into path-connected subsets with the property that two points in different subsets cannot be connected by a path in $S$.

(c) Which of the following loci in $\mathbb{R}^2$ are path-connected: $\{x^2 + y^2 = 1\}$, $\{xy = 0\}$, $\{xy = 1\}$?

I'm fine with parts (a) and (b). My questions and uncertainties are:

1. Artin doesn't give a metric on $\mathbb{R}^k$, but I assume when he says continuous, he means with respect to the standard Euclidean metric on $\mathbb{R}^k$. The metrics on $\mathbb{R}^k$, I believe, are all equivalent in the sense that they induce the same notion of continuity, so I believe it is fine to use the Euclidean metric. Some confirmation on this would be appreciated.

2. I'm a bit stumped on part (c). The first thing I did was plot each of the loci, but I don't have any intuition for which of these are path-connected from the plot or where the interval $[0,1]$ comes into play.

I'd appreciate some help on how to appropriate this problem and how to think about path-connectedness geometrically.

Answer 1

1. Yes, you can use the Euclidean metric.

2. Hints: $\{x^{2}+y^{2}=1\}$ is path connected because any two points on it can be joined by a circular arc. [Using polar form, we can take any two points $e^{i\theta_1}$ and $e^{i\theta_2}$ on the circle and define $X: [0,1] \to \{x^{2}+y^{2}=1\}$ by $X(t)=e^{it\theta_2+(1-t)i\theta_1}$].

$\{xy=0\}$ is the union of the two coordinate axes. It is path connected because any two points on it can be jointed by a single line segment or two line segments. (You can go from any point to $(0,0)$ and then to the other point).

$\{xy=1\}$ is not even connected, so it cannot be path connected. $\{y >-x\}$ and $\{y <-x\}%+$ form a separation of this set.

Answer 2

Path connectedness answers the question of whether you can draw a line between two arbitrary points in your space without lifting your pencil up, such that the line only passes through points in your space. So, for example, if you have two disjoint intervals, they are not path connected. If you have a circle, however, it is path connected - choose any two arbitrary points and your path is just the arc between them.

The convention is to consider a path as a function from $[0,1]$ to your space, but in reality, you can take any compact, connected interval in $\mathbb{R}$- say, $[22,2022]$. The important thing is compact and connected, however - it ensures that, since the function is continuous, its image is (a) connected (so, your pencil didn't lift up while drawing this path) (b) compact - so includes the endpoints (i.e. the points you're trying to find a path between).

And yes, it is fine to use the Euclidean metric here.