Sign up ×

Consider $\mathbb{R}^2$ with the metric: $$d((x_1,y_1),(x_2,y_2))= \begin{cases} |y_1-y_2| \text{ if } x_1=x_2 \\ 1+|y_1-y_2| \text{ if } x_1 \neq x_2 \end{cases}$$

Show that $E= \{(x,0) : -1 \leq x \leq 1\}$ is not compact.

I wanted to show this two different ways - the usual way of finding a finite subcover and then by showing its now totally bounded.

My problem here is trying to understand what an open cover would even look like. In the usual space we just draw disks (or balls in $\mathbb{R^n}$). In this metric space are all open sets just bars of infinite width and of height of at least $|y_2-y_1|$?

share|cite|improve this question

3 Answers 3

up vote 1 down vote accepted

Your space $E$ is isometric to $[-1,1]$ with the discrete metric $\rho(x,y) = \delta_{xy} = \begin{cases}1 & \text{if } x\neq y \cr 0 & \text{if } x=y\end{cases}$.

In particular every sequence in $E$ (or in any space with a metric defined like $\rho$) is bounded, but only the eventual constant sequences have convergent subsequences, so $E$ is not sequentially compact and thus not compact as the two notions coincide for metric spaces.

It's also easy to argue that $E$ is not totally bounded: You cannot cover it with finitely many $1$-balls (in the induced subspace metric) as each such ball is a singleton.

share|cite|improve this answer
I think Isometries are outside my scope of knowledge - the furthest I have gotten is Baby Rudin. – abet Oct 30 '12 at 5:35

$|y_2-y_1|$ is only defined in terms of two specific points, so it makes no sense to talk about bars with height at least that.

Consider a fixed point $(x, y)$. The only points that are close to it are points with the same $x$ value; everything else is at least distance 1 away. In effect, we've split up $\mathbb{R}^2$ into a collection of vertical lines, each of which has the usual metric on it.

So we can see that if $\{A_j\}$ is a set of open intervals in $\mathbb{R}$, then $\{x\} \times \cup\{A_j\}$ is an open set in our space. The open sets will consist of collections of these. In other words, for any $x$ value that is present at all, the corresponding $y$ values must form a union of open intervals in $\mathbb{R}$.

To make an open cover of $E$ that has no finite subcover, for every $0 \le x \le 1$ let $A_x = \{(x, y): -1 < y < 1\}$. Then $\cup A_x$ is an open cover of E. But there are $2^{\aleph_0}$ elements of this open cover and we can't remove even one. So E is not compact.

share|cite|improve this answer

Fix a point $p=\langle x_0,y_0\rangle$, and suppose that $\epsilon\le 1$; what points are in $B(p,\epsilon)$, the open $\epsilon$-ball centred at $p$? A point $q=\langle x,y\rangle$ is in $B(p,\epsilon)$ iff $d(p,q)<\epsilon$. If $x=x_0$, $d(p,q)=|y-y_0|$, so we must have $y_0-\epsilon<y<y_0+\epsilon$. If $x\ne x_0$, then $d(p,q)=1+|y-y_0|\ge 1\ge\epsilon$, so $q\notin B(p,\epsilon_0)$.

In other words, for $\epsilon\le 1$ the ball $B(p,\epsilon)$ is just the open vertical line segment

$$\big\{\langle x_0,y\rangle:y_0-\epsilon<y<y_0+\epsilon\big\}\;;$$

all points even moderately close to $p$ (meaning less than $1$ unit away) are on the same vertical line as $p$.

This means, for instance, that if $\langle q_n:n\in\Bbb N\rangle$ is a sequence converging to $p$, there must be an $m\in\Bbb N$ such that all of the points $q_n$ with $n\ge m$ lie on the line $x=x_0$: any point not on that line is at least $1$ unit away from $p$.

Now if $p=\langle x,0\rangle$, then $B(p,1)=\{\langle x,y\rangle:-1<y<1\}$. Thus, $$E=\bigcup_{x\in\Bbb R}B\big(\langle x,0\rangle,1\big)\;,$$ and $\mathscr{B}=\left\{B\big(\langle x,0\rangle,1\big):x\in\Bbb R\right\}$ is an open cover of $E$. Does $\mathscr{B}$ have any proper subcover at all? Or is every single member of $\mathscr{B}$ necessary in order to cover $E$?

share|cite|improve this answer

Your Answer


