Let $X = \mathbb{R}^2$ and $d: X \times X \to [0, \infty)$ defined by

$$d(x,y) = \begin{cases} \sqrt{(x_1-y_1)^2 + (x_2-y_2)^2} & \text{if $x = \lambda y$ for some $\lambda \in \mathbb{R}$} \\\sqrt{x_1^2 + x_2^2} + \sqrt{y_1^2+y_2^2} &\text{otherwise} \end{cases}$$

First I had to show that $E = \{x \in \mathbb{R}^2: x_1^2+x_2^2 =1\}$ is a bounded and closed set in $(X, d)$.

Then I was supposed to show that $E$ is not compact in two ways:

1) Find a sequence in $E$ with no convergent subsequence

2) Find an open cover of E which does not have a finite subcover

I'm confused about showing $E$ not compact because I thought that $E$ being bounded and closed meant it was sequentially compact which implied it was compact. Also, could someone provide a hint for finding a sequence in $E$ with no convergent subsequence.

  • $\begingroup$ Closed and bounded = compact depends on the metric space. A simple case of this failing is on the rationals. (e, pi) for example is closed and bounded on the rationals (but open on the reals). The purpose of this exercise is to show that on a non-compete metric space (that non converging metric space shows this space isn't complete) don't have the Heine Borel property. $\endgroup$
    – fleablood
    Mar 18 '17 at 5:59

It is unfortunate that you use $x$, and $y$ to indicate 2 different points on a 2D plane, and use subscripts to indicate what we usually consider the x-coordinate, and the y-coordinate.

On Compactness

The first part of this answer is brought to you by wikipedia.

Wikipedia's page on the Heine-Borel Theorem states that close and bounded is equivalent to compact only in a subset of Euclidean space. This means that the norm must be Euclidean or equivalent to Euclidean.

The metric given is not remotely close to Euclidean. As such, the norm is not Euclidean either. In fact, I cannot immediately (or even now) prove that $d$ is a metric at all. (It needs to satisfy the triangle inequality.) Edit: After 20 min of contemplation, I have determined that the proof of the triangle inequality for the metric $d$ is trivial.

Also according to that page, a subset of a metric space is compact if and only if it is complete and totally bounded. (More on this later.)

Sequence in $E$.

As for the second part of your question. I want to consider what $E$ is, and what the metric $d$ is.

E is the Euclidean unit circle.

The metric says: To find the distance between 2 points, take the Euclidean distance from the first point to the origin, and the Euclidean distance from the second point to the origin, and add them together. In the case that one point is is a multiple of the other, the distance between these two points is just normal Euclidean distance.

From this point, you can easily demonstrate that the distance between any 2 different points in the circle $E$ has a distance of 2.

Regarding Complete and Totally bounded

Since there are no Cauchy sequences in $E$ (since the distance between any two different points is 2), $E$ is trivially complete. (Except the sequence where every point is the same).

However, $E$ is not totally bounded. The only points covered by a open ball of radius $\epsilon<2$ around any point $x$ is $x$ itself. As such, you can't use a finite number of these balls to cover $E$.


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