Have I shown correctly that the set M is compact with respect to the metric $d(x,y):=\min\{|y-x|,1-|y-x|\}$ for $x,y \in M$?

Question

Take the half-open interval $M=[0,1)$ we define a distance $d:M\times M \rightarrow \Bbb R_0^+$ by

$d(x,y):=\min\{|y-x|,1-|y-x|\}$ for $x,y \in M$.

I want to show that $(M,d)$ is compact, given the hint that the sequence $(1-\frac{1}{n})$ converges to zero in this metric.(This is from a past exam paper by the way , it's not homework)

Here's what I'm thinking so far :

So I'm aiming to show that the set is sequentially compact so I can then say that it is compact.

So using the sequence mentioned above we take it that x and y are terms in the sequence and $\frac{1}{n}$ is the distance between them i.e. $\frac{1}{n}=|y-x|$. I need to show that it has a convergent subsequence which converges to a point in M.

So take $1-\frac{1}{2n}$ this is a subsequence of our original sequence which converges to 1.

However according to our distance function $d$ we take the minimum between $|y-x|$ and $1-|y-x|$ so if $1-\frac{1}{2n}$ converges to 1 at the same time we have $|y-x|=0$ taking the minimum of these gives zero, which is in our set, so our set is sequentially compact and so compact.

Could someone tell me please if this is correct, or how I can fix my argument?

Answer 1

My approch to this problem would be to note that the given metric is 'the same' as the standard metric $d_{st}(x,y)=|x-y|$ if $d_{st}(x,y) \le 0.5$.

Because under the standard metric the interval $[0,1]$ is compact, this means every sequence in $[0,1]$ has a (standard-)convergent subsequence.

Since $M \subseteq [0,1]$, any sequence of elements from $M$ has thus a subsequence with some standard-limit $t \in [0,1]$.

Now consider the two cases $t<1$ and $t=1$. The former can be handled by my first remark. For the latter, think of the hint given and how you may be able to generalize it.

Answer 2

To show that it is sequentially compact you have to take an arbitrary sequence and prove that it has a convergent subsequence. I think that the sequence they provide you with serves to build intuition as to what this space may look like. To that end consider the curve in the euclidean space $\mathbb{R^2}$ given by the polar equation $r=\sin 2\theta$ where $0 \leq \theta < \pi/2$. Said curve is closed and bounded and thus compact. Now find a homeomorphism between the space $M$ and this curve.