Show that $d_1(x, x')$ and $d_2(x, x')$ are both metrics and equivalent

Question

This is a problem that I'm trying to solve, and I want to know if my answer is correct. I have a tl;dr at the end.

Define $\phi(x) := (\cos(2\pi x), \sin(2\pi x)) \; (x \in [0, 1))$.

Question 1. Show that $\phi$ is a bijection between $[0,1)$ and $S^1 := \{ v \in \mathbb{R^2}: \Vert v\Vert_2 = 1 \}$, where $\Vert\cdot\Vert_2$ is the euclidean norm.

I do not have questions with this one, only with the next two.

Question 2. Show that $$ d_1(x, x') := \text{inf} \{ |x - x' + k| : k \in \mathbb{Z} \}, \; (x, x' \in [0,1))$$ defines a metric in $[0,1)$.

Solution.

1. \begin{align*} x = x' \implies \inf\{|k|: k \in \mathbb{Z} \} \implies k = 0 \implies d_1(x, x') = 0. \end{align*}

\begin{align*} d_1(x, x') = 0 \implies \exists k^* \in \mathbb{Z} \; \text{such that} \; |x - x' + k^* | = 0. \; \text{We know that} \; 1 > x \geq 0 \quad \quad \text{and} \; 1 > x' \geq 0 \implies -1 < -x' \leq 0 \implies -1 < x - x' < 1. \; \text{Now,} \; x - x' + k^* = 0 \implies 1 > x - x' = -k^* > -1 \implies k^* = 0 \implies x = x' \quad \text{(This form is the contrapositive of} \; x \neq x' \implies d_1(x, x') \neq 0). \end{align*}

2. \begin{align} d_1(x,x') = |x - x' + k^*| \; (k^* \in \mathbb{Z}) = |(-1)x - x' + k^*| = |x' - x + k^{**}| \geq \text{inf} \{|x - x' + k| : k \in \mathbb{Z} \} = d_1(x', x), \; \text{by the definition of infimum.} \; \text{Doing the same operations, but starting with} \; d_1(x', x) \; \text{implies} \; d_1(x, x') = d_1(x', x) \end{align}

3. \begin{align} d_1(x, x') \leq |x - x' + k| \; (\forall k \in \mathbb{Z}) = | x - x'' + k' + x'' - x' + k''| \leq |x - x'' + k'| + |x'' - x' + k''| \implies d_1(x, x') \leq |x - x'' + k'| + |x'' - x' + k''|. \; \text{Taking the infimum on both} \; k', k'' \; \text{we have} \; d_1(x, x') \leq d_1(x, x'') + d_1(x'', x'). \end{align}

Question 3. Show that $$d_2(x, x') := \Vert\phi(x) - \phi(x')\Vert_2, \; (x, x' \in [0, 1))$$ defines a metric in $[0,1)$ and that is equivalent to $d_1(x, x')$ in the sense that $d_1(x, x_n) \to 0 \iff d_2(x, x_n) \to 0 \; \forall x \in [0,1), \forall \{x_n\}_n \subset [0, 1)$.

Solution.

1.\begin{align} x = x' \implies \phi(x) = \phi(x') \implies \Vert 0\Vert_2 = 0. \end{align}

\begin{align} x \neq x' \implies \phi(x) \neq \phi(x') \; \text{(by injectivity of} \; \phi \implies \Vert\phi(x) - \phi(x')\Vert_2 \neq 0 \; \text{(by the fact that} \; \Vert\cdot\Vert_2 \; \text{is a norm)}. \end{align}

2. \begin{align} d_2(x, x') = \Vert\phi(x) - \phi(x')\Vert_2 = \Vert\phi(x') - \phi(x)\Vert_2 = d_2(x', x). \end{align}

3. \begin{align} d_2(x, x') = \Vert\phi(x) - \phi(x')\Vert_2 \leq \Vert\phi(x) - \phi(x'')\Vert_2 + \Vert\phi(x') - \phi(x'')\Vert_2 = d_2(x, x'') + d_2(x', x''). \end{align}

This proves that it's a metric. Let's prove that they are equivalent.

4. \begin{align} d_2(x_n, x) \to 0 \implies \cos(2\pi x_n) \to \cos(2\pi x) \; \text{and} \; \sin(2\pi x_n) \to \sin(2\pi x) \implies x_n \to x. \quad |x_n - x + k| \; (\forall k) \geq \inf \{|x_n - x + k| : k \in \mathbb{Z} \}. \; \text{In particular, this is true for} \; k = 0, \; \text{so} \; |x_n - x| \geq d_1(x_n, x), \; \text{and as} \; x_n \to x, d_1(x_n, x) \to 0. \end{align}

5. \begin{align} \text{(This one I'm having trouble justifying it)}. \; d_1(x_n, x) \to 0 \implies \quad \lim_{n \to \infty} \inf_k \{|x_n - x + k|\} = 0 \implies \inf_k \lim_{n\to \infty} |x_n - x + k| = 0 \implies k = 0 \implies x_n \to x \implies \cos(2\pi x_n) \to \cos(2\pi x) \; \text{and} \; \sin(2\pi x_n) \to \sin(2\pi x) \; \text{(by continuity of cos and sin)} \implies d_2(x_n, x) \to 0. \end{align}

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I don't know if you guys can help me with that, but it would be so nice if you could.

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tl;dr: I'm not quite sure about question 2 and the step 4 and 5 of question 3.

Answer 1

(This is more of a long comment)

Here's a general rule on writing mathematics: it is more important to correctly convey your thought process than to use correct notation. Usually correct notation is really helpful to that end, but not always. For example, you can take formalized euclidean geometry and try to prove theorems with a firrst-order language, but a drawing is much better accepted (in general) than that, even though it is not formal. That said, I can probably guess what you're trying to do on each of those steps, but it is not really clear what you're trying to explain with your solutions.

For example, in step 1 of Question 2, you never introduced $x,x'$. Why are they equal? Are they in $[0,1)]$? We can probably guess, but this is not acceptable. It would be much better to starting by writing "Let us prove that $d(x,x)=0$ for all $x$". Then you proceed with your proof.

Here is how I would write it.

Solution to Question 1.

In order to prove that $d_1$ is a metric, we need to prove that it satisfies three properties. First, note that $d_1(x,y)\geq 0$ for all $x,y\in[0,1)$, since it is defined as being the infimum of a family of non-negative numbers.

1. $d_1(x,x')=0$ if and only if $x=x'$ in $[0,1)$ In one direction, assume that $x=x'$. Since $0\in\mathbb{Z}$, then the definition of $d_1(x,x')$ yields $$d_1(x,x')\leq |x-x'+0|=|x-x|=|0|=0$$ so $d_1(x,x')=0$ because $d$ is non-negative.
   
   Conversely, suppose that $d_1(x,x')=0$. First, notice that $0\leq x,x'\leq 1$, so $$-1=0-1<x-x'\qquad\text{and}-1=0-1<x'-x$$ so $-1<x'-x<1$. This means that $|x'-x|<1$.

The equality $d_1(x,x')=0$ means that for every $\epsilon>0$ there exists $k^*\in\mathbb{Z}$ (depending on $\epsilon$) such that $|x-x'+k^*|\leq \epsilon$. Let us temporarily fix such $\epsilon$ and $k^*$. We then have $$|k^*|\leq |x'-x|+|x-x'+k^*|<|x'-x|+\epsilon$$ As this argument holds for all $\epsilon>0$, it in particular holds for $0<\epsilon<1-|x'-x|$, in which case we have $$|k^*|\leq |x-x'|+\epsilon<1$$ so $k^*=0$ because $k^*\in\mathbb{Z}$. It follows that $$|x-x'|=|x-x'+k^*|<\epsilon$$ whenever $\epsilon\in(0,1-|x'-x|)$. Taking the infimum on the right-hand side of the inequality above, for all such $\epsilon$, yields $$|x-x'|\leq 0$$ so $x=x'$, as we desired.

The other items should be similar. This solution is a little pedantic, but it should be as clear as possible. You can probably simplify it a little.

And in the solution above I already addressed one of the problems with you solution:

Problem 1: The metric $d$ is defined in terms of an infimum, and not a minimum. Therefore, if $d(x,y)=0$, we just have $$0=\inf\left\{|x-y+k|:k\in\mathbb{Z}\right\}$$ and infima are not necessarily attained.

You have two possible solutions for this problem:

• Use the definition of infimum throughout your solutions (this is the approach in my proposed solution); or
• Prove that the infimum is attained: prove that for all $x,y\in [0,1)$, there exists $k\in\mathbb{Z}$ such that $d(x,y)=|x-y+k|$. In fact, prove that you can always take $k=0$, $k=1$ or $k=-1$.

Problem 2: Step 4 is not correct: Why does $\cos(2\pi x_n)\to\cos(2\pi x)$ and $\sin(2\pi x_n)\to\sin (2\pi x)$ imply that $x_n\to x$ in the metric $d_1$? Why do you have a quantifier "$(\forall k)$" before an inequality?! (This is not a well-formed sentence!).

Step 5 is also not goot: You cannot change the order of limits and infima at will.

These two steps are indeed the hardest ones. Here is a hint: Prove the following characterization of convergence in $d_1$:

• If $x\neq 0$, then $x_n\to x$ in the metric $d_1$ if and only if $x_n\to x$ in the usual metric of $[0,1)$;
• If $x=0$, then $x_n\to 0$ in the metric $d_1$ if and only if $\min(|x_n|,|x_n-1|)\to 0$ in the usual metric of $\mathbb{R}$.

So for Step 4 you separate in two cases $x_n\to x$ in the metric $d_1$ when $x\neq 0$, or when $x=0$;

For step 5, also separate in cases when $x=0$ or $x\neq 0$. The argument should be similar to your solution of question 1, and using the characterization of convergence in $d_1$ above.