Show that if d is a metric, then d'=sqrt(d) is a metric

[docnet]

Homework Statement

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Relevant Equations

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$d'$ is a metric on $X$ because it satisfies the axioms of metrics:

Identity of indiscernibles:
$x=y\Longleftrightarrow d(x,y)=0\Longleftrightarrow \sqrt{d(x,y)}=\sqrt{0}$

Symmetry: $d(x,y)=d(y,x)\Longrightarrow \sqrt{d(x,y)}=\sqrt{d(y,x)}$

Triangle inequality: $d(x,z)\leq d(x,y)+d(y,z)\Longrightarrow \sqrt{d(x,z)}\leq \sqrt{d(x,y)+d(y,z)}\leq \sqrt{d(x,y)}+\sqrt{d(y,z)}$

Open sets for $d'$are the same as the open sets for $d$.

Given $x$ in $U$, let $\epsilon>0$ such that any $y$ that satisfies $d(x,y)< \epsilon$ is in the open set $U$ for $d$. Define $\delta=\sqrt{\epsilon}$. Then,
$d(x,y)<\epsilon\Longrightarrow \sqrt{d(x,y)}<\delta$.
So any $y$ in the open set $U$ for $d$ is in the open set $U$ for $d'$.

Given $x$ in $U$, let $\delta>0$ such that any $y$ that satisfies $\sqrt{d(x,y)}< \delta$ is in the open set $U$ for $'d$. Again, define $\delta^2=\epsilon$. Then,
$\sqrt{d(x,y)}<\delta \Longrightarrow d(x,y)<\epsilon$.
So any $y$ in the open set $U$ for $d'$ is in the open set $U$ for $d$.

 

[fresh_42]

Looks ok, except for some typos in the last part.

 

[FactChecker]

All true, but you should probably mention what properties of the square root function you are using in the last two inequalities of the triangle property proof. There are places where you are applying a property of the square root function that you should mention (and maybe prove).

 

[pasmith]

I think the last part is best done by noting that for every x_0 \in X and every r \geq 0 we have <br /> \{ x \in X : d(x,x_0) &lt; r \} = \{ x \in X: d&#039;(x,x_0) = \sqrt{d(x,x_0)} &lt; \sqrt{r}\}. So if a set contains a d-open ball of each of its points then it contains a d&#039;-open ball of each of its points and vice-versa.

 

[PeroK]

I would prefer something more explicit. Showing that $d'$ is a metric by almost never writing $d'$ seems not quite right to me. I feel I'm having to fill in too many blanks.

Also, for the open sets question, I'd like to see something more formulaic:

Let $U$ be open under $d$, then ... $U$ is open under $d'$.

Let $U be open under$d'$, then ...$U$is open under$d'##.

 

[docnet]

All true, but you should probably mention what properties of the square root function you are using in the last two inequalities of the triangle property proof. There are places where you are applying a property of the square root function that you should mention (and maybe prove).

Not sure what name this is called, but it seemed clear from

$\sqrt{d(x,y)+d(y,z)}\leq\sqrt{d(x,y)+d(y,z)+2\sqrt{d(x,y)d(y,z)}}=\sqrt{d(x,y)}+\sqrt{d(y,z)}$

I think the last part is best done by noting that for every x_0 \in X and every r \geq 0 we have <br /> \{ x \in X : d(x,x_0) &lt; r \} = \{ x \in X: d&#039;(x,x_0) = \sqrt{d(x,x_0)} &lt; \sqrt{r}\}. So if a set contains a d-open ball of each of its points then it contains a d&#039;-open ball of each of its points and vice-versa.

I agree with you. My main difficulty was explaining how $d(x,y)<r$ is the same as $d'(x,y)<\sqrt{r}$

I would prefer something more explicit. Showing that $d'$ is a metric by almost never writing $d'$ seems not quite right to me. I feel I'm having to fill in too many blanks.

Also, for the open sets question, I'd like to see something more formulaic:

Let $U$ be open under $d$, then ... $U$ is open under $d'$.

Let $U$ be open under $d'$, then ... $U$ is open under $d'$.

$d'$ is a metric on $X$ because it satisfies the axioms of metrics:

Identity of indiscernibles:
$x=y\Longleftrightarrow d(x,y)=0\Longleftrightarrow \sqrt{d(x,y)}=\sqrt{0}\Longleftrightarrow d'(x,y)=0$

Symmetry: $d(x,y)=d(y,x)\Longrightarrow \sqrt{d(x,y)}=\sqrt{d(y,x)}\Longleftrightarrow d'(x,y)=d'(y,x)$

Triangle inequality: $d(x,z)\leq d(x,y)+d(y,z)\Longrightarrow \sqrt{d(x,z)}\leq \sqrt{d(x,y)+d(y,z)}\leq \sqrt{d(x,y)}+\sqrt{d(y,z)}\Longleftrightarrow d'(x,z)\leq d'(x,y)+d'(y,z)$ (because the square root of the sum is less than equal to the sum of the square roots)Let $U$ be open under $d$, then for all $x$ in $U$, there is an $\epsilon>0$ such that any $y$ that satisfies $d(x,y)< \epsilon$ is in the open set $U$ for $d$. Define $\delta$ as $\sqrt{\epsilon}$, then any $y$ that satisfies $d(x,y)<\epsilon$ also satisfies $d'(x,y)< \delta$, and is in the open set $U$ for $d$.

Let $U$ be open under $d'$, then for all $x$ in $U$, there is a $\delta>0$ such that any $y$ that satisfies $d(x,y)< \delta$ is in the open set $U$ for $d'$. Define $\epsilon$ as $\delta^2$, then any $y$ satisfies $d'(x,y)< \delta$ also satisfies $d(x,y)<\epsilon$, and is in the open set $U$ for $d$.

 

[PeroK]

Let $U$ be open under $d$, then for all $x$ in $U$, there is an $\epsilon>0$ such that any $y$ that satisfies $d(x,y)< \epsilon$ is in the open set $U$ for $d$. Define $\delta$ as $\sqrt{\epsilon}$, then any $y$ that satisfies $d(x,y)<\epsilon$ also satisfies $d'(x,y)< \delta$, and is in the open set $U$ for $d$.

It's more about technique that correctness, but here's how I would do this:

Let $U$ be open under $d$ and $x \in U$. There exists $\epsilon>0$ such that $d(x, y) < \epsilon \ \Rightarrow \ y \in U$.

Now $d'(x, y) < \sqrt{\epsilon} \Rightarrow \ d(x, y) < \epsilon \ \Rightarrow \ y \in U$. And, as $x$ was arbitrary, we see that $U$ is open under $d'$.

 

[nuuskur]

#1 It's good practice to briefly mention what you are going to do before you start chasing the epsilons around. We take a subset of $X$ and show it is open with respect to $d$ if and only if it is open with respect to $\sqrt{d}$. Mentioning such things explicitly makes things easier for yourself and also helps convince your grader more.

 

[FactChecker]

Here are some steps where I think you could/should have mentioned some properties of the square root function and included some more detail.

Not sure what name this is called, but it seemed clear from

$\sqrt{d(x,y)+d(y,z)}\leq\sqrt{d(x,y)+d(y,z)+2\sqrt{d(x,y)d(y,z)}}=\sqrt{d(x,y)}+\sqrt{d(y,z)}$

$\sqrt{d(x,y)+d(y,z)}\leq\sqrt{d(x,y)+d(y,z)+2\sqrt{d(x,y)d(y,z)}}$ because the square root function is increasing and $d(x,y)+d(y,z) \leq d(x,y)+d(y,z)+2\sqrt{d(x,y)d(y,z)}$.

$\sqrt{d(x,y)+d(y,z)+2\sqrt{d(x,y)d(y,z)}}= \sqrt{(\sqrt{d(x,y)}+\sqrt{d(y,z)})^2} = \sqrt{d(x,y)}+\sqrt{d(y,z)}$

Another step could use some more explanation:

$d(x,y)=0\Longleftrightarrow \sqrt{d(x,y)}=\sqrt{0}$

Because the square root function is one-to-one.