- #1

- 587

- 247

- Homework Statement
- .

- Relevant Equations
- .

##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##.