Answer to Question #117946 in Differential Geometry | Topology for Sheela John

Question

Prove or disprove any metric defined on X(#0) induces a topology on X

Answer 1

Let $d$ is a metric on the set $X$ .

Claim:= The collection of all $\epsilon$ -ball $B_d(x,\epsilon)$ ,for $x\in X$ and $\epsilon$ >0 , is a basis for a topology on $X$

For each $x\in X$ , $x\in B(x,\epsilon)$ for any $\epsilon$ >0.

Before checking the second condition for a basis element ,we show that if $y$ is a point of the basis element $B(x,\epsilon)$ ,then there is a basis element $B(y,\delta)$ centered at y that is contained in $B(x,\epsilon)$ .

Define $\delta=\epsilon-d(x,y)>0 \ as \ d(x,y)<\epsilon.$

Then $B(y,\delta)\sub B(x,\epsilon)$ ,for if $z\in B(y,\delta)$ ,then $d(y,z)<\epsilon-d(x,y),$ from which we conclude that

$d(x,z)\leq d(x,y)+d(y,z)<\epsilon.$

Now to check the second condition for a basis ,let $B_1 \ and \ B_2 \$

be two basis elements and let $y\in B_1\cap B_2$ .We have just shown that we can choose positive numbers $\delta_1 \ and \ \delta_2$ so that $B(y,\delta_1)\sub B_1$ and $B(y,\delta_2)\sub B_2$ . let $\delta=min(\delta_1,\delta_2)$ .

Thus $B(y,\delta)\sub B_1\cap B_2$ .

As collection of all $\epsilon$ -ball $B(x,\epsilon)$ form a basis of $X$ .

Thus by definition of basis $X$ induce a Topology.