Question #117946
Prove or disprove any metric defined on X(#0) induces a topology on X
1
Expert's answer
2020-05-25T20:49:04-0400

Let dd is a metric on the set XX .

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

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

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

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

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

d(x,z)d(x,y)+d(y,z)<ϵ.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 yB1B2y\in B_1\cap B_2 .We have just shown that we can choose positive numbers δ1 and δ2\delta_1 \ and \ \delta_2 so that B(y,δ1)B1B(y,\delta_1)\sub B_1 and B(y,δ2)B2B(y,\delta_2)\sub B_2 . let δ=min(δ1,δ2)\delta=min(\delta_1,\delta_2) .

Thus B(y,δ)B1B2B(y,\delta)\sub B_1\cap B_2 .

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

Thus by definition of basis XX induce a Topology.


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