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

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 @$d@$ 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 @$\delta=\epsilon-d(x,y) \gt 0 \ as \ d(x,y) \lt \epsilon.@$

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

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

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.


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