Answer to Question #117901 in Differential Geometry | Topology for Sheela John

Question #117901
Prove or disprove any metric defined on X(#0) induces a topology on X
1
Expert's answer
2020-05-26T20:00:50-0400

Let "(X, \\rho)" be a metric space with metric "\\rho". Let's define the following:

"B_{\\varepsilon}(x) := y \\in X | \\rho(x,y) <\\varepsilon". Let's show, that for any metric space "(X, \\rho)" : "B_{\\varepsilon}(x) := y \\in X | \\rho(x,y) <\\varepsilon, x \\in X, \\varepsilon > 0" is a base for some topology. If we prove this, it will be enough, since any base generates topology on "X", that has, as open sets, all unions of elements of the base.

So, to prove that the defined above set is a base we will check if base criterias are true, namely:

1) "\\bigcup_{x,\\varepsilon}B_{\\varepsilon}(x) = X."

2) If "z \\in B_{\\varepsilon}(x) \\cap B_{\\delta}(y)," then there exist such element of our base "B_0", that "z \\in B_0" and "B_0 \\subset B_{\\varepsilon}(x) \\cap B_{\\delta}(y)".


1) For a given "x \\in X", "X = \\bigcup_{n=1}^{\\infin}B_n(x)". Therefore, "\\bigcup_{x,\\varepsilon}B_{\\varepsilon}(x) = X."

2) Let "z \\in B_{\\varepsilon}(x) \\cap B_{\\delta}(y)". Let "r := min(\\varepsilon - \\rho(x,z), \\delta - \\rho(y, z))." Then, "B_r(z) \\subset B_{\\varepsilon}(x) \\cap B_{\\delta}(y)".

So, we proved that the defined set is a base for some topology for "(X, \\rho)."


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