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# Topology - open or closed susbset

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For the following, I'm trying to decide (with proof) if A is a closed subset of Y with respect to the topology, T

(i) Y = N, T is the finite complement topology, A = {n e N | n^2 - 2011n+1 < 0}.

(ii) Y = R, T is the usual topology, A is the set of irrational numbers between 0 and 1.

(iii) Y = R, T is the discrete topology, A is the set consisting of all the numbers that do not have 5, 3, or 2 in any of their decimal expansions.

https://brainmass.com/math/geometry-and-topology/topology-open-or-closed-susbset-376660

#### Solution Preview

(i) A set B is open in the finite complement topology if NB is finite. As always, B is closed if the complement to B is open, that is, the set N(NB) = B is finite. The equation y=x^2-2011x+1 defines a parabola, which opens up. This parabola has two points of intersection with the x-axis, about a=0.0005 and b=2000 (approximately). The set of real solutions to the inequality
x^2-2011x+1<0
is then the interval (a,b), and the set of integer solutions ...

#### Solution Summary

This solution helps with a problem about geometry and topology. Step by step calculations are given.

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