Finite-Compliment Topology and intersection of interior

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SUMMARY

The discussion addresses the finite-complement topology on the real line, specifically demonstrating that the arbitrary intersection of the interiors of subsets does not equal the interior of the arbitrary intersection of those subsets. The counterexample provided is A(n) = R - {n}, which effectively illustrates that the intersection of the interiors, denoted as ∩int(A_n), does not equal int(∩A_n). This conclusion is established through the analysis of the properties of the finite-complement topology compared to the usual topology.

PREREQUISITES
  • Understanding of finite-complement topology
  • Familiarity with the concept of interior points in topology
  • Knowledge of set operations and intersections
  • Basic grasp of real analysis and topology notation
NEXT STEPS
  • Study the properties of finite-complement topology in detail
  • Explore examples of arbitrary intersections in various topological spaces
  • Learn about the differences between finite-complement topology and usual topology
  • Investigate the implications of interior points in different topological contexts
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This discussion is beneficial for students and researchers in mathematics, particularly those focusing on topology, as well as educators seeking to clarify concepts related to finite-complement topology and set operations.

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[SOLVED]Finite-Compliment Topology and intersection of interior

Homework Statement



Given topological space (R[tex]^{1}[/tex], finite compliment topology), find counter example to show that

Arbitary Intersection of (interior of subset of R[tex]^{1}[/tex]) is not equal to Interior of (arbitary intersection of subset of R[tex]^{1}[/tex]).

[tex]\bigcap^{\infty}_{n=1}int(A_{n})\neq int(\bigcap^{\infty}_{n=1}A_{n})[/tex]

Homework Equations



When we consider topological space (R[tex]^{1}[/tex], usual topology), it is easy to find out that [-1/n, 1/n] is the example.


The Attempt at a Solution



First, I thought what shapes the open of topological space (R[tex]^{1}[/tex], finite compliment topology) might have, and it seems to have some shape of real line having finite omissions.
And I tried [tex]A_{n}[/tex] = R-[-1/n, 1/n], and [-1/n, 1/n], but found out all of these are not counterexample..
 
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Look at random subsets of R, and study what their interiors look like.
 
A(n) = R-{n}

A(n)=R-{n} is the counterexample...
I tried this case many times but it was not the counterexample,
but after some fresh air, it is clear this is counter example..
 

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