Proving Convergence and Hausdorffness in the Finite Complement Topology on R

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The discussion focuses on the convergence of the sequence xn = 1/n in the finite complement topology on R (FCTR). It is established that R with FCTR is not a Hausdorff space, as any two neighborhoods U1 and U2 cannot be disjoint. Furthermore, it is proven that the sequence xn converges to every point in R, as the neighborhood of any point x in FCTR includes all points except xn, ensuring convergence to x.

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  • Understanding of finite complement topology (FCTR)
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Ka Yan
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Is my reasoning correct?

Original problem:
In the finite completement topology on R(denoted by FCTR here), to what point or poionts does the sequense xn = 1/n converge?

I firstly prove that R with FCT does not a Hausdorff.

Let Tf be FCTR, x1, x2 are two arbitrary points of R, U1, U2 are their neighborhoods, respectively.

Then R with Tf is not a Hausdorff, since:
U1, U2 are open in R, U1, U2[tex]\in[/tex]Tf. R-U1 is finite, and R-U2 is finite. Then if U1[tex]\cap[/tex]U2=[tex]\phi[/tex] (which is necessary for a Hausdorff space), then R-(U1[tex]\cap[/tex]U2) will be R. Whereas R-(U1[tex]\cap[/tex]U2) = (R-U1)[tex]\cup[/tex](R-U2), which is finite, by definition, and impossible to be R. Hence U1 and U2 are not disjoint. Thus R with FCT does not a Hausdorff.

Secondly I prove xn converge to every point of R.

Since for every x in R with FCT, the neighborhood of x is the set Ux=R-{xn}. And for every Ux, all xn are in Ux, thus xn converge to x. For the arbitrary of x, xn converge to every point of R.

Thx!
 
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