- 27

- 0

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!

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!

Last edited: