- #1
Ka Yan
- 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\inTf. R-U1 is finite, and R-U2 is finite. Then if U1\capU2=\phi (which is necessary for a Hausdorff space), then R-(U1\capU2) will be R. Whereas R-(U1\capU2) = (R-U1)\cup(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\inTf. R-U1 is finite, and R-U2 is finite. Then if U1\capU2=\phi (which is necessary for a Hausdorff space), then R-(U1\capU2) will be R. Whereas R-(U1\capU2) = (R-U1)\cup(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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