[Point-Set] Limit Point Help
Hello,
I'm taking Adv. Calculus (my uni's undergrad analysis course) this semester and took topology last semester. Since we just started continuity in this course, I was thinking of reproving a lot of stuff. Anyways on the last line of the board in the picture, I noticed something. Is the existence of a sequence contained in a set with a point removed that will also be eventually contained in any (open) neighborhood [1] the same as saying the any neighborhood has such a sequence [2] (swapping the for all and there exists).
I understand that in general, swapping for all and there exists changes the statements, but here, I was kind've wondering. After all, if we assume [1]. We can choose a sequence that satisfies [1]. So, any neighborhood must eventually contain this sequence, which gives us existence and thus [2]. However, if we assume [2], we only have that every neighborhood eventually contains such a sequence, not necessarily that there exists a sequence eventually contained in all of them (which is indeed [1] and what I made this post for).
My first approach for this direction was recognizing that all neighborhoods (by openness) contain an open ball centered around the point. So, choose such a ball from each neighborhood. By [2], each of the balls will eventually contain some sequence (that itself is contained by our set with the point removed). This is where I'm stuck, as one sequence may be eventually contained in a ball but that does not imply this same sequence will be eventually contained in the next smallest ball (only that it is eventually contained in all our larger balls and that there is SOME sequence contained in our next smallest ball). At this point, I feel that either [1] implies [2] but not the other way around, or that I'm missing something.
Thanks for the help in advance! P.S. When I say the set or point, I mean B and x resp. Also, N_x is my notation for the set of all neighborhoods (open sets containing x).
Hello,
I'm taking Adv. Calculus (my uni's undergrad analysis course) this semester and took topology last semester. Since we just started continuity in this course, I was thinking of reproving a lot of stuff. Anyways on the last line of the board in the picture, I noticed something. Is the existence of a sequence contained in a set with a point removed that will also be eventually contained in any (open) neighborhood [1] the same as saying the any neighborhood has such a sequence [2] (swapping the for all and there exists).
I understand that in general, swapping for all and there exists changes the statements, but here, I was kind've wondering. After all, if we assume [1]. We can choose a sequence that satisfies [1]. So, any neighborhood must eventually contain this sequence, which gives us existence and thus [2]. However, if we assume [2], we only have that every neighborhood eventually contains such a sequence, not necessarily that there exists a sequence eventually contained in all of them (which is indeed [1] and what I made this post for).
My first approach for this direction was recognizing that all neighborhoods (by openness) contain an open ball centered around the point. So, choose such a ball from each neighborhood. By [2], each of the balls will eventually contain some sequence (that itself is contained by our set with the point removed). This is where I'm stuck, as one sequence may be eventually contained in a ball but that does not imply this same sequence will be eventually contained in the next smallest ball (only that it is eventually contained in all our larger balls and that there is SOME sequence contained in our next smallest ball). At this point, I feel that either [1] implies [2] but not the other way around, or that I'm missing something.
Thanks for the help in advance! P.S. When I say the set or point, I mean B and x resp. Also, N_x is my notation for the set of all neighborhoods (open sets containing x).