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Homework Help: Topology: Understanding open sets

  1. Nov 29, 2017 #1
    1. The problem statement, all variables and given/known data
    We define ##X=\mathbb{N}^2\cup\{(0,0)\}## and ##\tau## ( the family of open sets) like this
    ##U\in\tau\iff(0,0)\notin U\lor \exists N\ni : n\in\mathbb{N},n>N\implies(\{n\}\times\mathbb{N})\backslash U\text{ is finite}##
    ##a)## Show that ##\tau## satisfies that axioms for open sets
    ##b)## Show that ##(0,0)## lies in the Closure of ##\mathbb{N}^2##
    ##c)## Describe closed sets in topology ##\tau##
    ##d)## show that there doesn't exists a sequence##(x_{n})_{n\in\mathbb{N}}\subset\mathbb{N}^2 \text{ for which }\lim x_{n}\xrightarrow[n->\infty]{X}(0,0)##.
    Assume that X is not first-countable
    ##e)##
    2. Relevant equations


    3. The attempt at a solution
    I'm having trouble visualizing the sets in ##\tau## I know from the first part that the every point which is not ##(0,0)## is open. Also I know that every union of such sets will also be open. Therefore ##\mathbb{N}^2## in itself is open. However I don't know how to visualize the other condition ##\exists N\ni : n\in\mathbb{N},n>N\implies(\{n\}\times\mathbb{N})\backslash U\text{ is finite}## I would really appreciate it if someone could explain to me how this sets look as I am unable to continue with the problem.
    Thank you
     
  2. jcsd
  3. Nov 29, 2017 #2

    fresh_42

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    What does ##\ni :## mean? And are you sure it is an OR in the defining property? That makes every set which doesn't contain ##(0,0)## an open set. Can you draw an open set which includes the origin, e.g. are ##\{(0,0),(1,1)\}## or ##\{(0,0),(1,\mathbb{N})\}## open?
     
  4. Nov 29, 2017 #3
    https://i.gyazo.com/a809d7be047d7ed855211b4d51372367.png
    a809d7be047d7ed855211b4d51372367.png
    This is the original picture of the problem and pretty I'm sure that I translated it correctly. The ##\ni## is the main part that is bothering me because I don't know exactly what it means
     
  5. Nov 29, 2017 #4

    WWGD

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    Well, let's see:
    So the open sets ##U ## are those which:
    1) Do not contain ## (0,0) ## or :
    2) The collection of pairs ## (n+k) \times \mathbb N - U ; k=0,1,2,...= \{ (n \times \mathbb N), (n+1) \times \mathbb N ,.....,(n+k) \times \mathbb N ,.... \} -U ## is finite . This means ##U## must contain EDIT all but finitely-many of the sets in each of the collections ## (n+k) \times \mathbb N; k=0,1,2,... ##
    Does this help?
     
  6. Nov 29, 2017 #5

    WWGD

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    Best I can tell
    It means something like " So that" , or " With the property that" . EDIT and so it seems like this may be:

    https://en.wikipedia.org/wiki/Cofiniteness#Cofinite_topology
     
    Last edited: Nov 29, 2017
  7. Nov 29, 2017 #6

    fresh_42

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    Never seen it before. It probably means ##\forall ##. It's a bit strange and should be explained somewhere in the book.
    Google translates "ali" by "but", so I assume it to mean "and" and not "or", which means that not all sets without the origin are open. Thus I read it as follows:
    $$
    U \in \tau \Longleftrightarrow (0,0) \notin U\; \wedge \; \exists N \in \mathbb{N}\; \forall n > N \, : \,\vert \, \{n\}\times \mathbb{N} \backslash U \,\vert < \infty
    $$
    I haven't checked whether it's a topology or not. I would feel more comfortable, if this guesswork above wouldn't had been necessary. Especially "ali" makes a major difference.
     
  8. Nov 29, 2017 #7

    WWGD

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    No, I think only those who contain ##(0,0)## and all-but-finitely-many elements in the set. Ze Risky topology you suggested in Random Thoughts. EDIT Then open sets are either those that contain all-but-finitely many elements of ##\mathbb N^2 ## or those that miss the origin. I think it is clear to see closedeness under union: 1) If neither set contains the origin, neither will the union; if each contains all-but-finitely-many, so will the union. 2) Clearly the whole space is there, as it contains all but finitely many ( none) . And clearly the empty set is also there. EDIT2: Still, don't b) and d) contradict each other?
     
    Last edited: Nov 29, 2017
  9. Nov 29, 2017 #8

    fresh_42

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    You are right and Google translate was wrong!
    I checked two other translation pages and they both had "ali = OR"
    Corrected condition:
    $$U\in \tau \Longleftrightarrow (0,0)\notin U \vee \exists N\in \mathbb{N} \; \forall n>N\, : \, |\,\{n\}×\mathbb{N} \,\backslash \,U\,| < \infty $$
     
  10. Nov 29, 2017 #9
    Thank you both
    Now I tried to continue what WWGD said. A set open if it does not contain ##(0,0)## or there exists some number ##N## so that for every number greater than N ##(\{n\}\times\mathbb{N})\backslash U ## is finite.
    So we need to check
    ##a)## whether any number of unions of such sets is again open
    ##b)## a finite number of intersection of such sets is again open
    So checking for intersections:
    Lets ignore the first part because if ##(0,0)\notin (A\wedge B)\implies (0,0)\notin A\cap B##
    Now lets say that there exist two numbers ##N,M## for sets ##U,V##such that ##\forall n>N\implies(\{n\}\times\mathbb{N})\backslash U\text{ is finite}## and ##\forall n>M \implies(\{n\}\times\mathbb{N})\backslash V\text{ is finite}## if we define ##k=\max\{M,N\}## the set we again get is open because both ##U## and ##V## can only miss finite many points therefore their intersection can only miss finite many points.

    I believe I understand now how this topology works and will probably be able to continue.

    Thank you both very much for the help
     
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