Results 1 to 9 of 9
Like Tree4Thanks
  • 1 Post By Prove It
  • 1 Post By HallsofIvy
  • 1 Post By Deveno
  • 1 Post By Deveno

Math Help - Topological space: Condition 3

  1. #1
    Super Member
    Joined
    Sep 2013
    From
    Portland
    Posts
    540
    Thanks
    73

    Topological space: Condition 3

    \for A,B \in \mathfrak{T}, A \cap B \in\mathfrak{T}

    What is a good way to prove this true? One video I saw said it can be proved by induction but I haven't done an induction proof on a set before. I thought maybe an element proof would be easier to work out.
    Last edited by Jonroberts74; August 9th 2014 at 03:29 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,835
    Thanks
    1604

    Re: Topological space: Condition 3

    Choose an element $\displaystyle \begin{align*} x \end{align*}$. If $\displaystyle \begin{align*} x \in A \end{align*}$ and $\displaystyle \begin{align*} A \subset \mathcal{I} \end{align*}$ then $\displaystyle \begin{align*} x \in \mathcal{I} \end{align*}$.

    But if $\displaystyle \begin{align*} x \in B \end{align*}$ and $\displaystyle \begin{align*} B \subset \mathcal{I} \end{align*}$, then $\displaystyle \begin{align*} x \in \mathcal{I} \end{align*}$.

    Since x is in both A and B, that means $\displaystyle \begin{align*} x \in A \cap B \end{align*}$, and since we know $\displaystyle \begin{align*} x \in \mathcal{I} \end{align*}$, that means $\displaystyle \begin{align*} A \cap B \subset \mathcal{I} \end{align*}$.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,458
    Thanks
    1868

    Re: Topological space: Condition 3

    How about using the basic definitions? " A\subseteq X is defined as "if x\in A then x\in X.


    Suppose x\in A\cap B then, by the definition of A\cap B, x\in A. Since A\subset T, ....
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Sep 2013
    From
    Portland
    Posts
    540
    Thanks
    73

    Re: Topological space: Condition 3

    Both of those are suitable, theres no leap to see its true.

    Thinking about the idea of proving it by some sort of induction is interesting though

    would that be along the lines

    for two such sets

    show its true for two sets intersecting

    then assume K sets intersecting A \cap B \cap......\cap K is true

    then proving K+1 sets having an intersection are in \mathfrak{T}

    A \cap B \cap......\cap K \cap K+1

    This seems maybe incorrect cause what is the K+1 set, the Kth set plus an added element or a set not equal to the Kth set but having intersection with the Kth set that is nonempty? (does it even matter if the intersection is non-empty considering the empty set is in \mathfrak{T} ? )
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Topological space: Condition 3

    If I may ask, what is $\mathfrak{T}$?

    If it is a topology, then it cannot be proven, it is axiomatically true. If it is some collection of sets, whether or not it is true depends on the collection.

    Both ProveIt's response and HallsOfIvy's response seem to indicate that they believe $A,B$ are subsets of $\mathfrak{T}$. I am not sure why they believe this to be true.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Sep 2013
    From
    Portland
    Posts
    540
    Thanks
    73

    Re: Topological space: Condition 3

    Quote Originally Posted by Deveno View Post
    If I may ask, what is $\mathfrak{T}$?

    If it is a topology, then it cannot be proven, it is axiomatically true. If it is some collection of sets, whether or not it is true depends on the collection.

    Both ProveIt's response and HallsOfIvy's response seem to indicate that they believe $A,B$ are subsets of $\mathfrak{T}$. I am not sure why they believe this to be true.
    yes it is a topology, a book I am reading on point-set topology recommended that it would be a good idea to prove it. I also watched a lecture on point-set topology that recommended proving it to oneself--stating that using an induction proof would be useful.

    So why do you say it cannot be proven?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Topological space: Condition 3

    If $\mathfrak{T}$ is some SPECIFIC topology, then it could be proven, but we would have to know what defines $\mathfrak{T}$.

    A topology on a set $X$ is, by definition:

    1. A subset of $\mathcal{P}(X)$ closed under finite intersections and arbitrary unions.
    2. Such a subset that contains $\emptyset$ and $X$ as elements.

    The elements of $\mathfrak{T}$ are called "open sets".

    Alternate definitions do exist, some texts define "neighborhood systems" or "closed sets" instead.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Sep 2013
    From
    Portland
    Posts
    540
    Thanks
    73

    Re: Topological space: Condition 3

    Quote Originally Posted by Deveno View Post
    If $\mathfrak{T}$ is some SPECIFIC topology, then it could be proven, but we would have to know what defines $\mathfrak{T}$.

    A topology on a set $X$ is, by definition:

    1. A subset of $\mathcal{P}(X)$ closed under finite intersections and arbitrary unions.
    2. Such a subset that contains $\emptyset$ and $X$ as elements.

    The elements of $\mathfrak{T}$ are called "open sets".

    Alternate definitions do exist, some texts define "neighborhood systems" or "closed sets" instead.
    My book also contains a 3rd condition

    Which is written as  U \cap V \in \mathfrak{T} whenever  U,V \in \mathfrak{T}

    and the lecture listed it as \for A,B \in \mathfrak{T}, A \cap B \in\mathfrak{T}

    but the intersection and union are covered in your first condition as opposed to my book which has them separately as the second and third conditions. So that makes sense
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Topological space: Condition 3

    Quote Originally Posted by Jonroberts74 View Post
    My book also contains a 3rd condition

    Which is written as  U \cap V \in \mathfrak{T} whenever  U,V \in \mathfrak{T}

    and the lecture listed it as \for A,B \in \mathfrak{T}, A \cap B \in\mathfrak{T}

    but the intersection and union are covered in your first condition as opposed to my book which has them separately as the second and third conditions. So that makes sense
    OK, but my point is this: you don't "prove these are true", you prove instead that they are true for some defined collection $\mathfrak{T}$.

    For example, the set of all unions of open intervals form a topology for the real line.

    If we take an arbitrary union of unions of open intervals, it's just a larger union, so is still in the collection.

    On the other hand, the intersection of any two open intervals is again a (possibly empty) open interval, and the intersection of two sets which are each unions of open intervals, is a union of the intersections.

    We can write the entire real line as: $\displaystyle \bigcup_{n \in \Bbb Z} (n-1,n+1)$, so it is in the set, and we can write $\emptyset = (a,a)$ for any real number $a$, which shows these two sets are in our collection.
    Last edited by Deveno; August 10th 2014 at 12:36 PM.
    Thanks from Jonroberts74
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compactness of topological space
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 24th 2012, 08:43 AM
  2. A is a subset of a topological space X
    Posted in the Differential Geometry Forum
    Replies: 8
    Last Post: August 28th 2011, 02:40 PM
  3. Topological space
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 2nd 2010, 09:23 PM
  4. discrete topological space
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: October 6th 2009, 03:17 AM
  5. a topological space
    Posted in the Differential Geometry Forum
    Replies: 8
    Last Post: March 8th 2009, 04:39 AM

Search Tags


/mathhelpforum @mathhelpforum