Topological Space

        In my previous blog  "TOPOLOGY - KING OF MATHEMATICS" , talked about geometric shapes and saw how "Distance" is the basic tool for the study of "Metric Space" and also saw how removing the notion of distance lead us to the more general study of spaces called "Topological Spaces".

       The goal of today's blog is to understand the Topological Space more formally and challenge our normal understanding to an another level.

      As in metric space main key-tool is distance, here it is neighbourhood. 

      First, we can start with what we mean by neighbourhood, its simply mean surrounding. If you choose a point and a r >0 radius circle around that point then that also a neighbourhood mathematically denoted as B(x,r) where x is point and r is my radius, this neighbourhood called r-neighbourhood. We can see one thing that if we choose r-neighbourhood the study came to study same as metric space. So, we can say that every metric space is a topological space, but we see that idea of neighbourhood is more general that we can take any kind of open, closed ball as well as any set of points that consider as general neighbourhood according to define. Noticeable thing here is balls in metric space are subset of my given set X. So, as a generalization we consider X with its subsets (that we called neighbourhood).

      Now our main tool is ready, but before going to definition we need some thing more.

      Here the concept of Open set comes. A set U in X is said to be open set in X if for all x ∈ U there exists a r>0 such that B(x,r) ⊂ U. In metric space open ball's are open sets that has some important properties,

       a) Union of any number of open sets in a metric space is an open set, 

       b)Intersection of finite number of open sets in a metric space is an open set.

      So lets generalize it, let X be any set and T is collection of subsets of X, if

           a) Φ ∈ T and X ∈ T.

          b) If Ai∈ T for any i then ∪ Ai∈ T.

         c) If Ai ∈ T for any i ∈I(finite set),then ∩ Ai∈ T.

          If above three properties satisfied we say (X,T) is a "Topological Space" and T is called the topology defined on X. We referred the subsets of X are in T as open sets.

          To challenge our understanding lets see some examples

 1) Let X= ℝ  and take our T as all open set in  ℝ

    >Lets verify the above stated properties first Φ ∈ T and ℝ ∈ T, as both are open set in ℝ

 and Union of any number of open sets in ℝ is an open set in ℝ,

Intersection of finite number of open sets in ℝ is an open set in ℝ.

So (ℝ,T) is a Topological Space.

 2) TRY this and say the answer, Let X= ℝ  and take our T as all open ball in  ℝ.

Hint - NO (why?)

3) Let X= ℝ  and take our T = { A⊂ X : A =Φ or X\A is countable}

    > Lets verify the above stated properties first Φ ∈ T and ℝ ∈ T as ℝ\ℝ = Φ is countable and let Ai ∈ T for any i =>  ℝ\ Ai is countable, take ∩( ℝ\ Ai) = ( ℝ\ ∪Ai) is countable intersection of countable set that is countable ∪ Ai ∈ T.
let Ai ∈ T for any i ∈ I (finite set), =>  ℝ\ Ai is countable, take  ∪ ( ℝ\ Ai) = ( ℝ\ ∩Ai) is finite union of finite set that is countable, ∩ Ai ∈ T
 So (ℝ,T) is a Topological Space. This topology called co-countable topology.

4)TRY this and say the answer, Let X= ℝ  and take our T = { A⊂ X : A =Φ or X\A is Finite}

5)TRY this and say the answer, Let X= ℝ  and take our T is all closed sets in ℝ.

6)TRY this and say the answer, Let X= ℝ  and take our T is all subsets in ℝ.

7)TRY this and say the answer, Let X= ℝ  and take our T ={ Φ , ℝ}.

      
         This above example shows the different kind of topology exist in ℝ, that blows up the idea we can define our neighbourhood many different way whereas it will be topological space or not that will be a different question. See this below,
          

              X = ℝ, consider the set T = { {3}, (10,15),  Φ , ℝ}

    > Lets verify the above stated properties first Φ ∈ T &  ℝ ∈ T,  and {3}∈ T, (10,15) ∈ T ,  {3}∪(10,15) ∈ T since for point 3 ∈{3}∈ T, {3}∩(10,15)= Φ ∈ T.
So, (ℝ,T) is a Topological Space.

    Let's try this one,

        Let X= ℂ  and take our T as all open set in  ℂ

    >   Lets verify the above stated properties first Φ ∈ T & ℂ ∈ T, , as both are open set in ℂ
    
     and Union of any number of open sets in ℂ is an open set in ℂ and Intersection of finite number of open sets in ℂ  is an open set in ℂ.

       TRY this, 

            Let X= ℂ  and take our T as all open set in  ℝ (ℝ + i*0 )

      >Lets verify the above stated properties first Φ ∈ T  as Φ is an open set in ℝ

 and Union of any number of open sets in ℝ is an open set in ℝ,

Intersection of finite number of open sets in ℝ is an open set in ℝ,

but is ℂ ∈ T ?

 If ℂ ∈ T, then for each point in ℂ there is an open set in T containing that point should belong to ℝ , try 1+ 5*i  ∈ ℂ, we can't find any open set in T such that 1+ 5*i is in there and that is open in ℝ.

     Try some crazy example and try it is neighbourhood and topological space or not.