Quick Revision: Basis and Subbasis (Part 1.2)

Defining Basis

Out study and examples so far gave use 3 insights about the proposed idea of basis for a topology on X:

  1. Property: It is a collection of subsets of X.
  2. Property: Union of elements in the collection is the whole of X.
  3. Property: All possible intersections is either empty or contained in the collection.
  4. Synthesizing Topology: We can synthesize a topology on X by taking all possible unions of sets in the collection.

Let’s try to frame a definition using these facts:

A X be a set. Then a collection of subsets of X \mathcal{B} is (almost) called a basis for a topology on X if

  1. Union of elements in the collection is the whole of X.
  2. All possible intersections is either empty or contained in the collection.

The topology generated by this basis is given by all possible union of elements of \mathcal{B}.

Why did I call it “Almost”? I’ll leave that as a mystery for now! I don’t think we need more examples, but let me give some!

Consider the first condition in the definition of basis: “Union of elements in the collection is the whole of X.” Which of the following options are equivalent to it?
“For every such that .”
True! The proof is easy, would you like to try to write it yourself?
Intersection of all elements are nonempty
Oh no! Wrong! Check the other option for details!
It’s a national secret.
There’s one true, but definitely not this option! 🙂 Consider a set that satisfies the condition “For every such that .” What can you say about union of sets in ?

Example 1

Consider the set of all real numbers, R and the collection B_{R_1}=\{\{1\},R\}. Clearly, it forms a basis (verify!) and the topology generated by it is \tau_{R_1} = \{\emptyset, \{1\},R\}.

In fact, this trick works with any set X with the collection \mathcal{B}_{X}=\{\{a\},X\} where a\in X.

Example 2

Consider any set X and consider the collection of all singleton subsets of X, say \mathcal{B}_{X_2}. Then \mathcal{B}_{X_2} generates discrete topology on X. (Verify!)

A Very Intriguing Example 3

That’s a lot of dramatic build up for this example – in fact, it helps us reframe the definition of basis and make it more general! Before we begin, note that our second condition insists that either possible intersections is empty or it is contained in our basis.

Consider the set X = {1, 2, 3, 4, 5}. Let \mathcal{B}_{X_3} = \{\{1\},\{2\},\{1,2,3,4\}\},\{1,2,5\}\}

Note that the condition 1 in definition is easily satisfied. But what about \{1,2,3,4\}\cap\{1,2,5\}\}=\{1,2\}\notin  \mathcal{B}_{X_3}! That means, second condition that either intersection is empty or is contained in \mathcal{B}_{X_3} seems to be false in this case!

But still, let’s try to find all possible unions and see if it forms a topology on X.

Does all possible unions form a topology on X?
Yes!
You got it right! Topology,
No!
Oh no! Wrong! Check the other option for details!
It’s a national secret.
It forms a topology! Check that option for details!

Absolutely curious, isn’t it? In fact, it’s time to rework our definition of basis a bit.

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