By Zehao Dong and Zihan Hu

1 Introduction

In Chapter 2, we studied the definition of sequence and the convergence of a sequence. Topological spaces provide a general framework for the study of convergence. However, instead of a distance function, we can think of the basic structure on a topological space as a collection of open sets.

Definition: In terms of open sets, a topological space is an ordered pair (X, τ ), where X is a set and τ is a collection of subsets of X, satisfying the following axioms.

1. The empty set and X itself belongs to τ .
2. Any union of members of τ belongs to τ .
3. The intersection of any finite number of members of τ belongs to τ .

Ex : Let X be set such that X = {a, b, c}. Then let τ be a collection of the subsets of X such that τ = {ø, {a, b}, {a, c}, X}. Then, even though the first two properties are satisfied, the intersection of {a, b} and {a, c} is {a}, which is not in τ . Therefore, τ is not a topology of the set X.

 

2 Relating to Last Week’s Reading

There are many ways of defining a topology on R. The standard topology on R is generated by the open intervals. we can define a topology τ on X = R by defining that T ⊆ R to be in τ if, for every point x ∈ T , there exists an e > 0 such that (x − e, x + e) ⊆ T . This topology is defined as the usual topology on R. For example: Let X = R and let τ = {∅, (1, 3), X}.

1. The subset {1, 2, 3} will not be in τ because it only contains the points 1, 2, and 3, but, for example, for point 1∈ T, there is not a value e > 0 so that (1 – e, 1 + e) is a subset of T. Hence, this T is not in τ.
2. The subset (1, 2) will be in τ because for example, for x = 3/2 ∈ T, there exists an e = 1/4 so that (3/2 -1/4 , 3/2 + 1/4) = (5/4, 7/4) is a subset of T. And we need to prove that for every x ∈ T, there exists an e > 0 such that (x − e, x + e) ⊆ T to prove that T∈ τ.
3. The subset [1, 2] will not be in τ for the similar reason as 1, for point 1∈ T, there is not a value e > 0 so that (1 – e, 1 + e) is a subset of T so this T is also not in τ.

This definition of usual topology can be used to explain convergence of a sequence and it is similar to the Theorem 2.2.3B in our reading last week.

By Theorem 2.2.3, a sequence (xn) converges to a real number x if, for every positive number e, there exists an N ∈ N such that whenever n ≥ N it follows that |xn − x| < e. This is an example of a distance function mentioned in our introduction, where |xn − x| represents the distance between xn and x on the line of R.

Theorem 2.2.3B is saying that a sequence (xn) converges to x if, given any e-neighborhood Ve(x) of x, there exists a point in the sequence after which all of the terms are in Ve(x). In other words, every s-neighborhood contains all but a finite number of the terms of (xn).

And to say a sequence (xn) in X converges in the topology τ to an element x ∈ X if, given any set T ∈ τ such that x ∈ T , the sequence (xn) is eventually in T .

Here we can see that convergence in topology is similar with Theorem 2.2.3B from topological aspect. We can treat Ve(x) as a of a type of T ∈ τ that contains the point x to which (xn) converges. Let T1 = (a, b) ∈ τ be an open interval such that x ∈ T1 and T2 = (c, d) ∈ τ be an open interval such that x ∈ T2 with a, b, c and d arbitrary.

Let τ = {∅, T1, T2, X}.

1. The empty set and X belongs to τ .
2. The union of members of τ belongs to τ. ex. T1 ∪ T2 ⊆ X ∈ τ.

3. The intersection of any finite number of members of τ belongs to τ . ex. T1 ∩ T2 = x ∈ X ∈ τ and T1 ∩ X = T1 ∈ τ .

So τ will be a topology because it satisfies all the three properties of topology mentioned above. And the sequence (xn) in X converges in the topology τ to an element x ∈ X if, given any set T ∈ τ such that x ∈ T , the sequence (xn) is eventually in T .

 

3 Examples of Topologies

For X = R, we have two extreme examples:

1. Trivial Topology: τ = {ø, R}.
2. Discrete Topology: τ = P(R) (i.e. The power set of R, which contains all possible subsets of R).

Notice that both topologies satisfy the properties mentioned before. Both topologies contains ø and the entire set X which equals to R.

For the trivial topology, the intersection of the two elements is ø, which is in τ , and the union of the two elements is R, which is also in τ .

For the discrete topology, since it contains all subsets of R, the intersection or union of any elements inside the power set of R will still be in τ .

4 Examples of a Sequence In R That Converges In Some Of the Topological Space

Before digging into examples, let us review the definition of convergence of a sequence in topological space.

Definition: Let (X, T ) be a topological space and xn ∈ X a sequence. We say that the sequence xn converges to x0 ∈ X if for every open set U ⊆ X which contains x0 there exists an n0 ∈ N such that for all n ≥ n0 the points xn lies in U .

Then, consider the following three sequences.

1. an = 1/n.
2. bn = n.
3. cn = 1.

We know that in the usual topology, an converges to 0, cn converges to 1, while bn diverges. However, this is not the case in other topologies.

In trivial topology {ø, R}, the only open set that contains the limit of the three sequences is R. Since for all of the three sequences, xn ∈ R, ∀n ∈ N, all of them converge in the trivial topology.

In discrete topology, notice that T = {0} is in the topology, and since an converges to 0, by definition, because an never enters the set T (for all n, an ≠ 0), an does not satisfy the condition of being eventually inside every set T in τ which contains the point 0, so an fails to converge to 0 in the discrete topology. bn still diverges in the discrete topology. Because cn is contained in every subset in the discrete topology that contains 1, cn converges in the discrete topology.

References

https://en.wikipedia.org/wiki/Topologicalspace

Click to access 04%20-%20Congergent%20sequences.pdf

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