Sierpinski Space

Sierpinski Space

M. VELRAJAN 

    

1. What is the smallest topological space which is neither indiscrete nor discrete?

To answer this question we start with a small topological space. 

So we start with a set with only one point.

Consider the set X consisting of only one point, say {0}. 

The discrete topology in X = {0} is same as the indiscrete topology {∅, X}.

And it is the only possible topology in X.

So, next we consider a set with 2 points.

Consider the set X consisting of two points, say {0, 1}. 

Besides the discrete topology {∅, X, {0}, {1}} and indiscrete topology {∅, X} 

the only possible topologies in X are {∅, X, {0}} and {∅, X, {1}}.

It is clear that the topological spaces X with {∅, X, {0}} and X with {∅, X, {1}}

are homeomorphic under the homeomorphism of interchanging of 0 and 1. 

So, as two finite homeomorphic spaces have the equal number of open sets, 

there are only 3 non homeomorphic topologies in X = {0, 1}, namely, 

the discrete topology {∅, X, {0}, {1}}, the indiscrete topology {∅, X} and the

topology {∅, X, {0}}.

The space X = {0, 1} with the topology {∅, X, {0}} is called the Sierpinski space.

Ofcourse, the space X = {0, 1} with the topology {∅, X, {1}} is also be called as

the Sierpinski space.

But by Sierpinski space we mean 

the space X = {0, 1} with the topology {∅, X, {0}}. 

And we denote the Sierpinski space by 𝓢.

So, the Sierpinski Space is the smallest topological space which is neither

discrete nor indiscrete.

2.   We denote the space X = {0, 1} with the discrete topology {∅, X, {0}, {1}} 

by 2.

 

3.  We know that, if  𝛕1 and 𝛕2 are two topologies on a set X then 

𝛕1 = 𝛕2 if and only if the identity map f : (X, 𝛕1) → (X, 𝛕2) is a homeomorphism.

Hence the identity map f : 𝓢 → 2 is not a homeomorphism.

Since {1} is open in 2 and {1} is not open in 𝓢 and since f ({1}) = {1}, 

f is not continuous.

In a discrete space X all the subsets of X are open, hence 

any map from X to any other space Y is continuous.

Hence the identity map g = f -1 : 2 → 𝓢 is continuous.

So, the identity map g : 2 → 𝓢 is an example of a bijection which is continuous but its inverse g -1 = f is not continuous.

4.  The proper subsets of the Sierpinski space 𝓢 are {0} and {1}.

The subspace topologies of {0} and {1} inherited from Sierpinski space 𝓢 are

{∅ ∩ {0}, {0, 1} ∩ {0}, {0} ∩ {0}} = {∅, {0}} and

{∅ ∩ {1}, {0, 1} ∩ {1}, {0} ∩ {1}} = {∅, {1}} respectively.

Consider the restriction map f| {0} : {0} → 2. 

Since f| {0}(0) = f(0) = 0 we get (f| {0})-1(∅) = ∅, (f| {0})-1({0}) = {0}, (f| {0})-1({1}) = ∅, (f| {0})-1({0, 1}) = {0}. 

Since ∅ and {0} are open in the subspace {0}, f| {0} : {0} → 2 is continuous.

Now consider the restriction map f| {1} : {1} → 2. 

Since f| {1}(1) = f(1) = 1 we get (f| {1})-1(∅) = ∅, (f| {1})-1({0}) = ∅, (f| {1})-1({1}) = {1}, (f| {1})-1({0, 1}) = {1}. 

Since ∅ and {1} are open in the subspace {1}, f| {1} : {1} → 2 is continuous.

Thus for all proper subsets A of the Sierpinski space 𝓢, f|A : A → 2 is continuous. But f : 𝓢 → 2 is not continuous.

Hence the identity map f : 𝓢 → 2 is a counterexample to show that the converse of the following is not true.

“If X and Y are topological spaces and f : X → Y is continuous then for all proper subsets A of the space X, f|A : A → Y is continuous”.

5.  Suppose f : X → {0, 1} is a map from a set X into the set {0, 1}.

If U = f -1({1}) then f = 𝛘U, the characteristic function of the subset U of X,

defined by 𝛘U(x) = { 0, if x ∉ U 1, if x ∈ U .

Hence the characteristic function of the subsets U of X are the only functions from

X into {0, 1}.

6.  Let (X, 𝛕) be a topological space. 

Then, by 5., the characteristic function of the subsets U of X are the only functions

from X into the Sierpinski space 𝓢.

For any subset U of the space X, 

the characteristic function 𝛘U of U from X into 𝓢 is continuous 

if and only if 𝛘U- 1(∅) = ∅, 𝛘U- 1({0}) = X - U, 𝛘U- 1({0, 1}) = X are open in X

if and only if X - U is open in X

if and only if U is closed in X.

Hence the set 𝓢 X of all continuous functions from X into 𝓢 is given by

𝓢 X = {𝛘U / U is closed in X}

     = {𝛘U / X - U is open in X}

     = {𝛘X - U / U is open in X}.

Since U1 = U2 if and only if X - U1 = X - U2, 

U → 𝛘X - U is a bijection from the topology 𝛕 on X onto 𝓢 X.

Thus, there is a bijection between the topology 𝛕 on the topological space X

onto the set 𝓢 X of all continuous functions from X into the Sierpinski space 𝓢

And 𝓢 X can be described as 𝓢 X = {𝛘X - U / U is open in X} and

𝛕 = {U / 𝛘X - U ∈ 𝓢 X} = {X - U / 𝛘U ∈ 𝓢 X}.

7.  All the sequences (an) in the Sierpinski space 𝓢 are convergent and 

converge to 1.

(Since an is either 0 or 1 and {0, 1} is the only neighbourhood of 1.) 

In particular, the constant sequence 0, 0, 0, . . .  converges to both 1 and 0. 

Oh, a sequence converges to two points.

But the constant sequence 1, 1, 1, . . . converges to 1 only.

The sequence 0, 1, 0, 1, 0, 1, . . . converges to 1 but not converges to 0 

(since {0} is neighbourhood of 0).

8. Since {0} is a neighbourhood of 0 not containing 1, 

the Sierpinski space 𝓢 is T0.

Since {0, 1} is the only neighbourhood of 1, the Sierpinski space 𝓢 is not T1 and

not T2 or not Hausdorff.

So, the Sierpinski space 𝓢 is a counterexample for 

“Every T1 space is T0, but not conversely.”

9. If E is the Euclidean space then for any x ∈ E, 

the slice E × {x} parallel to the first factor is closed in E × E. But,

the slice 𝓢 × {0} parallel to the first factor is not closed in the product 𝓢 × 𝓢.

For,  if A = 𝓢 × {0} then the complement of A is Ac = {(0, 1), (1, 1)}.

The only open set containing (1, 1) in 𝓢 × 𝓢 is {0, 1} × {0, 1} = 𝓢 × 𝓢 and it is not

contained in Ac. Hence Ac is not open in 𝓢 × 𝓢. 

Thus,

Though the Sierpinski Space is the smallest one, 

as example and counterexample Sierpinski Space has a 

significant role in topology.

Besides this the Sierpinski space also plays a role in the 

Theory of Computation.