Introduction to Topology(Exercises and Solutions)

TOPOLOGY

The study of topology and its spaces is an important aspect of mathematics,topological spaces like other mathematical spaces have axioms that must be satisfied for a topological space to hold.

Today i will be treating those axioms with solution to exercises from the book "Topology without tears" by Sydney A. Morris.

I will be looking at the definitions of topology and topological spaces.

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What is Topology and Topological Space?

Now coming to think of it,what is a topology?

A topology is a non-empty set X,and a collection $\mathcal{T}$ of subsets of of X satisfying the following three axioms:

(i) X and the empty set $\phi$, belong to $\mathcal{T}$.

(ii) The union of any(finite or infinite) number of sets in $\mathcal{T}$ belongs to $\mathcal{T}$.

(iii) The intersection of any two sets in $\mathcal{T}$ belongs to $\mathcal{T}$.

Once the above axioms are satisfied then we say $\mathcal{T}$ is a topology on X and the pair $(X,\mathcal{T})$ is a topological space.

Now lets take a look at some exercises from the book "Topology without tears" by Sydney A. Morris,so that we now see the clear picture of what topology and topological spaces are all about.

Exercises From "Topology Without Tears"

 Let $X=\{a,b,c,d,e,f\}$. Determine whether or not each of the following collections of subsets of X is a topology on X.

(1) $\mathcal{T}_1=\{X,\phi,\{a\},\{a,f\},\{b,f\},\{a,b,f\},\}$

(2) $\mathcal{T}_2=\{X,\phi,\{a,b,f\},\{a,b,d\},\{a,b,d,f\}\}$

(3) $\mathcal{T}_3=\{X,\phi,\{f\},\{e,f\},\{a,f\}\}$

Solution

(1) we are going to use the axioms of topology to proof the statement.

$\mathcal{T}_1=\{X,\phi,\{a\},\{a,f\},\{b,f\},\{a,b,f\},\}$.

(i) Now we begin with axiom I, which states that, the set X and the empty set $\phi$ must belong to $\mathcal{T}_1$ which is satisfied.

(ii) Axiom II states that, the union of any finite or infinite number of sets in $\mathcal{T}_2$ belongs to $\mathcal{T}_2$.

Now lets check if this is true.

-if we take the union of X and all other sets of $\mathcal{T}_1$ will yield X,which is in $\mathcal{T}_1$.

e.g if we take the union of

$X\cup{\phi}=X\in{\mathcal{T}}_1$

$X\cup{\{a\}}=X\in{\mathcal{T}}_1$

$X\cup{\{a,f\}}=X\in{\mathcal{T}}_1$

the union of $\phi$ with all elements of $\mathcal{T}_1$ will be the individual set or element and is in $\mathcal{T}_1$.

and also

$\{a\}\cup\{a,f\}=\{a,f\}\in\mathcal{T}_1$

$\{a\}\cup\{b,f\}=\{a,b,f\}\in\mathcal{T}_1$

$\{a,f\}\cup\{b,f\}=\{a,b,f\}\in\mathcal{T}_1$
$\{a,f\}\cup\{a,b,f\}=\{a,b,f\}\in\mathcal{T}_1$

$\{b,f\}\cup\{a,b,f\}=\{a,b,f\}\in\mathcal{T}_1$

(iii) Now we proof the third axiom which states that, the intersection of any two sets in $\mathcal{T}_1$ belong to $\mathcal{T}_1$.

The intersection of X with all elements of $\mathcal{T}_1$ will give you the individual element or set.

e.g

$X\cap\phi=\phi\in{\mathcal{T}}_1$

$X\cap{\{a\}}=\{a\}\in{\mathcal{T}}_1$

$X\cap{\{a,f\}}=\{a,f\}\in{\mathcal{T}}_1$

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also the intersection of the empty set $\phi$ with all elements of $\mathcal{T}_1$ will be the empty set $\phi$ which is in $\mathcal{T}_1$.

Now

$\{a\}\cap\{a,f\}=\{a\}\in\mathcal{T}_1$

$\{a\}\cap\{b,f\}=\phi\in\mathcal{T}_1$

$\{a\}\cap\{a,b,f\}=\{a\}\in\mathcal{T}_1$

$\{a,f\}\cap\{b,f\}=\{f\}\notin\mathcal{T}_1$

Hence $\mathcal{T}_1$ is not a topology on $X$ and does not form a topological space.

(2)$\mathcal{T}_2=\{X,\phi,\{a,b,f\},\{a,b,d\},\{a,b,d,f\}\}$

We use the axioms of topology to prove the above statement.

(i)Axiom I of topology is clearly satisfied as $X$ and $\phi$ belong to $\mathcal{T}_2$

(ii) Axiom II states that the union of any finite or infinite number of sets in $\mathcal{T}_2$ belong to $\mathcal{T}_2$.

Taking the union of $X$ with all the elements or sets of $\mathcal{T}_2$ will yield X.

If we also take the union of the empty set $\phi$ with all the elements or sets of $\mathcal{T}_2$,this will yield the individual of sets or elements in $\mathcal{T}_2$.

i.e

$\phi\cup\{a,b,f\}=\{a,b,f\}\in\mathcal{T}_2$

$\phi\cup\{a,b,d\}=\{a,b,d\}\in\mathcal{T}_2$

Now taking

$\{a,b,f\}\cup\{a,b,d\}=\{a,b,d,f\}\in\mathcal{T}_2$

$\{a,b,f\}\cup\{a,b,d,f\}=\{a,b,d,f\}\in\mathcal{T}_2$

(iii) Axiom III states that the intersection of 2 sets in $\mathcal{T}_2$ belongs to $\mathcal{T}_2$.

-intersecting $X$ with all the elements of $\mathcal{T}_2$ will yield the individual element in $\mathcal{T}_2$ which will clearly be in $\mathcal{T}_2$.

-intersecting $\phi$ with all the elements of $\mathcal{T}_2$ will yield $\phi$ which belong to $\mathcal{T}_2$.

-Now intersecting

$\{a,b,f\}\cap\{a,b,d\}=\{a,b\}\notin\mathcal{T}_2$

Hence $\mathcal{T}_2$ is not a topology on $X$ and hence does not form a topological space.

(3) $\mathcal{T}_3=\{X,\phi,\{f\},\{e,f\},\{a,f\}\}$

we use the axioms of topology to prove.

(i) axiom I is clearly satisfied as $X$ and $\phi$ clearly belong to $\mathcal{T}_3$.

(ii) for axiom II, we check for the following

-The union of $X$ and all the sets or elements of $\mathcal{T}_3$ will yield $X$ which is in $\mathcal{T}_3$.

-The union of $\phi$ with all elements of $\mathcal{T}_3$ will yield the elements of $\mathcal{T}_3$.

-Now

$\{f\}\cup\{e,f\}=\{e,f\}\in\mathcal{T}_3$

$\{f\}\cup\{a,f\}=\{a,f\}\in\mathcal{T}_3$

$\{e,f\}\cup\{e,f\}=\{a,e,f\}\notin\mathcal{T}_3$

hence $\mathcal{T}_3$ is not a topology on $X$ and hence does not form a topological space